PhysicsPhysics QuestionsOscillations Questions for CBSE Class 11th

Oscillations Questions for CBSE Class 11th

  1. The time period of oscillation of a particle that executes SHM is 1.2s. The time starting from extreme position, its velocity will be half of its velocity at mean position is
  2. A particle is executing simple harmonic motion between extreme positions given by (-1,-2,-3)cm and (1,2,1)cm. Its amplitude of oscillation is 6cm
  3. A particle is moving in a circle of radius R = 1 m with constant speed v = 4 m/s . The ratio of the displacement to acceleration of the foot of the perpendicular drawn from the particle onto the diameter of the circle is
  4. A particle executes SHM on a straight line. At two positions, its velocities are u and v while accelerations are α and ß respectively [ß > α > 0] . The distance between these two positions is
  5. If the displacement x and velocity (v) of a particle executing simple harmonic motion are related through the expression 9 v 2 = 36 − x 2 , here x is in meter and V is in m/s. Then time period of vibration is
  6. A particle executes SHM with amplitude 0.5 cm and frequency 100s -1 . The maximum speed of the particle is ( in m/s)
  7. Identify the function which represents a periodic motion.
  8. A particle is executing SHM along a straight line and its position at time t is given by x = ( 4 c o s ⁡ 2 t + 3 s i n ⁡ 2 t ) cm. Then the particle arrives at its extreme position after a time of
  9. A particle is executing simple harmonic motion along X-axis with its equilibrium position at the origin. Variation of its velocity with displacement is as shown in the graph. Then maximum acceleration of the particle is
  10. The equation of motion of a particle is d 2 y dt 2 + Ky = 0 , where K is positive constant. The time period of the motion is given by
  11. A body is moving in a room with a velocity of 20 m/s perpendicular to the two walls separated by 5 meters. There is no friction and the collisions with the walls are elastic. The motion of the body is
  12. A simple pendulum performs simple harmonic motion about.x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a 2 will be
  13. The phase difference between two SHM y 1 = 10 sin ( 10 πt + π 3 ) and y 2 = 12 sin ( 8 πt + π 4 ) at t = 0 . 5 s is
  14. The function sin 2 ( ωt ) represents:
  15. A spring has length l and force constant k it is cut into two springs of length l 1 and l 2 such that l 1 =nl 2 (n = an integer). Mass ‘m’ suspended from l1oscillates with time period…
  16. The acceleration due to gravity on a planet is 3/2 times that on the earth. If length of a seconds pendulum on earth is 1m, length of seconds pendulum on surface of planet is
  17. A particle executes SHM along a straight line 4cm long. When the displacement is 1cm its velocity and acceleration are numerically equal. The time period of SHM is
  18. The bob of a simple pendulum of mass 100g is oscillating with a time period of 1.42s. If the bob is replaced by another bob of mass 150 g but of same radius, the new time period of oscillation
  19. For a particle executing SHM, the kinetic energy (K) is given by K = K 0 cos 2 ωt . The equation for its displacement is
  20. The maximum acceleration of a particle executing simple harmonic motion is ‘ α ’ and the maximum velocity is ‘ β . The frequency of the SHM is given by the expression
  21. Assertion: The periodic time of a hard spring is less as compared to that of a soft spring. Reason: The periodic time depends upon the spring constant, and spring constant is large for hard spring
  22. Assertion: The graph of total energy of a particle in SHM w.r.t., position is a straight line with zero slope. Reason: Total energy of particle in SHM remains constant throughout its motion
  23. The displacement y of a particle executing periodic motion is given by y = 4 cos 2 ( t 2 ) sin ( 1000 t ) . This expression may be considered to be a result of the superposition of how many independent harmonic motions
  24. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
  25. The potential energy of a particle of mass I kg in motion along the x-axis is given by: U = 4 ( 1 – cos 2 x ) , where x is in metres. The period of small oscillation (in sec) is
  26. At time t = 0, one particle is at maximum positive amplitude and the other particle is at half of the positive amplitude. Their amplitudes and time periods T are same. If they are approaching, find the time by which they cross each other:
  27. A particle is subjected to two simple harmonic motions in the same direction having equal amplitude and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, what is the phase difference between the two simple harmonic motions?
  28. A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0. l0 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:
  29. A particle is moving along the x-axis under the influence of a force given by F = -5x + 15. At time t = 0, the particle is located at x = 6 and is having zero velocity. It takes 0.5 seconds to reach the origin for the fist time. The equation of motion of the particle can be represented by
  30. A particle is executing a motion in which its displacement as a function of time is given by x = 3 sin ( 5 πt + π 3 ) + cos ( 5 πt + π 3 ) where x is in m and t is in s. Then the motion is
  31. Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed …. oscillations.
  32. A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period T ‘ = 2 T . The density of the pendulum bob is equal to (density of water = 1)
  33. A brass cube of side a and density σ is floating in mercury of density ρ . If the cube is displaced a bit vertically, it executes S.H.M. Its time period will be
  34. A particle performs S.H.M. of amplitude A along a straight line. When it is at a distance 3 2 A from mean position, its kinetic energy gets increased by an amount 1 2 mω 2 A 2 due to an impulsive force. Then its new 2 amplitude becomes
  35. Which of the following is correct about a SHM, along a straight line?
  36. The potential energy of a particle executing SHM changes from maximum to minimum in 5 s. Then the time period of SHM is:
  37. A particle is executing SHM according to the equation x = A cos ωt . Average speed of the particle during the interval 0 ≤ t ≤ π 6 ω
  38. If displacement x and velocity y are related as 4 v 2 = 25 – x 2 in a SHM Then time period of given SHM is (consider SI units)
  39. A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is I cm, the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. Then find out the frequency of SHM.
  40. A particle is vibrating simple harmonically with an amplitude ‘a’ . Tlte displacement of the particle when its energy is half kinetic and half potential is
  41. For a particle in S.H.M., if the amplitude of displacement is ‘a’ and the amplitude of velocity is ‘v’ the amplitude of acceleration is
  42. A particle executes S.H.M between x = -A to x = +A. The time taken for it in going from 0 to A 2 is T 1 and from A 2 to A is T 2 . Then
  43. A particle executing S.H.M. of amplitude 4 cm and T = 4 sec. The time taken by it to move from positive extreme position to half the amplitude is
  44. The equation of motion of a particle executing simple harmonic motion is a + 16 π 2 x = 0 . In this equation, a is the linear acceleration in m / s 2 of the particle at a displacement x in meter. The time period in simple harmonic motion is:
  45. A small body of mass 1 kg is executing SHM with amplitude 0.1 m and period 0.4 s. The maximum force acting on the body is:
  46. A particle is executing a simple harmonic motion of period 2 s. When it is at its extreme displacement from its mean position, it receives an additional energy equal to what it had in its mean position. Due to this in its subsequent motion:
  47. A particle slides back and forth between two inclined friction less planes as shown in Fig.Which of the following statements are correct with respect to the motion of particle? (i) The motion is oscillatory (ii) The motion is simple harmonic (iii) It his the initial height, period = 4 sin θ 2 h g (iv) If h is the initial height, period = 2 h g sin θ
  48. A simple pendulum of length / is vibrating with an amplitude A.Its energy is E. Consider the following situations: (i) Amplitude is doubled keeping length the same (ii) Length is doubled, keeping amplitude the same. Then energy of the pendulum in the two situations will be:
  49. A pendulum is undergoing SHM. The velocity of the bob in the mean position is V 0 . If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be:
  50. Two pendulums suspended from same point having lengths 2m and 0.5m. If they displaced slightly and released, then they will be in same phase, when small pendulum will have completed :
  51. A second’s pendulum is mounted in a rocket. Its period of oscillation will decrease when the rocket is:
  52. A frictionless inclined surface having inclination 30 0 has a 2.5 kg mass held by a spring which is fixed at the upper end. If the mass is taken 2.5 cm up along the surface, tension in the spring reduces to zero. On releasing the mass now, its angular frequency of oscillations will be:
  53. Two SHMs with same frequency act on a particle at right angles, i.e., along x and/-axes. If their amplitudes are equal and the phase difference between them is ( π / 2 ), then the resultant is:
  54. A particle is subjected to two SHMs, X 1 = A 1 sin ωt and X 2 = A 2 sin ωt + π 4 . The resultant SHM will have an amplitude of:
  55. The period of oscillation of mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be :
  56. Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants k 1 and k 2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is:
  57. Length of a spring of force constant k in its unstretched condition is l. The spring is cut into two parts which have their unstretched lengths in the ratio l 1 : l 2 = q : 1 . Force constants of the two parts k 1 and k 2 are then:
  58. Two blocks each of mass m are connected with springs each of force constant fr as shown in the Fig. Initially the springs are relaxed. The mass A is displaced to the left and ^B to the right by the same amount and released, then the time period of oscillation is:
  59. What happens to the energy of a particle, in SHM, with time in the presence of damping forces?
  60. In case of a forced vibration, the resonance wave becomes very sharp when the
  61. Two simple harmonic motions given by x 1 = Acosωt and x 2 = Asinωt are superposed on a single particle. Then the particle starts its motion from a point whose distance from equilibrium position is
  62. A particle is oscillating along a straight line with time period 6 sec and amplitude ‘A’. Then the time taken by the particle to directly move from x = – A 2 to x = + A 2 is
  63. A body performs SHM along the straight line segment ABCDE with C as the mid point of segment AE (A and E are the extreme position for the SHM). Its kinetic energies at B and D arc each one fourth of its maximum value. If length of segment AE is 2R, then the distance between B and D is
  64. A particle is moving along x-axis and its position is given by x = 2 cos 2 4 πt   cm . Then
  65. A particle is executing simple harmonic motion along a straight line with amplitude 2 cm. Maximum restoring force acting on the particle is 2N. Find the distance of the particle from the equilibrium position when restoring force acting on the particle is 1N.
  66. A spring is fixed to the ground as shown below. A block of mass m is fixed to the spring. The spring is compressed by 5cm. Imagine that we can remove the spring suddenly and the block will fly off with the speed it has at that instant. The spring has to be removed at such a position that the distance travelled upwards by the block relative to that position is maximum. Find that position.
  67. The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motion are 1m and 6.28 s, respectively. Then its angular frequency and acceleration at a displacement of 0.5 m are respectively
  68. The equation of motion of a particle in SHM is a + 4 x = 0 . Here ‘a’ is linear acceleration of the particle at displacement ‘x’ in metre. Its time period is
  69. The velocity of a particle in SHM at the instant when it is 0.6 cm away from the mean position is 4 cm/s. If the amplitude of vibration is 1 cm then its velocity at the instant when it is 0.8 cm away from the mean position is (in cm/s)
  70. The time period of oscillation of a simple pendulum is √2 s. If its length is decreased to half of initial length, then its new period is
  71. A simple pendulum has a time period T 1 when on the earth’s surface and T 2 when taken to height R above the earth’s surface, where R is the radius of the earth. The value of T 2 T 1 is
  72. A particle of mass 2g is initially displaced through 2cm and then released. The frictional force constant due to air on it is 12×10 -3 N / m. The restoring force constant is 50×10 -3 N /m.If it is in oscillatory motion, its time period is
  73. A particle starts its SHM from mean position at t = 0. If its time period is T and amplitude A then the distance travelled by the particle in the time from t = 0 to t = 5 T 4 is
  74. A particle performing SHM with a frequency of 5 Hz and amplitude 2 cm is initially at positive extreme position. The equation for its displacement (in metre) is
  75. A particle oscillates as per the equation x = ( 7 cos 0 · 5 πt ) m , the time taken by the particle to move from the mean position to a point 3.5 m away is
  76. If the displacement x and velocity v of a particle executing SHM are related as 4v 2 = 25 – x 2 . Then its maximum displacement in metre (x, v are in SI) is
  77. Find the average kinetic energy of a simple harmonic oscillator if its total energy is 10 joule and minimum potential energy is 2 joule.
