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Q.

Maximum kinetic energy of a particle of mass 1 kg in SHM is 8 J. Time period of SHM is 4 s. Maximum potential energy during the motion is 10 J. Then,

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a

amplitude of oscillations is approximately 2.53 m

b

minimum kinetic energy of the particle is 2 J

c

maximum acceleration of the particle is approximately 6.3 m/s2

d

minimum potential energy of the particle is 2 J

answer is A, B, C.

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Detailed Solution

Maximum kinetic energy = energy of oscillation in SHM

8=12kA2

  kA2=16 …… (i) 

Further,  2πmk=4

  1k=4π2 or k=π24 … (ii)

From Eq. (i) and (ii), we get

k=24 N/m

And A=2.53 m

Maximum acceleration of the particle will be

amax=ω2A=kmA=251(2.53) =6.3 m/s2

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