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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. The amplitude of oscillation in meter is equal to

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a

0.05

b

0.06

c

0.01

d

None of these

answer is B.

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Detailed Solution

E=12mω2A2⇒E=12m(2πf)2A2⇒A=12πf2Em Putting E=K+U we obtain, A=12π25π2×(0.5+0.4)0.2⇒A=0.06m

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