  78. The displacement of a particle of mass 3g executing simple harmonic motion is given by x=3sin(0.2t) in SI units. The kinetic energy of the particle at a point which is at a displacement equal to 1/3 of its amplitude from its mean position is
  79. A spring of natural length 80cm with a load has a length of 100cm. If it is slightly pulled down and released its period will be
  80. The period of a simple pendulum is found to be increased by 50% when the length of the pendulum is increased by 0.6m. The initial length is
  81. The equation of motion of a particle started at t=0, is given by y = 5 sin 20 t + π 3 cm . The least time after which acceleration becomes zero is
  82. A simple pendulum of length l is connected to the ceiling of a vehicle that is moving down along a smooth inclined plane 4 in 5. Then its period of oscillation is
  83. The velocities of a body executing SHM at displacement ‘a’ and ‘b’ are ‘b’ and ‘a’ respectively. The amplitude of SHM is
  84. The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when
  85. In the system shown, the ratio of energy stored in the spring s 1 to that of spring s 2 in equilibrium is
  86. Average velocity of a particle executing SHM in one complete vibration is
  87. A particle of mass 10g is placed in a potential field given by V = ( 50 x 2 + 100 ) J / Kg . The frequency of oscillation in cycle/sec is
  88. The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the figure. y -projection of the radius vector of rotating particle P is
  89. A particle of mass m = 2kg executes SHM in xy plane between points A and B under the action of a force F = F x i ^ + F y j ^ . Minimum time taken by the particle to move from A to B is 1 second. At t = 0, the particle is at x = 2 and y = 2. Then F x as function of time t is
  90. The total energy of a particle executing SHM is 48 Joule. Then find its potential energy when its displacement is half of the amplitude.
  91. The instantaneous displacement of a simple pendulum oscillator is given by x = Acos ω t + π 4 . Its speed will be maximum at time
  92. A pendulum oscillates in a vertical plane. When the bob is at extreme position, its acceleration is 5 m / s 2 . If g = 10   m / s 2 , angular amplitude of oscillation of the pendulum is
  93. A simple pendulum has length 5 m and amplitude of oscillation is 10 cm. If g = 10    m / s 2 , the maximum velocity of the bob is
  94. Five identical springs are used in the three configurations as shown in figure. The ratio of frequencies of oscillations in the given three configurations respectively is
  95. A particle of mass 0.1 kg is executing simple harmonic motion with amplitude 20 cm along a straight line. Its maximum and minimum potential energies are 0.8 J and 0.6 J respectively. Then period of oscillation of the particle is
  96. A cubical body of density σ , side L floats in a liquid of density ρ ( ρ < σ . If it is slightly displaced downwards and released, the time period will be
  97. The displacement of two identical particles executing SHM are represented by equations x 1 = 4   sin 10 2 t + π / 6   a n d   x 2 =   5 2 cos ω t . For what value of ω energy of both the particles is same?
  98. A student says that he had applied a force F = – k/x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether r is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle
  99. A particle moves on the x-axis according to the equation x = A + B sinwt. The motion is harmonic with amplitude
  100. The motion of a particle is given by X = A sinωt + B cosωt the motion of the particle is
  101. The displacement of particle is represented by the equation . The motion of the particle is
  102. Two mutually perpendicular linear simple harmonic motions of equal amplitude and frequency are imposed on a particle along x and y axis respectively. The initial phase difference between them is π/2. The resultant path followed by the particle is:
  103. A particle is subjected to two S H Ms x 1 = A 1 sin wt and x 2 = A 2 sin (wt + ). The resultant S H M will have an amplitude of
  104. The maximum displacement of an oscillating particle is 0.05m. If its time period is 1.57s (i) What is the velocity at the mean position ? (ii) What is its acceleration at the extreme position ?
  105. The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motion are 1m and 6.28s, respectively. Find its (i) angular frequency, (ii) acceleration and (iii) velocity at a displacement of 0.5m.
  106. The equation motion of a particle in S.H.M is . In the equation ‘a’ is the linear acceleration (in m/sec 2 ) of the particle at a displacement ‘x’ in meter. The time period of S H M in seconds is
  107. Two simple harmonic oscillators with amplitudes in the ratio 1 : 2 are having the same total energy. The ratio of their frequencies is
  108. A particle is moving in a circle of radius R = 1 m with constant speed v = 4 m/s. The ratio of the displacement to acceleration of the foot of the perpendicular drawn from the particle onto the diameter of the circle is
  109. A body of mass 36gm moves with S.H.M of amplitude A = 13 cm and period T = 12sec. At a time t = 0 the displacement is x = +13cm. The shortest time of passage from x = +6.5cm to x = – 6.5cm is
  110. A 20g particle moves in SHM with a frequency of 3 oscillations per second and amplitude of 5 cm. Through what total distance does the particle move during one oscillation? What is its average speed ?
  111. The equation for the displacement of a particle executing SHM is cm. Find (i) the velocity at 3cm from the mean position, (ii) acceleration at 0.5s after leaving the mean position.
  112. A body executing SHM at a displacement ‘x’ its PE is E 1 , at a displacement ‘Y’ its PE is E 2 . The P.E at a displacement (x + y) is
  113. Two SHMs are represented by the equations . Their amplitudes are in the ratio
  114. A particle of mass 4 gm is moving simple harmonically with a period 8 sec and amplitude 8cm. What is its velocity when the displacement is 8cm
  115. The displacement of a particle from its mean position (in meter) is given by . The motion of the particle is
  116. When two displacements represented by and are super- imposed the motion is
  117. For a particle executing SHM. The KE ‘K’ is given by K = K 0 cos 2 wt. The maximum value of PE is
  118. A simple pendulum is of length 50cm. Find its time period and frequency of oscillation. (g = 9.8 m/s 2 ).
  119. On a planet, a body is let fall freely from a height 8m reaches the ground in 2 sec. If the length of a simple pendulum is 1m on that planet, its time period is
  120. If the value of ‘g’ on the surface of the moon is g / 6, the time period of a pendulum on the surface of the moon w.r.t time period on the earth will be
  121. A ball of mass 5kg hangs from a spring of force constant k N/m and oscillates with certain time period. If the ball is removed, the spring shortens by
  122. If the period of oscillation changes from 2 sec to 1.5 sec, and the length of simple pendulum is 1m, then the change in length is
  123. The pendulum of a certain clock has time period 2.04s. How fast or slow does the clock run during 24 hour ?
  124. Two springs of force constant 1000 N/m and 2000 N/m are stretched by same force. The ratio of their respective potential energies is
  125. The potential energy of a particle oscillating on x-axis is given as U=20+(x–2) 2 where ‘U’ is in joules and ‘x’ is in metres. Its mean position is
  126. The displacement of two identical particles executing SHM are represented by equations . For what value of , energy of both the particles is same.
  127. A load of mass ‘M’ is attached to the bottom of a spring of mass ‘M/3’ and spring constant ‘K’. If the system is set into oscillation, the time period of oscillation is
  128. Three masses 0.1 kg, 0.3 kg and 0.4 kg are suspended at the end of the spring. When the 0.4 kg is removed, the system oscillates with a period of ‘2’ sec. When 0.3 kg mass is also removed, the system will oscillate with a period
  129. A spring of spring constant 5 × 10 3 N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by 5 cm is
  130. A mass of 2kg oscillates on a spring with force constant 50 N/m. By what factor does the frequency of oscillation decrease when a damping force with constant b = 12 is introduced?
  131. A uniform circular disc of mass 12kg is held by two identical springs as shown in the figure. When the disc is pressed down slightly and released, it executes SHM with a time period of 2 sec. The force constant of each spring is:
  132. Figure shows the displacement –time graphs of two simple harmonic motions I and II from the graph it follows that
  133. A 4kg roller is attached to a massless spring of spring constant k = 100N/m. It rolls without slipping along a frictionless horizontal road. The roller is displaced from its equilibrium position by l0cm and then released. Its maximum speed will be
  134. One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x o from the black. The spring is then compressed by 2x o and released. The time taken to strike the wall is
  135. A 2kg object attached to a spring moves without friction and is driven by an external force, given by . Force constant of the spring is 20 N/m. Amplitude of motion and time period of oscilation respectively are
  136. If the differential equation of a damped oscillator is then its amplitude A 0 vary with time ‘t’ as A 0 (t) =
  137. If vibrating loaded spring (mass ‘m’) is in a liquid then restoring force (–kx) and viscous force (–bv) are acting. The differential equation for its one dimensional motion is (k = spring constant, b = damping constant)
  138. Maximum velocity of a particle executing SHM is V. Then average speed of the particle over half cycle is
  139. A particle is vibrating in a simple harmonic motion with amplitude 3 cm. When the restoring force acting on the particle is half of the maximum possible restoring force, find the ratio of kinetic energy of the particle to its potential energy.
  140. A chimpanzee swinging on a swing in a sitting position, stands up suddenly, the time period will
  141. Out of the following functions representing motion of a particle which represents SHM (I) y = sin ωt – cos ωt (II) y = sin 3 ωt (III) y = 5 cos( 3 π 4 – 3 ωt ) (IV) y = 1 + ωt + ω 2 t 2
  142. Statement A: The time period of a simple pendulum of infinite length is infinite. Statement B: The time period of a simple pendulum is directly proportional to the square root of length.
  143. Two simple harmonic motions of angular frequency 100 and 1000 rad s – 1 have the same displacement amplitude. The ratio of their maximum acceleration is
  144. A particle executing SHM while moving from one extremity is found at distance x 1 , x 2 and x 3 from the centre at the end of three successive seconds. The time period of oscillation is (where θ = cos – 1 ( x 1 + x 3 2 x 2 ) )
  145. A particle is moving along the x-axis under the influence of a force given by F = – 5 x + 15 . At time t = 0, the particle is located at x = 6 and is having zero velocity. It takes 0.5 seconds to reach the origin for the first time. The mass of the particle is
  146. The displacement of a particle from its mean position (in metre) is given by y = 0 . 2 sin ( 10 πt + 1 . 5 π ) cos ( 10 πt + 1 . 5 π ) . The motion of particle is
  147. A particle is executing linear SHM. The average kinetic energy and average potential energy, over a period of oscillation, respectively are K av and U av . Then,
  148. For a particle executing S.H.M., the kinetic energy K is given by K = K 0 cos 2 ωt . The maximum value of potential energy is
  149. The metallic bob of a simple pendulum has the relative density ρ .The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by
  150. An object is attached to the bottom of a light vertical spring and set vibrating with time period 628 millisecond. The maximum speed of the object is 15 cm/sec and the amplitude in centimeters is
  151. A particle is executing simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/sec. The distance of the particle from the mean position when its speed becomes 5 cm/s is
  152. A particle in S.H.M. is described by the displacement function x(t) = a cos ( ωt + θ ).If the initial (t = 0) position of the particle is I cm and its initial velocity is π cm / s . The angular frequency of the particle is π rad / s , then its amplitude is
  153. A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is l0 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is
  154. Two particles P 1 and P 2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time after which the separation between the particles become 6 cm.
  155. A particle performs SHM with a time period T and amplitude a. The magnitude of average velocity of the particle over the time interval during which it travels a distance a 2 from the extreme position is
  156. A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?
  157. A particle is moving in a circle with uniform speed. Its motion is
  158. For a particle executing simple harmonic motion, determine the ratio of average acceleration of particle from extreme position to equilibrium position to the maximum acceleration.
  159. Which of the following expression does not represent SHM?
  160. A simple harmonic oscillation has got a displacement 0.02 m and acceleration 2.0 m/s 2 at any time. The angular frequency of the motion is:
  161. A particle performs SHM along a straight line with the period I and amplitude A. The mean velocity of the particle averaged over the time interval during which it travels a distance A/2 starting from the extreme position is:
  162. A particle executes simple harmonic “motion between x= – A and x = + A. The time taken by it to go from O to A 2 is T 1 and to go from A/2 to A is T 2 . Then:
  163. The maximum velocity of a particle executing simple harmonic motion with an amplitude 7mm, is 4.4 m/s. The period of oscillation is :
  164. Two simple harmonic motions are represented by the equations y 1 = 0 . 1 sin 100 πt + π 3 and y 2 = 0 . 1 cos πt . The phase difference of the velocity of particle l, with respect to the velocity of particle 2 is :
  165. Equation of a simple harmonic motion is given as x = 3 sin 20 πt + 4 cos 20 πt where x is in cms and t in seconds. The amplitude is:
  166. A point mass oscillates along the x-axis according to the law x = x 0 cos ωt – π 4 . If the acceleration of the particle is written as a = A cos ωt + δ , then :
  167. Out of the following functions representing motion of a particle which represents SHM : (i) y = sin ωt – cos ωt (ii) y = sin 3 ωt (iii) y = 5 cos 3 π 4 – 3 ωt (iv) y = 1 + ωt + ω 2 t 2
  168. A given pendulum has a spherical solid bob of aluminium.Its period is I sec. Now this bob is replaced by a spherical solid bob of the same diameter but made of iron. Density of aluminium is 2.7 g/cc and that of iron is 9.8 g/cc. The period of oscillation of the pendulum will now be:
  169. Two simple pendulums of lengths I m and 16 m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is:
  170. The length of a simple pendulum is increased by 44%. What is the percentage increase in its time period?
  171. A particle of mass m is released from rest and follows a parabolic path as shown in Fig. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?
