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Question

An object of mass 0.2kg executes simple harmonic motion along the X-axis with a frequency of 25πHz. At the position, x=0.04m, the object has a kinetic energy of 0.5J and potential energy of 0.4J, the amplitude of oscillation in the meter is equal to :

A
0.05
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B
0.06
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C
0.01
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D
None of these
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Solution

The correct option is B 0.06
E=12mω2A2

E=12m(2πf)2A2

A=12πf2Em

Putting E=K+U we obtain,

A=12π(25π)2×(0.5+0.4)0.2

A=0.06 m

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