The potential energy of a particle (U_x) exec | vedclass

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Question

(A) The potential energy $(U)$ of a particle executing $S.H.M.$ with displacement $y$ from the mean position is given by the formula:$U = \frac{1}{2}

Vedclass · Accepted Answer

$U_x = \frac{1}{2} k (x - a)^2$

Vedclass · Answer

(A) The potential energy $(U)$ of a particle executing $S.H.M.$ with displacement $y$ from the mean position is given by the formula:$U = \frac{1}{2} m \omega^2 y^2 = \frac{1}{2} k y^2$where $k = m \omega^2$ is the force constant.If the displacement $y$ is expressed as $(x - a)$, where $x$ is the position and $a$ is the mean position, the potential energy becomes:$U_x = \frac{1}{2} k (x - a)^2$Thus, option $A$ represents the potential energy of a particle in $S.H.M.$ relative to a shifted equilibrium position $a$.

The potential energy of a particle $(U_x)$ executing $S.H.M.$ is given by

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