  172. A second’s pendulum is working in a lift. If the lift begins to fall freely towards the earth, what will be the time period of the pendulum in this case?
  173. The vertical extension in a light spring by a weight of 1 kg suspended from the lower end is 9.8 cm. The period of oscillation of the spring is:
  174. A particle at the end of a spring executes simple harmonic motion with a period t 1 ,while the corresponding period for another spring is t 2 .If the period of oscillation with the two springs in series is T, then :
  175. A force F 1 when it acts on a body causes it to execute SHM of time period T 1 whereas another force F 2 when actions on the same body causes it to execute SHM of time period T 2 .If both forces act on the body simultaneously and in a manner that they are always acting in the same phase, the time period of SHM will be:
  176. Two springs of force constants k and 2k arc connected to a mass mas shown below : The frequency of oscillation of the mass is :
  177. A mass is suspended separately by two different springs in successive order, then time period is T 1 and T 2 respectively.If it is connected by both springs as shown in Fig. then time period is T 0 . The correct relation is
  178. Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k,, the frequency of oscillation of the block is :
  179. An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The rnass is released with the spring initially unstretched. Then the maximum extension in the spring is
  180. A block of mass M kept on smooth horizontal surface are connected by four massless springs whose force constants are 2k, 2k, k and 2krespectively as shown in the figure. If the mass M is displaced in the horizontal direction, then the frequency of the system is
  181. Two springs of force constants K and 2K are connected to a mass as shown below. The frequency of oscillation of the mass is
  182. A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 4.0 x 10 – 3 m. What must be the least period of these oscillations, so that the object is not detached from the platform? (Take g = 10 m / s 2 )
  183. A particle executes simple harmonic motion along a straight line with maximum velocity 4 m/s and maximum acceleration 2 π m / s 2 . If the particle starts its motion from mean position, then the minimum time after which its speed will be 2 m/s is
  184. Maximum acceleration of the bob of a simple pendulum oscillating with amplitude A is 6m/s 2 and that of another pendulum having same amplitude ‘A’ is 3 m/s 2 . A third pendulum is formed by joining the strings of the pendulum and its bob is allowed to oscillate with same amplitude ‘A’. Then maximum acceleration of this bob is
  185. A particle executes simple harmonic motion with time period 4 second. If it starts its motion from extreme position, then the time interval after which its kinetic energy will be equal to its potential energy for the first time is
  186. Potential energy of a particle of mass 80 gm varies with its distance (x) from a fixed point on its line of motion according to the graph shown in the figure. Then time period of oscillation of the particle is
  187. A solid thin metallic disc of radius R is hinged at point O on its rim and freely suspended as shown in the figure. The lower endge of the disc is slightly pulled to the right and released. Then time period of oscillation of the disc is
  188. What will be time period of the displaced body of mass m?
  189. A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is:
  190. A horizontal platform is made to execute SHM of amplitude a in the vertical direction. An object placed on the platform will lose contact with it, when the frequency of oscillation exceeds:
  191. A particle moving with SHM from an extremity of the path towards centre is observed to be at distances x 1 , x 2 and x 3 from the centre at the ends of three successive seconds. The period of SHM is:
  192. Two simple harmonic motions of angular frequencies 100 and 1000 rad/s have the same displacement amplitude. The ratio of their maximum accelerations is :
  193. The displacement of a particle executing SHM at any time t (seconds) is x=0.01sin100π(t +0.05)then its time period will be
  194. Amplitude of oscillation of a particle that executes SHM is 2 cm. Its displacement from its mean position in a time equal to1/6 th of its time period is
  195. A particle executing SHM has amplitude of 1m and time period π sec. Velocity of particle when displacement is 0.8m is
  196. A simple harmonic oscillator is of mass 0.100 kg. It is oscillating with a frequency of 5 π Hz . If its amplitude of vibration is 5 cm, then force acting on the particle at its extreme position is
  197. A pendulum of length L swings from rest to rest n times in one second. The value of acceleration due to gravity is
  198. The amplitude of damped oscillator becomes 1/3 in 2 s. Its amplitude after 6 s is 1/n times the original. The value of n is
  199. A particle is subjected to two SHMs x 1 = A 1 sinωt and x 2 = A 2 sin ωt + π 4 . The resultant SHM will have an amplitude of
  200. The period of oscillation of a particle in SHM is 4 sec and its amplitude of vibration is 4 cm. The distance of the particle 0.5 s after passing the mean position is
  201. A body executing SHM has a maximum velocity of 1ms -1 and a maximum acceleration of 4ms -1 . Its time period of oscillation is
  202. An oscillating mass spring system has mechanical energy 1 joule, when it has an amplitude 0.1m and maximum speed of 1ms- 1 . The force constant of the spring is (in Nm -1 )
  203. A seconds pendulum is suspended from the roof of a lift. If the lift is moving up with an acceleration 9.8 m/s 2 , its time period is1 s
  204. A seconds pendulum is shifted from a place where g = 9.8 m/s 2 to another place where g = 9.82 m/s 2 . To keep period of oscillation constant its length should be
  205. A body dropped from a height h onto the floor makes elastic collision with the floor. The frequency of oscillation of its periodic motion is
  206. A particle performing SHM having amplitude ‘a’ possesses velocity ( 3 ) 1 / 2 2 times the velocity at the mean position. The displacement of the particle shall be
  207. A body is executing SHM. At a displacement x its PE is E 1 and at a displacement y its PE is E 2 . The PE at displacement (x+y) is
  208. A body of mass 0.5 kg is performing SHM with a time period π/2 seconds. If its velocity at mean position is 1ms –1 , then restoring force acting on the body at a phase angle 60° from extreme position is
  209. When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2.0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity of it in m/s is (g = 10m/s 2 )
  210. A block of mass M suspended from a spring oscillates with time period T. If spring is cut in to two equal parts and same mass M is suspended from one part , new period of oscillation is
  211. Two masses m 1 and m 2 are suspended from a spring of spring constant ‘k’. When the masses are in equilibrium, m 1 is gently removed. The angular frequency and amplitude of oscillation of m 2 will be respectively
  212. A simple pendulum with a brass bob has a period T. The bob is now immersed in a nonviscous liquid and oscillated. If the density of the liquid is 1/8th of brass, the time period of the same pendulum will be
  213. A pendulum clock is taken 1 km inside the earth from mean sea level. Then the pendulum clock
  214. A particle of mass m is attached to a spring of spring constant ‘k’ and has a natural frequency ω o . An external force F(t) proportional to cos ⁡ ωt ω ≠ ω o is applied to the oscillator. The maximum displacement of the oscillator will be proportional to
  215. A sewing machine needle is oscillating simple harmonically in a vertical line of length 6 cm with a frequency 30 per minute. The displacement of the needle 1/6s after crossing the mean position is
  216. A particle in SHM is described by the displacement function x ( t ) = Acos ⁡ ( ωt + ϕ ) . If the initial (t = 0) position of the particle is 1 cm, its initial velocity is p cm s – 1 and its angular frequency is π rad s -1 , then the amplitude of its motion is
  217. The equation of the displacement of two particles making SHM are represented by y 1 = a sin ⁡ ( ωt + ϕ ) & y 2 = acos ⁡ ( ωt ) . Phase difference between velocities of two particles is
  218. A 20 g particle is executing SHM between the limits (5, 0, 0)cm and (15, 0, 0)cm. The total distance covered during one oscillation is
  219. A cylinder of mass M and radius .H is resting on a horizontal platform which is parallel to X-Y plane) with its axis fixed along the Y-axis and free to rotate about its axis. The platform is given a motion in X-direction given by x = A cos ωt . There is no slipping between the cylinder and platform. The maximum torque acting on the cyliner during its motion is
  220. For a particle executing simple harmonic motion, the kinetic energy K is given by K = K 0 cos 2 ⁡ ωt The maximum value of potential energy is
  221. The total energy of the body executing S.H.M. is E, Then the kinetic energy when the displacement is half of the amplitude is
  222. In a S.H.M. when the displacement is one half the amplitude, what fraction of the total energy is kinetic ?
  223. A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be
  224. Two simple pendulums of lengths 1 m and 16 m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is
  225. The kinetic energy of a particle, executing S.H.M. is 16 J, when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillations is
  226. A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the-particle to move from x = + 2 cm to x = + 4 and back again is given by
  227. Two pendula of length 121 cm and 100 cm start vibrating. At some instant the two are in mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position
  228. Two simple pendulums of length l m and 16 m start at t =0 from mean position. The minimum time after which they will be again in phase will be
  229. A pendulum suspended from the ceiling of the train has got a time period of two second when the train is at rest. What will be the time period of the pendulum if the train accelerates at 10 m/s 2 ?
  230. The total energy of a particle, executing simple harmonic motion is
  231. A heavy brass-sphere is hung from a spiral spring and it executes vertical vibrations with period T. The ball is now immersed in non-viscous liquid with a density one-tenth that of brass. When set into vertical vibrations with the sphere remaining inside the liquid all the time, the period will be
  232. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω 0 . . An external force F(f) proportional to cos ωt ω ≠ω 0 is applied to the oscillator. The time displacement of the oscillator will be proportional to
  233. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω 0 . . An external force F(f) proportional to cos ωt ω ≠ ω 0 is applied to the oscillator. The time displacement of the oscillator will be proportional to
  234. A spring when stretched up to 2 cm has potential energy U. If it is stretched up to 10 cm, potential energy would be
  235. Two sprin8s, of force constants k 1 , and k 2 , are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k 1 and k 2 , are made four times their original values, the frequency of oscillation becomes
  236. A particle at the end of a spring executes simple harmonic motion with a period t 1 , while the corresponding period for another spring is t 1 If the period of oscillation with the two springs in series is T , then
  237. An elastic spring, suspended vertically gets extended in length by i, when a mass M is attached at other end. What will be the period of vibration of the spring when a mass m is attached at the lower end ?
  238. The frequency of oscillation of the springs shown in fig (4) will be
  239. A particle executing S.H.M. between x = – A and x = + A. {. The time taken for it to go from 0 to A /2 is T 1 and to go from A /2 to A is T 2 .Then
  240. A body of mass M is released from a height h to a scale pan hung from a spring as shown in fig (9). The spring constant of the spring is k, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is
  241. Two identical springs of constant.k are connected in series and parallel as shown in fig. {g). A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be
  242. A particle of mass 0.3 kg is subjected to a force F = -kx with k = 15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
  243. A particle executing SHM of amplitude 4 cm and T = 4 s. The time taken by it to move from positive extreme position to half the amplitude is
  244. Two particles are executing SHMs. The equations of their motions are y 1 = 10 sin ωt + π 4 and y 2 = 5 sin ωt + 3 π 4 What is the ratio of their amplitudes?
  245. A horizontally placed spring mass system has time period T. The same system is now placed on a car moving with acceleration a in horizontal direction. Then,
  246. A body is vibrating in simple harmonic motion. If its acceleration is 12 cms – 2 at a displacement 3 cm from the mean position, then time period is
  247. The velocity-time graph of a harmonic oscillator is shown in the figure below. The frequency of oscillation is
  248. Which of the following quantities is always negative in SHM?
  249. A block of mass m, attached to a spring of spring constant k, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed v when the spring is at its natural length. Before coming to an instantaneous rest, if the block moves a distance x from the mean position, then
  250. The equation of motion of a particle is x = a cos ( αt ) 2 . The motion is
  251. A solid cylinder of mass M and radius R is connected to spring of force constant k as shown in fig. Cylinder rolls without slipping on the horizontal surface. Its time period of oscillation is
  252. The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos ⁡ ( ω t ) where x = displacement at time t ω =frequency of oscillation Which one of the following graphs shows correctly the variation a with t?
