Questions
A particle moves with a simple harmonic motion in a straight line. In the first second starting from rest it travels a distance and in the next second it travels a distance in the same direction. The amplitude of the motion is
detailed solution
Correct option is C
Let the acceleration be f; f=−ω2xTherefore, distance of the particle from the centre at any time t is given by x=rcos(ωt) where r is the amplitude when t=1s,x=r−a∴ (r−a)=rcosωcosω=r−ar ........(i) When t=2s,x=r−a−b, therefore r−a−b=cos2ω∴ r−a−b=r2cos2ω−1 ………..(ii)Substituting the value of cosω from Eq. (i) in Eq. (ii), we get r−a−b=r2(r−a)2r2−1=2(r−a)2r−r∴ r(3a−b)=2a2⇒r=2a23a−bSimilar Questions
A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is . What must be the least period of these oscillations, so that the object is not detached from the platform
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