The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when (K is a positive constant)

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U=−KX22

U=KX2

U=K

U=KX

answer is A.

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Detailed Solution

F=− kx ⇒dW=Fdx=− kxdxSo ∫ 0 WdW=∫ 0 x−kx dx ⇒W=U=−12kx2

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