  253. The phase difference between the displacement and acceleration of a particle executing simple harmonic motion is
  254. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s – 2 at a distance of 5 m from the mean position. The time period of oscillation is
  255. If a dielectric slab is just made to enter within a parallel plate capacitor, then the subsequent motion of the dielectric slab is
  256. The speed of a particle executing SHM is given by the equation v 2 = 144 − 9 x 2 then the wrong statements among the following is
  257. A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β . Then, its time period of vibration will be
  258. A particle is executing SHM along a straight line. Its velocities at distances x 1 and x 2 from the mean position are V 1 and V 2 , respectively. Its time period is
  259. A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s . When the mass m is increased by 1 k g , the time period of oscillations becomes 5 s . The value of m in kg is
  260. A body executes sample harmonic motion under the action of a force F 1 with a time period (4/5) sec. If the force is changes to F 2 it executes SHM with time period (3/5) sec. if both the forces F 1 a n d F 2 act simultaneously in the same direction on the body, its time period (in seconds) is:
  261. When two displacements represented by y 1 = asin ( ωt ) and y 2 = b cos ( ωt ) are superimposed the motion is
  262. A particle of mass m oscillates along x -axis according to equation x = a sin ω t . The nature of the graph between momentum and displacement of the particle is
  263. A mass M, attached to a horizontal spring, executes SHM with amplitude A 1 . When the mass M passes through its means position then a smaller mass m is placed over it and both of them move together with amplitude A 2 . The ratio of A 1 A 2 is
  264. A spring of force constant k is cut into lengths of ratio 1: 2: 3 . They are connected in series and the new force constant is k’. Then they are connected in parallel and force constant is k”. Then k’ : k” is
  265. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
  266. The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :
  267. A uniform rod AB of length L is pivoted at one end A and hangs vertically. The time period of small oscillations about an axis passing through A and perpendicular to the rod is
  268. Two pendulums of lengths 121cm and 100cm start vibrating. At some instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position?
  269. Four massless springs whose force constants are 2k, 2k, k and 2 k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of the system.
  270. The displacement of a particle executing simple harmonic motion is given by x = a sin ω t + a cos ω t The total energy of the particle is
  271. A particle is moving in a circle with uniform speed. Its motion is
  272. In the figure all springs are identical having spring constant K and mass m each. The block has also mass m. The frequency of oscillation of the block is
  273. The distance covered by a particle undergoing SHM in one time period is (amplitude = A),
  274. What is the minimum coefficient of static friction μ s for which the 1kg block will not slip relative to the lower block if the lower block is displaced by 50 mm to the right from equilibrium position and released ? Take g = 10 m / s 2 .
  275. A thin rod AB of mass m and length l is hinged at end A and suspended freely. A particle of mass m is attached at the lower end B. The lower end of the rod is slightly pulled to one side and released. Then frequency of oscillation of the rod is
  276. A body is hanging from a rigid support by an inextensible string of length l . It is struck by an identical body of mass m with horizontal velocity u = 2 g l . The collision is perfectly inelastic. The increase in tension will be (Just after the collision)
  277. The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is
  278. The minimum time required for a boat to cross a river is 30 min and the time required for the boat to cross the same river in the shortest possible path is 50 min. In both cases velocity of boat relative to still water is ‘V’ and velocity of water is ‘u’. Then u v is
  279. The Time period of oscillation of a simple pendulum whose length is very large (>> Radius of earth R) is T. What will be the time period of a simple pendulum whose length is equal to the radius of earth(R) ?
  280. A particle is executing SHM of amplitude A and time period T along a straight line. Then time required for the particle to directly move from x = − A 2 to x = + A 2 is
  281. A particle is executing SHM along a straight with an amplitude of 5 cm and time period 2 sec. Find the acceleration of the particle when its velocity is equal to half of its maximum velocity.
  282. A thin rod of length ‘l’ is hinged at the upper end ‘O’ as shown in figure. The lower end is slightly pulled to the right and released and the rod starts oscillating in the plane of the figure. Then its time period of oscillation is
  283. A block is suspended by a spring and its time period of small vertical oscillation is found to be T. Now the spring is cut into two segments the ratio of whose length is 1 : 2. The segments are arranged in parallel combination and the same block is suspended from the combination. If the block is now allowed to oscillate vertically, find its new time period.
  284. The total energy of a body executing SHM is E. Then the kinetic energy when its distance from equilibrium position is half of the amplitude, is
  285. The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minutes will be 1/x times the initial amplitude, where x is
  286. Two springs are made to oscillate simple harmonically due to the same mass individually. The time periods obtained are T 1 and T 2 . If both the springs are connected in series and then made to oscillate by the same mass, the resulting time period will be
  287. A small mass executes linear SHM about O with ampli­tude a and period T. Its displacement from O at time T/8 after passing through O is
  288. A smooth inclined plane having angle of inclination of 30 0 with the horizontal has a 2.5 kg mass held by a spring which is fixed at the upper end. If the mass is taken 2.5 cm up along the surface of the inclined plane, the tension in the spring reduces to zero, If the mass is now released, the angular frequency of oscillation is
  289. A body of mass 1 kg is executing SHM. Its displacement y (cm) at t seconds is given by y = 6 sin 100 t + π 4 . Its maximum kinetic energy is
  290. The period of small oscillations of a particle of mass m in a potential energy field given by U = U 0 1 – cos α x is
  291. A hollow sphere is filled with water. It is hung by a long thread to make it a simple pendulum. As the water flows out of a hole at the bottom of the sphere, the frequency of oscillation will
  292. A simple pendulum of length 1m is taken to height R (radius of the earth) from the earth’s surface. The time period of small oscillation of the pendulum at that point is
  293. A simple pendulum of length 90 cm is oscillating with an angular amplitude of 4 o ,Then the maximum velocity of its bob is ( Take   g = 10 m / s 2 )
  294. The displacement equation of a particle is is x = ( 3 sin 2 t + 4 cos 2 t ) c m . Then magnitude of maximum acceleration of the particle is
  295. The time period of a simple pendulum suspended from the ceiling of a lift is T. A block is also suspended by a spring from the ceiling of the lift and the block is oscillating vertically with time period T. Now the lift starts moving upward with constant acceleration ‘a’. If the new time period of the simple pendulum and spring – block be T P and T S respectively, then
  296. In the arrangement shown, the block is slightly displaced to the right and then released. Then time period of oscillation of the block is
  297. Two simple harmonic motions given by y 1 = 10 sin 100 πt and y 2 = 5 cos 100 πt − π / 6 are super posed on a particle. The amplitude of the resulting motion of the particle is
  298. A block is suspended by a spring and allowed to oscillate vertically with small amplitude. Time period of oscillation is T. Now the spring is cut into two parts whose lengths are in the ratio 2 :3. The two parts of the spring are attached to the same block as shown in figure. The block is now slightly pushed to the right and released. The new time period of oscillation is
  299. A particle is executing SHM with time period π sec along a straight line. At t=0, it has started its motion from one of its extreme position. Then the minimum time after which its kinetic energy will be equal to its potential energy is
  300. A particle is executing two different simple harmonic motions, mutually perpendiuclar, of different amplitudes and having phase difference of π 2 . The path of the particle will be
  301. A block is suspended by a spring and the elongation produced in the spring is found to be x 0 . Then the block is further pulled downward by a distance x 0 and released. Then acceleration of the block in its lower most position is
  302. Two particles P and Q start from origin at the same time and execute simple harmonic motion along x-axis with same amplitude but with periods 3 sec and 6 sec respectively. The ratio of the speeds of P and Q when they meet for the first time after start is
  303. A particle moves in a circular path with a uniform speed. Its motion is
  304. A particle moves on the x-axis according to the equation x = x 0 sin 2ωt. The motion is simple harmonic.
  305. At two particular closest instants of time t 1 and t 2 , the displacements of a particle performing SHM are equal. At these instants.
  306. A particle moves in the x–y plane according to the equation; . Motion of the particle is:
  307. The function represents
  308. The composition of two simple harmonic motions of equal periods at right angles to each other and with a phase difference of α, results in the displacement of the particle along a
  309. A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be
  310. The time period of a particle performing linear SHM is 12s. What is the time taken by it to make a displacement equal to half its amplitude from equilibrium position?
  311. The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4 cm. Its initial displacement is 2 cm towards positive extreme position. Its equation for displacement is
  312. The angular velocity and amplitude of a simple pendulum are ‘ ω ’ and A respectively. At a displacement ‘x’. If its KE is ‘T’ and PE is ‘U’ then the ratio of ‘T’ to ‘U’ is
  313. Which of the following expressions for force acting on a particle corresponds to simple harmonic motion along a straight line, where x is the displacement and a,b,c are positive constants ?
  314. A simple harmonic oscillator is of mass 0.1kg. It is oscillating with a frequency of . If its amplitude of vibration is 5cm find the force acting on the particle at its extreme position.
  315. A particle is vibrating in SHM with an amplitude of 4cm. At what displacement from the equilibrium position it has half potential and half kinetic
  316. The displacement of a particle executing SHM is given by y = 10 sin (3t + )m and ‘t’ is in seconds. The initial displacement and maximum velocity of the particle are respectively
  317. A particle moves according to the law . The distance covered by it in the time interval between t = 0 to t = 3 sec is
  318. For a body in S.H.M the velocity is given by the relation v = m/sec. The maximum acceleration is
  319. Acceleration displacement graph of a particle executing S.H.M is as shown in given figure. The time period of its oscillation is (in sec)
  320. An object executing SHM has an acceleration of 0.2m/s 2 at a distance of 0.05m from the mean position. Calculate the period of SHM.
  321. The velocity of a particle under going SHM at the mean position is 2ms –1 . Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude.
  322. A block is placed on a horizontal platform. The system is making horizontal oscillations about a fixed point with a frequency of 0.5Hz. The maximum amplitude of oscillation if the block is not to slide on the horizontal platform, is (Coefficient of friction between the block and the platform = 0.4)
  323. A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them is:
  324. A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position is
  325. The periods of a pendulum on two planets are in the ratio 3 : 4. The acceleration due to gravity on them are in the ratio
  326. Maximum restoring force on the bob of a simple pendulum of mass 100 gm when its amplitude is 10 0 is
  327. The period of a simple pendulum measured inside a stationary lift is ‘T’. If the lift starts moving upwards with a acceleration g / 3. What will be the time period
  328. A mass of 2kg attached to a spring of force constant 800N/m makes 100 oscillation. The time taken is (in seconds)
  329. The time period of simple pendulum is ‘T’. When the length increases by 10 cm, its period is T 1 . When the length is decreased by 10 cm, its period is T 2 . Then the relation between T, T 1 and T 2 is
  330. A mass ‘M’ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period ‘T, if the mass is increased by ‘m’, time period becomes , then the ratio of (m / M ) is
  331. T 1, T 2 and T 3 the time periods of a given pendulum on the surface of the earth, at a depth ‘h’ in a mine and at an altitude ‘h’ above the earth’s surface respectively, then
  332. The elongation of a spring of length ‘L’ and of negligible mass due to a force is ‘x’. The spring is cut into two pieces of length in ratio 1 : n. The ratio of the respective spring constants is
  333. Springs of spring constants K, 2K, 4K, 8k, …. 2048 K are connected in series. A mass ‘m’ is attached to one end the system is allowed to oscillation. The time period is approximately
  334. If a spring of force constant ‘K’ is cut into three equal parts, then the force constant of each part will be
  335. A spring balance has a scale that reads 0 to 20 kg. The length of the scale is 10 cm. A body suspended from this balance, when displaced and released, oscillates with period of . What is the mass of the body
  336. A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be
  337. In forced oscillations displacement equation is then amplitude ‘A’ vary with forced angular frequency and natural angular frequency ‘ ’ as (b = damping constant)
  338. A particle performs harmonic oscillations along a straight line with a period T and amplitude a. The mean velocity of the particle averaged over the time interval during which it travels a distance starting from the extreme positions is
  339. A particle mass ‘m’ free to move in x-y plane is subjected to a force whose components are F x = –kx and F y = –ky, where ‘k’ is a constant. The particle is released when t = 0 at the point (2,3). The particle is in simple harmonic motion along the line represented by the equation.
  340. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45 0 , then
  341. If the differential equation of a damped oscillator is then energy of damped oscillator vary with time ‘t’ is E(t) =
  342. A particle executes simple harmonic motion of period T and amplitude l along a rod AB of length 2l. The rod AB itself executes simple harmonic motion of the same period and amplitude in a direction perpendicular to its length. Initially, both the particle and the rod are in their mean positions. The path traced out by the particle will be
  343. If the differential equation of a damped oscillator is then for small damping is
  344. If the differential equation of a damped oscillator is then angular frequency of damped oscillator is
  345. A block is suspended by a spring and the deflection produced in the spring is 2 cm when the block is slightly pulled downward and released, its time period is found to be 1 second. When another block is suspended by the same spring, its elongation is found to be 8 cm. Now if this block is slightly pulled downward and released, what will be its time period?
  346. In the arrangement shown, the blocks are pulled apart by 3 cm and released. If maximum velocity of 1 kg block is 1 m/s, what is the maximum velocity of 2 kg block?
