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Question: The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when ...

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

A

U=−KX22U = - \frac{KX^{2}}{2}

B

U=KX2U = KX^{2}

C

U=KU = K

D

U=KXU = KX

Answer

U=−KX22U = - \frac{KX^{2}}{2}

Explanation

Solution

F=−kxF = - kx ⇒ dW=Fdx=−kxdxdW = Fdx = - kxdx

So ∫0WdW=∫0x−kxdx\int_{0}^{W}{dW} = \int_{0}^{x}{- kxdx}⇒ W=U=−12kx2W = U = - \frac{1}{2}kx^{2}