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Q.

An object of mass 0.2 kg executes simple harmonic along X-axis with frequency of 25πHz. 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

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a

0.05

b

0.06

c

0.01

d

None of the these

answer is B.

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Detailed Solution

E = 12mω2A2 ⇒ E = 12m(2πf)2A2⇒A = 12πf2EmPutting values we obtain,A = 12π(25π)2×(0.5+0.4)0.2 ⇒ A = 0.06 m

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