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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 mTotal Energy = Kinetic energy + Potential EnergyT.E=K.E+P.E⟹T.E=0.5 J+0.4 J⟹T.E=0.9 JT.E=12mω2a2[ω=2πf(f→frequencyanda→amplitude)]⟹0.9=12×0.2×(2π×25π)2×a2a=  ⎷0.9[12×0.2×(2π×25π2)]⟹a=0.06 m

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