  347. A particle is executing simple harmonic motion according to the equation x = ( 5 cos ⁡ 100 t + 12 sin ⁡ 100 t ) cm where t is time in second. Then maximum velocity of the particle is
  348. Two simple harmonic motions given by x 1 = 3 cos ⁡   10 t ​  c m  and  x 2 = 4   sin ⁡ 10 t   c m are superposed on a particle. Then amplitude of motion of the particle is
  349. The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be
  350. Displacement between maximum potential energy position and maximum kinetic position for a particle executing S.H.M. is
  351. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes S.H.M. of time period T. If the mass is increased by m, the time period becomes 5 T 3 . Then the ratio of m M is
  352. The total energy of a particle ,executing simple harmonic motion is( x is displacement)
  353. There is a body having mass m and performing SHM with amplitude a. There is a restoring force F = -Kx, where x is the displacement. The total energy of body depends upon
  354. The kinetic energy and the potential energy of a particle executing SHM are equal. The ratio of its displacement and amplitude willl be
  355. When a mass is hung by a spring, the spring stretches 6.0 cm. Determine its time period of vibration if it is then pulled down a little and released.
  356. A block of mass M is suspended by a spring and allowed to oscillate vertically with time period 1 second wen another block of mass m is attached to the block of mass M and allowed to oscillate vertically, the time period is found to be 2 sec then, M/m is equal to
  357. A particle is executing simple harmonic motion along a straight line with amplitude 2 cm. When the particle is at a distance of 1 cm from its equilibrium position, its velocity is equal to 4   c m s – 1 . Then time period of oscillation is
  358. A particle executing SHM with amplitude A has total energy E. At what distance from the mean position its kinetic energy will be 5 E 9 ?
  359. A particle doing simple harmonic motion, amplitude = 4 cm, time period = 12 sec. The ratio between time taken by it in going from its mean position to 2 cm and from 2 cm to extreme position is
  360. A simple pendulum has time period T 1 . The point of suspension is now moved upward according to equation y = k t 2 where k = 1 m / sec 2 . If new time period is T 2 , then ratio T 1 2 T 2 2 will be
  361. A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U ( x ) = k [ x ] 3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is
  362. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to
  363. The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If it is a second’s pendulum on earth)
  364. Two simple harmonic motions are represented by the equations y 1 = 0 . 1 sin ( 100 πt + π 3 ) a n d y 2 = 0 . 1 cos πt . . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is
  365. A particle executes simple harmonic motion with a frequency f. The frequency with which its kinetic energy oscillates is
  366. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω 0 . An extemal force F(t) proportional to cos ω t ( ω ≠ w 0 ) is applied to the oscillator. The time displacement of the oscillator will be proportional to
  367. Two bodies M and N of equal masses are suspended from two separate massless springs of force constants k 1 and k 2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude M to that of N is
  368. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t 0 in air. Neglecting frictional force of water and given that the density of the bob is ( 4 3 ) × 1000 kg / m 3 . Which relationship between t and t 0 is true?
  369. A particle at the end of a spring executes simple harmonic motion with a period t 1 , while the corresponding period for another spring is t 2 . If the period of oscillation with the two springs in series is T, then
  370. Assertion: Simple harmonic motion is not a uniform motion. Reason: It is the projection of uniform circular motion.
  371. Assertion: In simple harmonic motion, the velocity is maximum when acceleration is minimum Reason: Displacement and velocity of SHM. differ is phase by π 2 .
  372. A particle is performing simple harmonic motion. If its velocities are v 1 and v 2 at the displacements from the mean position arc y 1 and y 2 respectively, then its time period is
  373. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes( 5 4 T ). The ratio of m M is
  374. The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 cm/s. The distance of the particle from the mean position at which the speed of the particle becomes 8 3 cm/s will be
  375. Average velocity of a particle executing SHM in one complete vibration is
  376. The displacement of a particle executing simple harmonic motion is given by y = A 0 + Asin ωt + B cos ωt . Then the amplitude of its oscillation is given by
  377. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
  378. A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β . Then, its time period of vibration will be:
  379. A particle is executing SHM along a straight line. Its velocities at distances x 1 and x 2 from the mean position are V 1 and V 2 respectively. Its time period is
  380. When two displacements represented by y 1 = a sin ( ωt ) and y 2 = b cos ( ωt ) are superimposed the motion is
  381. The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are
  382. Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is
  383. The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be
  384. The displacement of a particle along the x-axis is given by x = a sin 2 ωt . The motion of the particle corresponds to
  385. Statement I: Two unequal springs of same material are loaded with same load. The longer one will have larger value of time period. Statement II: The concept will follow, if we made a experiment to measure.
  386. Statement I: A small body of mass 0.1 kg is undergoing SHM of amplitude 1.0 m and period 0.2 s. The maximum value of the force acting on it is 98.7 N. Statement II: Maximum force acting on it is given by F = mω 2 r .
  387. Statement I: A man with a wristwatch spring wound on his hand falls from the top of a tower. The watch will show this correct time. Statement II: The acceleration due to gravity have no effect on time period of watch at the time of falling.
  388. Statement I: Sine and cosine functions are periodic functions Statement II: Sinusoidal functions repeat its values after a definite interval of time.
  389. Statement I : For a simple pendulum the graph between g and T 2 is hyperbola. Statement II : T = 2 π g l
  390. Statement I: In a simple harmonic motion the kinetic and potential energies become equal when the displacement is 1 2 times the amplitude. Statement II: In SHM kinetic energy is zero when potential energy is maximum.
  391. Statement I: Damped vibrations indicate loss of energy. Statement II: The loss may be due to friction, air resistance etc.
  392. Statement I: The simple harmonic motion is to and fro and periodic. Statement II: The motion of the earth is periodic.
  393. Statement I: The bob of a simple pendulum is a ball full of water. If a fine hole is made at the bottom of the ball, then the time period will no more remain constant. Statement II: The time period of simple pendulum does not depend upon mass.
  394. Statement I: A hole were drilled through the centre of earth and a ball is dropped into the hole at one end, it will not get out of other end of the hole. Statement II: It will come out of the other end normally.
  395. Statement I: The time period of a pendulum, on a satellite orbiting the earth in infinity. Statement II: Time period of a pendulum is inversely proportional to square root of acceleration due to gravity.
  396. Which one of the following equations of motion represents simple harmonic motion?
  397. The particle executing simple harmonic motion has a kinetic energy K 0 cos 2 ωt . The maximum values of the potential energy and the total energy are respectively
  398. A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ . If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T then:
  399. A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is
  400. The speed (v) of a particle moving along a straight line, when it is at a distance (x) from a fixed point on the line, is given by: v 2 = 144 – 9 x 2 . Column-I Column-II i. Motion is simple harmonic of period p. 2 π 3 units ii. Maximum displacement from the fixed point is q. 12 units iii. Maximum velocity of the particle r. 27 units iv. Magnitude of acceleration at a distance (S) 3 units from the fixed point is s. 4 units Now, match the given columns and select the correct option from the codes given below.
  401. A horizontal spring-block system of mass 2 kg executes S.H.M. When the block is passing through its equilibrium position, an object of mass I kg is put on it and the two move together. The new amplitude of vibration is (A being its initial amplitude)
  402. A particle of mass m is executing oscillations about the origin on the X-axis with amplitude A. Its potential energy is given as U ( x ) = βx 4 where β is a positive constant. The x-coordinate of the particle, where the potential energy is one-third of the kinetic energy, is
  403. A body is executing a simple harmonic motion such that its potential energy is U 1 at x and U 2 at y. When the displacement is x+y , the potential energy will be
  404. A mass of 0.98 kg attached to a spring of constant K = 100 N m – 1 is hit by a bullet of 20 gm moving with a velocity 30 ms – 1 horizontally. The bullet gets embedded and the system oscillates with the mass on horizontal friction less surface. The amplitude of oscillations will be
  405. A particle moves along a straight line to follow the equation ax 2 + bv 2 = k , where a, b and k are constants and x and v are x-coordinate and velocity of the particle respectively. Find the amplitude.
  406. A particle free to move along the x-axis has potential energy given by U ( x ) = K [ 1 – exp ( – x 2 ) ] for – ∞ < x < + ∞ where t is a positive constant of appropriate dimensions. Then
  407. Find the distance covered by a particle from time t = 0 to t = 6 sec, when the particle follows the movement according to y = a cos ( π 4 ) t ,
  408. A particle executes simple harmonic motion according to the displacement equation y = 10 cos ( 2 πt + π 6 ) cm ; , where t is in seconds. The velocity of the particle at t = 1 6 seconds will be:
  409. Two S.H.Ms. s 1 = a sin ωt and s 2 = b sin ωt are superimposed on a particle. The s 1 and s 2 are along the directions which make angle 37° with each other.
  410. A charged particle is deflected by two mutually perpendicular oscillating electric fields such that the displacement of the particle due to each one of them is given by x = A sin ( ωt ) and y = A sin ( ωt + π 6 ) respectively. The trajectory followed by the charged particle is
  411. The equation of the resulting oscillation obtained by the summation of two mutually perpendicular oscillations with the same frequency f 1 = f 2 = 5 Hz and same initial phase δ 1 = δ 2 = 60 0 is, (Given, their amplitudes are A 1 = 0 . 1 m and A 2 = 0 . 05 m
  412. The displacement of a particle varies according to the relation x = 4 ( cos πt + sin πt ) The amplitude of the particle is
  413. A particle is acted simultaneously by mutually perpendicular simple harmonic motion.x = a cos ωt and y = a sin ωt .The trajectory of motion of the particle will be
  414. The displacement of a particle varies with time as x = 12 sin ωt – 16 sin 3 ωt (in cm). If its motion is S.H.M., then its maximum acceleration is
  415. Two pendulum start oscillating in the same direction at the same time from the same mean position. Their time periods are respectively 2 s and 1.5 s. The phase difference between them, when the smaller pendulum has completed one vibration, will be
  416. Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is 5 T 4 , where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.
  417. A simple pendulum hung from the ceiling of a train moving at constant speed has a period T. If the train starts accelerating or decelerating, then what will be the effect on time period of pendulum?
  418. Two simple pendulums whose lengths are 100 cm and 121 cm are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again?
  419. A spring of spring constant K is cut into n equal parts, out of which r parts are placed in parallel and connected with mass m as shown in figure. The time period of oscillatory motion of mass m is
  420. A force of 6.4 N stretches a vertical spring by 0. I m. The mass that must be suspended from the spring so that it oscillates with a time period of π 4 second.
  421. Two equal masses are suspended separately by two springs of constant K 1 and K 2 . If their oscillations satisfy the condition that their maximum velocities are equal, then the ratio of the amplitudes of the oscillations of the masses respectively is
  422. The frequency of oscillation ( 10 π ) (in Hz) of a particle of mass 0.1 kg which executes SHM along x-axis. The kinetic energy is 0.3 J and potential energy is 0.2J at position x = 0.02 m. The potential energy is zero at mean position. Find the amplitude of oscillations (in meters)
  423. A particle of mass m is executing SHM about the origin on x-axis with frequency ka πm , where k is a constant and a is the amplitude. Find its potential energy, if x is the displacement at time t:
  424. A body is executing SHM under the action of a force whose maximum magnitude is 50 N. The magnitude of force acting on the particle at the time when its energy is half kinetic and half-potential is (Assume potential energy to be zero at mean position).
  425. A linear harmonic oscillator of force constant 2 × 10 6 Nm – 1 and amplitude 0.01 m has a total mechanical energy 160 J. Among the following statements, which are correct? i. Maximum PE is 100 J ii. Maximum KE is 100 J iii. Maximum PE is 160 J iv. Minimum PE is zero
  426. The total mechanical energy of a particle executing simple harmonic motion is E. When the displacement is half the amplitude its kinetic energy will be
  427. A particle of mass l0 gm is placed in a potential field given by V = (50 x 2 + 100) J/kg. The frequency of oscillation in cycle/sec is
  428. A 4 kg particle is moving along the x-axis under the action of the force F = – ( π 2 16 ) xN . At t = 2 sec, the particle passes through the origin and at t = l0 sec its speed is 4 2 m/s. The amplitude of the motion is:
  429. Total energy of a particle executing S.H.M. is (x is displacement from mean position):
  430. An object of mass 0.2 kg executes simple harmonic along X-axis with frequency of 25 π Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J amplitude of oscillation in meter is equal to
  431. A body is executing Simple Harmonic Motion. At a displacement x its potential energy is E 1 and at a displacement y its potential energy is E 2 . The potential energy E at displacement (x +y) is
  432. A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement x. Which of the following statement is true?
  433. The KE and PE of a particle executing SHM with amplitude A will be equal when its displacement is
  434. The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when
  435. Match the columns Column-I Column-II i. Zero gravity p. an SHM is definitely not possible ii. No restoring force q. an SHM is definitely possible iii. Mechanical energy remains conserved r. an SHM may be possible iv. displacement of a particle from a reference level is given by y = sin 2 ωt s. Can’t say Now, select the correct option from the codes given below.
