The P.E. of a particle executing SHM at a dis | vedclass

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Question

(A) The potential energy $(PE)$ of a particle executing simple harmonic motion $(SHM)$ is defined as the work done against the restoring force to disp

Vedclass · Accepted Answer

$\frac{1}{2}m{\omega ^2}{x^2}$

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(A) The potential energy $(PE)$ of a particle executing simple harmonic motion $(SHM)$ is defined as the work done against the restoring force to displace the particle from its equilibrium position.The restoring force is given by $F = -kx$,where $k$ is the force constant.The potential energy is calculated by integrating the work done: $PE = \int_{0}^{x} F dx = \int_{0}^{x} kx dx = \frac{1}{2}kx^2$.We know that the angular frequency $\omega$ is related to the force constant $k$ and mass $m$ by the relation $\omega^2 = \frac{k}{m}$,which implies $k = m\omega^2$.Substituting this value of $k$ into the expression for potential energy,we get:$PE = \frac{1}{2}m\omega^2x^2$.

The $P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is

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