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Question

# A particle executes simple harmonic motion of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?

A
0.81A
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B
0.71A
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C
0.41A
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D
0.91A
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Solution

## The correct option is B 0.71ADistance form position can be represented as x=Asinwt....(i)By Differentiating wrt t, we get V=Awcoswt....(ii)we know that, KE=12mv2 PE=12KX2 where k=m2(w2)Then, when KE=PE12mv2=12KX2⇒V2=(WX)2...(iii)From (i), (ii) and (iii)(AWcoswt2)=(WAsinwt)2⇒(coswt)2=(sinwt)2⇒tan2wt=1So, 1×1=A/sinwt/=A1√2=0.71A

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