  436. Time period (T) and amplitude (A) are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase 3 π 2 and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other.
  437. The phase difference between two particles executing SHM of the same amplitudes and frequency along same straight line while passing one another when going in opposite directions with equal displacement from their respective stating point is 2 π 3 . If the phase of one particle is π 6 , find the displacement at this instant, if amplitude is A.
  438. The phase difference between the displacement and acceleration of a particle executing simple harmonic motion is
  439. A particle executes SHM of amplitude A and time period T. The distance travelled by the particle in the duration its phase changes from π 12 to 5 π 12 .
  440. Two simple harmonic motions are represented by equations y 1 = 4 sin ( 10 t + ∅ ) ⇒ y 2 = 5 cos 10 t What is the phase difference between their velocities?
  441. Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same shaight line. The maximum distance between the two particles is a 2 . The phase difference between the particle is:
  442. A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x = 2 cm to x = +4 cm and back again is given by
  443. At t = 0, a particle of mass m starts moving from rest due to a force F = F 0 sin ( ωt ) i ^
  444. The displacement of a particle is represented by the equation y = sin 3 ωt . The motion is
  445. The function sin 2 ( ωt ) represents:
  446. The function sin 2 ( ωt ) represents:
  447. A plate oscillated with time period T. Suddenly, another plate put on the first plate, then time period
  448. If a spring has time period T, and is cut into n equal parts, then the time period of each part will be
  449. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m / s 2 at a distance of 5 m from the mean position. The time period of oscillation is
  450. A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass rn is increased by I kg, the time period of oscillations becomes 5 s. The value of m in kg is
  451. A simple pendulum hangs from the ceiling of a car. If the car accelerates with a uniform acceleration, the frequency of the simple pendulum will
  452. Which of the following function represents a simple harmonic oscillation
  453. A point performs simple harmonic oscillation of period T and the equation of motion is given by x = asin ( ωt + π 6 ) . After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
  454. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:
  455. The resultant of two rectangular simple harmonic motions of the same frequency and unequal amplitudes but differing in phase by π 2 is
  456. A particle is executing SHM. At a point x = A 3 , kinetic energy of the particle is K, where A is the amplitude. At a point x = 2 A 3 , kinetic energy of the particle will be:
  457. The speed of a particle moving along a straight line, when it is at a distance (x) from a fixed point on the line is given by v 2 =108-9x 2 (in CGS unit), Then
  458. How long after the beginning of motion is the displacement of a harmonically oscillating point equal to one half of its amplitude, if the period is 24 sec and initial phase is zero?
  459. The number of harmonic components in the oscillation represented by y – 4 cos 2 2t sin 4t and their corresponding angular frequencies are:
  460. If time taken by a particle executing SHM, from mean position to extreme position is 6 s, the frequency of SHM is :
  461. A particle executes SHM. Its velocities are v 1 and v 2 at displacements x 1 and x 2 from mean position respectively. The frequency of oscillation will be:
  462. The time period and the amplitude of a simple pendulum are 4 second and 0.20 metre respectively. If the displacement is 0.1 m at time t = 0, the equation of its displacement is represented by:
  463. A particle moves along a circle of radius 5 cm with a period of 6 sec in the clockwise direction. At t = 0 the particle is at position A as shown in the Fig.The expression for the SHM of the x-projection of the radius vector of the particle is
  464. A small mas.s executes linear simple harmonic motion about a point O with amplitude a and period T. Its displacement from O at time T/8 after passing through O is:
  465. If a simple harmonic motion is represented by d 2 x dt 2 + αx = 0 ; its time period is :
  466. Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?
  467. A coin is placed on a horizontal platform which under goes vertical simple harmonic motion of angular frequency ω ). The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time:
  468. A particle starts SHM from the mean position. Its amplitude is A and,total energy E. Atone instant its kinetic energy is 3 E 4 .Its displacement at that instant is :
  469. The displacement of a particle varies according to the relation. x = 4 cos πt + sin πt The amplitude of the particle is:
  470. The displacement of a particle is represented by the equation y = sin 3 ωt . The motion is :
  471. Velocity-time graph of a particle performing SHM is as shown in Fig.Amplitude of oscillation is then nearly:
  472. A particle is performing simple harmonic motion. Equation of its motion is x = 5 sin 4 t – π 6 , x being the displacement from mean position. Velocity (in ms -1 )of the particle at the instant when its displacement is 3, will be:
  473. An object of mass 4 kg is moving along X-axis and its potential energy as a function of x can be expressed as U(x) = 4 (1- cos 2x) joule then time period for small oscillations is:
  474. Fig. shows an object of mass 50 gm placed on a rough horizontal surface. The surface together with the block on it is performing simple harmonic motion along the horizontal with a time period 0.5 sec. Coefficient of friction between the block and to surface is 2 5 . Maximum amplitude of such that the object does not slip along the surface is: (g=10 m/s 2 )
  475. Displacement of a particle executing SHM can be expressed as x = a sin ωt + δ At t = 0, it is observed that the particle is at x = a 2 and, moving along positive r-direction. Then δ could be:
  476. Time period of a particle in SHM is 8 sec. At t: 0, it is at mean position. Ratio of the distances travelied by it in the first and second, seconds is:
  477. Two simple harmonic motions are represented by y 1 = 5 sin 2 πt + 3 cos 2 πt y 2 = 5 sin 2 πt + π 4 The ratio of the amplitude of two SHM’s is:
  478. A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is
  479. A simple pendulum performs simple harmonic motion about x – 0with an amplitude a and time period T. The speed of the pendulum at x = a 2 will be :
  480. If amplitude of a particle executing SHM is doubled, which of the following quantities will be doubled? (i) Time period (ii) Maximum velocity (iii) Maximum acceleration (iv) Total energy
  481. The x-t graph of a particle undergoing simple harmonic motion is shown in the Fig. The acceleration of the particle at t = 4 3 sec is :
  482. A particle moves in a straight line. If u is the velocity at a distance x from a fixed point on the line and v 2 = a – bx 2 , where a and b arc constants, then: (i) the motion continues along the positive r-direction only (ii) the motion is simple harmonic (iii) the particle oscillates with a frequency equal to b 2 π (iv) the total energy of the particle is ma
  483. Function x = A sin 2 ωt + B cos 2 ωt + C sin ωt cos ωt represents SHM : (i) for any value of A, B and C (except C = 0) (ii) if A=- B; C = 2B, amplitude = B 2 (iii) if A = B; C = 0 (iv) if A = B; C = 2B, amplitude = B
  484. The period of a simple pendulum of length L at aplace where the acceleration due to gravity is g is T. What is the period of a simple pendulum of the same length at a place where the acceleration due to gravity is 1.02 g?
  485. A simple pendulum performs SHM about x = 0 with an amplitude A and time period T. The speed of the pendulum at x = A/2 will be:
  486. The bob of a simple pendulum executes simple harmonic motion in water with a period r,while the period of oscillation of the bob is t o in air. Neglecting frictional force of water and given that the density of the bob is 4 3 × 1000 kg / m 3 .What relationship between t and t o is true?
  487. Angular amplitude of a simple pendulum is θ 0 . Maximum tension in the string for small θ 0 will be:
  488. Two spherical bob of masses m A and m B are hung vertically from two strings of lengths I A and l B respectively. They are executing SHM with frequencies f A and f B such that f A = 2 f B .Then :
  489. Two pendulums of lengths 1.69 m and 1.44 m start swinging together. After how many vibrations will they again start swinging together?
  490. Two pendulums of time periods 3 s and 7 s respectively start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase:
  491. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s 2 at a distance of 5 m from the mean position. The time period of oscillation is
  492. The time period of a second’s pendulum is 2 sec; the spherical bob which is empty from inside has a mass of 50 g. This is now replaced by another solid bob of same radius but having a different mass of 100 g. The new time period will be:
  493. Mkg weight is suspended from a weightless spring and it has time period T. If now 4 M kg weight is suspended from the same spring, the new time period will be:
  494. A mass of I kg attached to the bottom of a spring has a certain frequency of vibration. The following mass has to be. added to it in order to reduce the frequency by half:
  495. A weightless spring which has a force constant fr oscillates with frequency f when a mass m is suspended from it. The spring is cut into two halves and a mass 2 m is suspended from one of the halves. The frequency of oscillation will now be:
  496. A force of 6.4 N stretches a vertical spring by 0.1 m. The mass that must be suspended from the spring so that it oscillates with a period of ( π / 4 ) sec is:
  497. A light spiral spring supports 200 g weight at its lower end; it oscillates with a period of 1 sec. How much weight (in g) must be removed from the lower end to reduce the period to 0.5 sec?
  498. Masses m and 3m are attached to the two ends of a spring of spring constant k.If the system vibrates freely, the period of oscillation will be:
  499. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes, 5 T 3 ,then the ratio of m M is :
  500. Time period of a spring mass system is T. If this spring is cut into two parts whose lengths are in the ratio 1 : 3 and the some mass is attached to the longer part, the new time period will be :
  501. Two blocks, each of mas s m are connected by a massless spring of spring constant fr and the system is placed on a friction less surface as shown. Time period of system for small oscillations will be:
  502. The system shown in Fig. is initially in equilibrium. When the 500 gm mass is removed, the system oscillates with a time period 4 sec. On removing the 700 gm mass also, time period of oscillation will be:
  503. On a smooth inclined plane, a body of mass m is attached between two massless springs. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body is:
  504. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young’s modulus of the wire arc A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to:
  505. The displacement of an object attached to a spring and executing – simple harmonic motion is given by x = 2 × 10 – 2 cos πt m .The time at which the maximum speed first occurs is:
  506. A spring elongated by length l when a mass ‘M’ is suspended to it. Now a small mass ‘m’ is attached and then released, its time period of oscillation is:
  507. The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be :
  508. Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in Fig.) If the mass M is displaced in the horizontal direction, then the frequency of the system:
  509. A body of mass 100 gm is executing simple harmonic motion. Its displacement x (in cm) at time t (in second) is given by x = 10 cos ( 50 t – π / 3 ) . Then maximum restoring force acting on the body is
  510. A mass of 2 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in Fig. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion so that the mass gets detached from the pan ? (Take g = 10 m/s 2 )
  511. A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass rt is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of min kg is :
  512. A uniform rod of length 2.0 m is suspended through an end and is set oscillating with small amplitude under gravity. The time period of oscillation is approximately
  513. Time period of a simple pendulum of length l is T 1 and time period of a uniform rod of the same length l pivoted about one end and oscillating in a vertical plane is T 2 . Amplitude of oscillations in both the cases is small. Then T 1 T 2 is :
  514. A spring of force constant fr is cut into lengths of ratio 1: 2: 3. They are connected in series and the new force constant is k’. Then they are connected in parallel and force constant is k” . Then k’ : k” is :
  515. A particle is subjected to two mutually perpendicular SHMs such that x = 2 sin ωt and y = 2 sin ωt + π / 4 . . The path of the particle will be:
  516. A particle is subjected to two simple harmonic motions given by, y 1 = 10 sin ωt , y 2 = 5 sin ωt + π . The maximum speed of the particle is:
  517. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45 o , then: (i) the resultant amplitude is ( 1 + 2 ) a (ii) the phase of the resultant motion relative to the first is 90 0 (iii) the energy associated with the resulting motion is 3 + 2 2 times the energy associated with any single motion (iv) the resulting motion is not simple harmonic
  518. When two displacements represented by y 1 = a sin ωt and y 2 = b cos ωt are superimposed the motion is :
  519. A small sphere is suspended by a string of length l . The sphere is making oscillatory motion between extreme positions A and B then
  520. A particle executes SHM with a time period of 4 s. Find the time taken by the particle to go directly from its mean position to half of its amplitude.
