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Q.

Kinetic energy of a particle of mass 40 gm executing simple harmonic motion along a straight line is given by K=161−0.25x2  joule where x is in cm. Then frequency of oscillation of the particle is

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a

5002πHz

b

500πHz

c

250πHz

d

1000πHz

answer is A.

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Detailed Solution

Kinetic energy K=12mω2A21−x2A2Comparing this equation with the given equation we get, A2=10.25⇒A=2 cmand 12mω2A2=16⇒12×401000.ω2.21002=16⇒ω2=16×50×2500 ω=10002 rad/s∴ f=ω2π=5002π Hz
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