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Standard XI

Physics

COE in SHM

A particle is...

Question

A particle is executing linear simple harmonic motion about the origin x = 0. Which of the graphs shown in the figure represents the variation of the potential energy function U(x) versus displacement x?

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Solution

The correct option is D

Here, $F = - k x \Rightarrow \frac{- d U}{d x} = - k x \Rightarrow \int_{0}^{U} d U = \int_{0}^{x} k x d x ∴ U = \frac{1}{2} k x^{2}$

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A particle is executing linear simple harmonic motion about the origin x = 0. Which of the graphs shown in the figure represents the variation of the potential energy function U(x) versus displacement x?

The correct option is D Here, F=−kx⇒−dUdx=−kx⇒∫U0dU=∫x0kxdx∴U=12kx2

The correct option is D

Here, $F = - k x \Rightarrow \frac{- d U}{d x} = - k x \Rightarrow \int_{0}^{U} d U = \int_{0}^{x} k x d x ∴ U = \frac{1}{2} k x^{2}$