  521. A mass m is suspended separately by two different springs of spring constant K 1 and K 2 gives the time period t 1 and t 2 respectively. If the same mass m is connected by both springs as shown in the figure, then time period t is given by the relation
  522. For a particle executing S.H.M. the displacement x is given by x = A cos ωt . Identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x
  523. A hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flows out of the hole at the bottom, the period of oscillation will
  524. A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be
  525. A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 3.92 x 10 – 3 m. What must be the least period of these oscillations, so that the object is not detached from the platform
  526. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
  527. The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the figure. y-projection of the radius vector of rotating particle P is
  528. Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k, the frequency of oscillation of block is
  529. A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g = l0 m / s 2 )
  530. A bead of mass m can slide on a frictionless wire as shown in figure. Because of the given shape of the wire, near P, the bottom point, it can be approximated as parabola. Near P, the potential energy of the bead is given U = c x 2 where c is a constant and x is measured from P. The bead, if displaced slightly from point P will oscillate about P. The period of oscillation is
  531. Two particles of same time period (T) and amplitude undergo SHM along the same line with initial phase of π 6 . lf they start at the same instant and at same point along opposite directions. Find the time after which they will meet again for the first time:
  532. A uniform stick of mass M and length L is pivoted at its centre. Its ends are tied to two springs each of force constant K. In the position shown in figure, the strings are in their natural length. When the string is displaced through a small angle θ and released. The stick:
  533. Two very small balls, each having mass m, are attached to two massless rods of length l. Now these rods are joined to form V like figure having angle 60°. This assembly is now hinged in a vertical plane so that it can rotate without any friction about a horizontal axis (perpendicular to the plane of figure) as shown in the figure. The period of small oscillation of this assembly is
  534. One end of an ideal spring is connected with a smooth block with the other end with rear wall of driving cabin of a truck as shown in figure. Initially, the system is at rest. If truck starts to accelerate with a constant acceleration ‘a’ , then the block (relative to truck ):
  535. A pendulum has period T for small oscillations. An obstacle is placed directly beneath the pivot, son that only the lowest one quarter of the string can follow the pendulum ob when it swings in the left of its resting position as shown in the figure. The pendulum is released from rest at a certain point A. The time taken by its to return to that point is:
  536. Two identical rods each of length l and mass m are welded together at right angles and then suspended from a knife-edge as shown. Angular frequency of small oscillations of the system in its own plane about the point of suspension is
  537. A uniform square plate of side ‘a’ is hinged at one of its corners. It is suspended such that it can rotate about horizontal axis. The time period of small oscillation about its equilibrium position.
  538. A block P of mass m is placed on a frictionless other block Q of same mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure. μ s is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude l. The maximum value of the friction force between P and Q is:
  539. A particle is performing a linear simple harmonic motion. If the acceleration and the corresponding velocity of the particle are a and v respectively, which of the following graphs is correct?
  540. Two particles A and B execute simple harmonic motion according to the equations y 1 = 3 sin ωt and y 2 = 4 sin ( ωt + π 2 ) + 3 sin ωt . The phase difference between them is
  541. A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be
  542. Three simple harmonic motions of equal amplitudes A and equal time periods along the same line combine. The phase of the second motion is 60° ahead of the first and phase of the third motion is 60° ahead of the second. The amplitude of resultant motion is
  543. The resulting amplitude A’ and the phase of the vibrations S = A cos ( ωt ) + A 2 cos ( ωt + π 2 ) + A 4 cos ( ωt + π ) + A 8 cos ( ωt + 3 π 2 ) = A ‘ cos ( ωt + δ ) then A’ and δ are respectively.
  544. Two light strings, each of length l, are fixed at points A and B on a fixed horizontal rod xy. A small bob is tied by both strings and in equilibrium, the strings are making angle 45° with the rod. If the bob is slightly displaced normal to the plane of the strings and released then period of the resulting small oscillation will be:
  545. A simple pendulum of length 1 m is allowed to oscillate with amplitude 2°. It collides elastically with a wall inclined at l° to the vertical. Its time period will be: (use g = π 2 )
  546. A U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now m gm of a liquid of density d is poured into it. The column of liquid in this tube will oscillate with a period T such that
  547. In case of a simple pendulum, time period versus length is depicted by
  548. A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration 3 g m / s 2 . The period of small oscillations of the pendulum about its equilibrium position is ( g = π 2 m / s 2 ):
  549. Two pendulums have time periods T and 5 T 4 . They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation?
  550. A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be
  551. The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is
  552. The period of a simple pendulum, whose bob is a hollow metallic sphere is T. The period is T 1 when the bob is filled with sand, T 2 when it is filled with mercury and T 3 when it is half filled with mercury. Which of the following is true?
  553. In a horizontal spring-mass system, mass m is released after being displaced towards right by some distance at t = 0 on a frictionless surface. The phase angle of the motion in radian when it is first time passing through the equilibrium position is equal to
  554. A system is shown in the figure. The time period for small oscillations of the two blocks will be
  555. ln the figure shown a block of mass m is attached at ends of two springs. The other ends of the spring are fixed. The mass m is released in the vertical plane when the springs are relaxed. The velocity of the block is maximum when
  556. In the figure all springs are identical having spring constant k and mass m each. The block also has mass m. The frequency of oscillation of the block is
  557. A block of mass m is at rest on an another block of same mass as shown in figure. Lower block is attached to the spring then the maximum amplitude of motion so that both the block will remain in contact is
  558. Two identical springs are attached to a small block P. The other ends of the springs are fixed at A and B. When P is in equilibrium the extension of top spring is 20 cm and extension of bottom spring is l0 cm. The period of small vertical oscillations of P about its equilibrium position is (use g = 9.8 m / s 2 )
  559. Figure shows a system consisting of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically down from its equilibrium position and then released, the period of its vertical oscillation is
  560. Four massless springs whose force constants are 2k,2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of oscillation of the system is
  561. One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x 0 from the black. The spring is then compressed by 2 x 0 and released. The time taken to strike the wall is
  562. In figure S 1 and S 2 are identical springs. The oscillation frequency of the mass m is f. If one spring is removed, the frequency will become
  563. A particle is executing simple harmonic motion along a straight line with amplitude 2 cm. Maximum restoring force acting on the particle is 2N. Find the distance of the particle from the equilibrium position when restoring force acting on the particle is 1N.
  564. In simple harmonic motion how many times potential energy is equal to kienetic energy during one complete period?
  565. A particle is moving on x-axis has potential energy U = 2 − 20 x + 5 x 2 Joules along x-axis. The particle is released at x = 3. The maximum value of ‘ x.’ will be: [x is in meters and U is in joules]
  566. What will be the force constant of the spring system shown in the figure?
  567. A block is suspended by a spring from the root of a stationary lift and its time period for small vertical oscillation is 2 sec. If the lift starts moving upward with constant acceleration of g/2, the new time period of oscillation is
  568. When a block A is suspended by a spring and allowed to oscillate vertically, its time period of oscillation is found to be 3 sec. When another block B is suspended by the same spring, its time period of oscillation is found to be 4 sec. Now the two blocks A and B are tied together, suspended by the same spring and allowed to oscillate vertically. Then time period of the composite block is
  569. Two particles start moving simultaneously from the same point and execute simple harmonic motion along the same line. Distances of the particle from their equilibrium position are x 1 = 4 cos 4 πt   cm   and   x 2 = 4 cos 4 πt + π 3 cm . Then maximum separation between the particles is
  570. Kinetic energy of a particle of mass 40 gm executing simple harmonic motion along a straight line is given by K = 16 1 − 0 .25 x 2 joule where x is in cm. Then frequency of oscillation of the particle is
  571. A block is suspended by a spring and the elongation of the spring is found to be 10 cm. The block is now pulled downward by further 5 cm and released. Then frequency of oscillation of the block is (Take g = 10 m/s 2 )
  572. Force on a particle is given by: F = -kx n . For what values of n will the motion be oscillatory?(k is a positive constant)
  573. What will be time period of the displaced body of mass m?
  574. A simple harmonic motion is given by y = 5 ( sin 3 πt + 3 cos 3 πt ) What is the amplitude of motion if y is in cm :
  575. If a body is executing simple harmonic motion, then:
  576. The equation x = Asin² (wt+ π/6) represents
  577. A simple pendulum has a time period T 1 when on the earth’s surface and T 2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T 2 /T 1 is
  578. If the displacement of a particle executing SHM is given by, y = 0.30 sin (220t +0.64) in meters, then the frequency and the maximum velocity of the particle is
  579. On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K, the period of oscillation of the body (assuming the springs as mass less) is
  580. The displacement of a particle executing SHM is given by x = 0.01 sin 100π (t+0.05). The time period is:
  581. The displacement of a particle in SHM is x = 3 sin ⁡ ( 20 πt ) + 4 cos ⁡ ( 20 πt ) cm . Its amplitude of oscillation is
  582. A particle moves according to the equation x = acos ⁡ πt 2 . The distance covered by it in the time interval between t = 0 to t = 3 s is
  583. The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4 cm. Its initial displacement is 2 cm towards positive extreme position. Its equation for displacement is
  584. The period of oscillation of a particle in SHM is 4s and its amplitude of vibration is4 cm. The distance of the particle 1 3 s after passing the mean position is
  585. A body executing SHM has a maximum velocity of 1ms-1 and a maximum acceleration of 4ms -2 . Its amplitude in metres is
  586. For a body in SHM the velocity is given by the relation V= 144 − 16 x 2 ms − 1 . The maximum acceleration is
  587. The maximum velocity of a particle executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is
  588. A small body of mass 10 gram is making harmonic oscillations along a straight line with a time period of π /4 and the maximum displacement is 10cm. The energy of oscillator is
  589. At what displacement is the KE of a particle performing SHM of amplitude 10cm is three times its PE ?
  590. The average kinetic energy of a simple harmonic oscillator is 2 joule and its total energy is 5 joule. Its minimum potential energy is
  591. Two springs of force constants 1000 N/m and 2000 N/m are stretched by same force. The ratio of their respective potential energies is
  592. A spring when loaded has a potential energy ‘E’. Then ‘m’ turns out of ‘n’ turns are removed from the spring. If the same load is suspended, then the energy stored in the spring is
  593. In a spring block system if length of the spring is reduced by 1 %, then time period
  594. A body is executing SHM. If the force acting on the body is 6N when the displacement is 2cm, then the force acting on the body at a displacement of 3 cm is
  595. A freely falling body takes 2 seconds to reach the ground on a planet when it is dropped from a height of 8m. If the period of a simple pendulum is 2 seconds on that planet then its length is
  596. A seconds pendulum is shifted from a place where g = 9.8 m/s 2 to another place where g = 9.78 m/s 2 . To keep period of oscillation constant its length should be
  597. The pendulum of the grandfather’s clock takes 1s to oscillate from one end to another a distance of 10 cm. Considering it a simple pendulum, find its maximum velocity
  598. A particle executes SHM with an amplitude of 0.2 m. Its displacement when its phase is 90° is
  599. The time period of oscillation of a particle that executes SHM is 1.2s. The time starting from mean position at which its velocity will be half of its velocity at mean position is
  600. An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 cm/s and the period is 628 milli seconds. The amplitude of the motion in centimetres is
  601. The acceleration of a particle in SHM is 0.8ms -2 , when its displacement is 0.2 m. The frequency of its oscillation is
  602. The equation for the displacement of a particle executing SHM is x = 5sin (2πt)cm Then the velocity at 3cm from the mean position is(in cm/s)
  603. The ratio of velocities of particle in SHM at displacements A/3 and 2A/3 is
  604. The equation of motion of a particle in SHM is a + 16π 2 x = 0 . Here ‘a’ is linear acceleration of the particle at displacement ‘x’ (a, x are in SI). Its time period is
  605. A body of mass 1/4 kg is in SHM and its displacement is given by the relation x = 0 . 05 Sin 20 t + π 2 m . If t is in seconds, the maximum force acting on the particle is
  606. The amplitude of oscillation of particle in SHM is √3cm. The displacement from mean position at which its potential and kinetic energies are in the ratio 1 : 2 is
  607. In a spring block system if length of the spring is increased by 4 %, then time period
  608. A spring of force constant k is cut into two equal parts, and mass ‘m’ is suspended from parallel combination of the springs. Then frequency of small oscillations is
  609. The length of a pendulum changes from 1 m to 1.21m. The percentage change in its period is
  610. The mass and diameter of a planet are twice those of the earth. The period of oscillation of a pendulum on this planet will be, if it is a seconds pendulum on the earth 2.4 s
  611. Three simple pendulums have lengths of strings 49 cms, 48 cms and 47 cms with diameters of the bobs 2 cms, 4 cm and 6 cm respectively. Then their periods are in the ratio
  612. The amplitude of a damped oscillator becomes 1 27 th of its initial value after 6 minutes. Its amplitude after 2 minutes is
  613. The amplitude of vibration of a particle is given by a M = a 0 aω 2 – bω + c where a 0 , a, b and c are positive. The condition for a single resonant frequency is
  614. Displacement time equation of a particle executing SHM is, x = 10 sin π 3 t + π 6 cm . The distance covered by particle in 3s is
  615. The period of a particle in SHM is 8 s. At t = 0, it is at the mean position. The ratio of the distances travelled by it in the 1st and the 2nd second is
  616. A particle executes SHM with a time period of 16 s. At time t = 2s, the particle crosses the mean position while at t = 4s, its velocity is 4ms –1 . The amplitude of motion in metre is
  617. Two simple harmonic motions are represented by the equations, y 1 = 0 . 1 sin 100 πt + π 3 and y 2 = 0 . 1 cosπt .The phase difference of velocity of particle 1 with respect to the velocity of particle 2 is
  618. A body is executing SHM under the action of a force whose maximum magnitude is 50 N. The magnitude of force acting on the particle at the time when its energy is half kinetic and half potential is
  619. A body of mass ‘m’ is suspended to an ideal spring of force constant ‘k’. The expected change in the position of the body due to an additional force ‘F’ acting vertically downwards is
  620. A block of mass 1 kg is connected with a massless spring of force constant 100 N/m .At t = 0, a constant force F = 10 N is applied on the block. The spring is in its natural length at t = 0. The speed of particle at x = 6 cm from mean position is
  621. A mass M is suspended from a spring of negligible mass . The spring is pulled a little and then released so that the mass executes simple harmonic motion with time period T. If the Mass is increased by ‘m’ then the time period becomes 5 T 4 the ratio m M is
  622. The metallic bob of a simple pendulum has the relative density ρ . The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is
  623. Four simple harmonic vibrations x 1 = 8 sin ⁡ ( ωt ) , x 2 = 6 sin ⁡ ωt + π 2 , x 3 = 4 sin ⁡ ( ωt + π ) and x 4 = 2 sin ⁡ ωt + 3 π 2 are superimposed on each other. The resulting amplitude is…. units.
