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Question

(C) For a particle performing Simple Harmonic Motion ($S$.$H$.$M$.),the potential energy $E$ at displacement $x$ is given by $E = \frac{1}{2} k x^2$,w

Vedclass · Accepted Answer

$\frac{2 E}{F}+x=0$

Vedclass · Answer

(C) For a particle performing Simple Harmonic Motion ($S$.$H$.$M$.),the potential energy $E$ at displacement $x$ is given by $E = \frac{1}{2} k x^2$,where $k$ is the force constant.The restoring force $F$ acting on the particle at displacement $x$ is given by $F = -k x$.From the expression for potential energy,we have $2 E = k x^2$.Dividing this equation by the expression for force $F = -k x$,we get:$\frac{2 E}{F} = \frac{k x^2}{-k x}$Simplifying the expression,we get:$\frac{2 E}{F} = -x$Rearranging the terms,we obtain:$\frac{2 E}{F} + x = 0$Thus,the correct relation is $\frac{2 E}{F} + x = 0$.

$A$ particle performing $S$.$H$.$M$. when displacement is '$x$',the potential energy and restoring force acting on it are denoted by '$E$' and '$F$' respectively. The relation between $x, E$ and $F$ is

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