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Question

An object of mass 0.2kg executes simple harmonic oscillations along the x-axis with a frequency of (25/π)Hz. At the position x=0.04m, the object the kinetic energy of 0.5 J and potential energy of 0.4J. The amplitude of oscillations is _________m.

A
0.06 m
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B
6 m
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C
600 m
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D
60 m
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Solution

The correct option is A 0.06 m
Total Energy = Kinetic energy + Potential Energy
T.E=K.E+P.E
T.E=0.5 J+0.4 J
T.E=0.9 J
T.E=12mω2a2[ω=2πf(ffrequencyandaamplitude)]
0.9=12×0.2×(2π×25π)2×a2
a=  0.9[12×0.2×(2π×25π2)]
a=0.06 m

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