  624. The displacement of a particle executing S.H.M from its mean position is given by x = 0 .5 sin ⁡ ( 10 πt + ϕ ) cos ⁡ ( 10 πt + ϕ ) . The ratio of the maximum velocity to the maximum acceleration of the body is given by
  625. Four springs of constant as shown are attached to a pair of masses m each as shown. The time period will be 2π times
  626. A horizontal spring block system of mass M executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass m is put on it and the two move together. Find the new amplitude
  627. For a particle executing SHM the displacement x is given by x = Acos ω t. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x –
  628. Two identical springs are fixed at one end and masses M and 4M are suspended at their other ends as shown in figure. They are both stretched down from their mean position and let go simultaneously. If they are in the same phase every 4 seconds, the spring constant k is
  629. Two pendulums of lengths 1.44 and 1 metre length start swinging together. After how many vibrations of the longer pendulum will they again start swinging together ?
  630. The acceleration of a particle, starting from ‘rest, varies with time according to the relation a = − sω 2 sin ⁡ ωt The displacement of this particle at a time f will be
  631. A simple pendulum performs simple harmonic motion about X=0 with an amplitude A and time period T. The speed of the pendulum at X = A /2 will be
  632. A simple harmonic oscillator has an amplitude a and time period T. The time required to travel from x = a to x = a / 2 is
  633. A particle starts s with S.H.M. from the mean position . Its amplitude is A and its time periods T, At one time, its speed is half that of the maximum speed. What is thie distance of the particle from mean position at that instant ?
  634. A block P of mass m is placed on a frictionless horizontal surface. Another block Q of same mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure, µ s is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The maximum value of the friction force between P and Q is
  635. A linear harmonic oscillator of forte constant 2×10 6 N/m and amplitude of 0 . 01 m has a total mechanical energy of 160 J. Its
  636. The angular frequency of a simple pendulum is ω rad/sec. Now if the length is made one fourth of the original length, the angular frequency becomes
  637. A particle is executing simple harmonic motion with an amplitudes a. When its kinetic energy is equal to its potential energy, its distance from the mean position is
  638. A weighted glass tube floats in a liquid with len8lh /of ‘ the tube immersed in the liquid. It is pushed down through a certain distance y and released. The frequency of oscillations will be
  639. A simple pendulum of length / and bob mass m is displaced from its equilibrium position O to a position P so that the height of P above O is h, It is then released. What is the tension in the string when the bob passes through the equilibrium position O ? Neglect friction, v is the velocity of the bob at O.
  640. The length of a simple pendulum is increased by 44%. What is the percentage increase in its period ?
  641. A body performs S.H.M. Its K.E. K varies with time t, as indicated in graph [fig. (5)]
  642. A pendulum has period T for small oscillations. An obstacle is placed directly beneath the pivot, so that only the lowest one quarter of the string can follow the pendulum bob when it swings in the left of its resting position as shown in fig. (4). The pendulum is released from rest at a certain point A. The time taken by it to return to that point is
  643. Two bodies M and N of equal masses arc suspended from two separate mass less springs of spring constants k 1 and k 2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of M to that of N is
  644. A spring has a certain mass suspended from it and its period for vertical oscillations is T 1 The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now T 2 The ratio of T 2 / T 1 is
  645. A spring has a force constant k and a mass m is suspended from it. The spring is cut in half and the same mass is suspended from one of the halves. If the frequency of oscillation in the first case is α , then the frequency in the second case will be
  646. What will be the force constant of the spring system shown in fig. (7)
  647. In arrangement given in fig. (8), if the block of mass m is displaced, the frequency is given by
  648. A body of mass m hangs from three springs, each of spring constant ,k as shown in fig. (11). If the mass is slightly displaced and let go, the system will oscillate with the time period
  649. Two masses ml and m 1 are suspended together by a mass less spring of constant k When the masses are in equilibrium, m, is removed without disturbing the system. Then the angular frequency of osculation of m 2 is
  650. A mass M is attached to a horizontal spring of force constant,k fixed one side to a rigid support as shown in fig. (13). The mass oscillates on a friction less surface with time period T and amplitude A. When the mass is in equilibrium position another mass m is gently placed on it. What will be the new time period of oscillations ?
  651. A trolley of mass m is connected to two identical springs, each of force constant k as shown in fig. (12). The trolley is dissipated from its equilibrium position by a distance x and released. The trolley executes S.H.M. of period T. After some time it comes to rest due to friction, The total energy dissipated as heat is (assumed the damping force to be weak)
  652. A block of mass M is suspended from a light spring of force constant k. Another mass m moving upwards with velocity v hits the mass M and gets embedded in it. What will be the amplitude of oscillation of the combined mass ?
  653. A particle of mass m executes simple harmonic motion with amplitude ‘ a ‘ and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is
  654. A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000 N/m. The spring compresses by
  655. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t o in air. Neglecting frictional force of water and given that the density of the bob is (4 / 3) x 1000 kg / m 3 . What relationship between t and t o is true ?
  656. If a simple harmonic motion is represented by d 2 x dt 2 + αx = 0 its time period is
  657. Two simple harmonic motions are represented by the equations y 1 = 0 ⋅ 1 sin ⁡ 100 πt + π 3 and y 2 = 0 ⋅ 1 cos ⁡ ( 100 πt ) The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is
  658. The displacement of an object attached to a spring and executing simple harmonic motion is given by X = 2 × 10 − 2 cost πt metre. The time at which the maximum speed first occurs is
  659. The function sin 2 ωt represents
  660. The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would
  661. On a smooth inclined plane, a body of mass M is attached b€tween two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body (assuming the springs as mass less) is
  662. Starting from the origin a body oscillates simple harmonically with period of 2 s. After what time will its kinetic energy be 75% of the total energy
  663. The displacement of a particle varies according to the relation x = 4 ( cos ⁡ πt + sin ⁡ πt ) The amplitude of the particle is
  664. A body executes simple harmonic motion. The potential energy (P.8.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which one of the following statements is true ?
  665. A mass m is attached to two springs of same force constant in accordance with the following four configurations. Let T 1 .,, T 2 , T 3 and T 4 be the time periods respectively. In which combination, the time period is largest
  666. Two particles A and B of equal masses are suspended from two mass less springs of spring constant k 1 and k 2 respectively. If the maximum velocities during oscillation, are equal, the ratio amplitudes of A and B is
  667. Infinite springs with force constants k,2 k, 4ft and I k,… respectively are connected in series. The effective force constant of the spring will be
  668. Two springs are connected to a block of mass M placed on a friction less surface as shown in fig. (1). If both the springs have a spring constant k, the frequency of oscillation of the block is
  669. An ideal spring with spring constant.k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unscratched. Then the maximum extension in the spring is
  670. A sphere of radius r is placed on a concave surface of radius of curvature 5 r, which itself is placed on a horizontal table as shown in fig. (3). When the sphere is displaced from the equilibrium position it describes S.H.M. of period
  671. A particle executes S.H.M. of amplitude A. If I 1 , and T 2 , are the times taken by the particle to traverse from 0 to A/2 and from A I 2 to A respectively, then(T 1 / T 2 ) will be equal to
  672. The time period of a particle in simple harmonic motion is 8 second. At t =0, it is at the mean position. The ratio of the distances traveled by it in the first and second second is
  673. When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2 .0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity in it in m/s is [Acceleration due to gravity = 10 m / s 2 )
  674. Two identical blocks A and B, each of mass ‘m resting on smooth floor are connected by a light spring of natural length I and the spring constant k, with spring at its natural length. A third identical block C (mass m) moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is
  675. A Toy gun has a spring of force constant ,k. After changing the spring by compressing it through a distance of & the toy releases a shot of mass m vertically upwards. Then the shot will travel a vertical height is equal to
  676. A system shown in fig. (5) consists of mass less pulley, a spring of force constant k and a block of mass m. If block is just slightly displaced vertically down from its equilibrium position and released. The time period of vertical oscillation is
  677. In fig. (6), the spring has a force constant k The pulley s light and smooth, the spring and string are light. The suspended block has a mass m. If the block is slightly displaced from its equilibrium position and then released, the period of its vertical oscillation is
  678. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to mass less spring of spring constant.k. A mass m hangs freely from the free end of the spring, The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to
  679. A spring of force-constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of
  680. Two blocks A and B each of mass m are connected by a mass less spring of natural length I and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in fig. (7). A third identical block C, also of mass m, moving on the floor with a speed v along the line joining A and 8, and collides with A. Then
  681. For a particle executing S.H.M., the displacement x is given by x= α cos ωt Identify the graph (fig. 10), which represents the variation of potential energy (P.E.) as a function of time , and displacement x
  682. A mass m is attached to four springs of spring constants 2.k,2k,-k and k as shown in fig. (2). The mass s capable of oscillating on friction less horizontal floor. If it is displaced slightly and released then the frequency of resulting S.H.M. would be
  683. A mass m is suspended by means of two coiled springs which have the same length in unstretched condition fig. Their force constants are k 1 and k 2 k 1 and k 2 respectively. When set into vertical vibrations, the period will be
  684. A particle of mass m is attached to three springs A, B and C of equal force constant k [Fig. (8)]. If the particle is pushed towards any one of the springs, then the time period of its oscillation will be
  685. A block of mass m is attached to two unstretched springs of spring constants k, each as shown. The block is displaced towards right through a distance x and is released. The speed of the block as it passes through the mean position will be
  686. One end of a long metallic wire of length L is tied to the ceiling, the other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross- section and Young’s modulus of the wire are A and Y, respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to
  687. In a simple pendulum, the period of oscillation T is related to length of the pendulum l as
  688. Under the action of a force F = – kx 3 , the motion of a particle is (k = a positive constant)
  689. The time period of a simple pendulum inside a stationary lift is 5 s. What will be the time period when the lift moves upward with an acceleration g 4 ?
  690. The ratio of frequencies of two pendulums are 2 : 3, then their lengths are in ratio
  691. The acceleration a of a particle undergoing SHM is shown in the figure. Which of the labelled points corresponds to the particle being at – x max ?
  692. A body oscillates in SHM according to the equation (in SI unit), x = 5 cos ( 2 πt + π 4 ) . Its instantaneous displacement at t = 1s is
  693. Two particles are executing simple harmonic motion. At an instant of time t, their displacements a y 1 = A cos ( ωt ) and y 2 = A sin ( ωt ) Then, the phase difference between y 1 and y 2 is
  694. A particle executing simple harmonic motion along Y-axis has its motion described by the equation y = A sin ( ω t) + B. The amplitude of the simple harmonic motion is
  695. A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos ω t and y = a sin ω t. The trajectory of motion of the particle will be
  696. A particle of mass 2 kg moves in simple harmonic motion and its potential energy U varies with position x as shown in the figure. The period of oscillation of the particle is

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