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(B) Given: Displacement $= x$,Potential Energy $(P.E.) = E$,Restoring Force $= F$. For a particle in $S.H.M.$,the restoring force is given by $F = -kx

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

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(B) Given: Displacement $= x$,Potential Energy $(P.E.) = E$,Restoring Force $= F$. For a particle in $S.H.M.$,the restoring force is given by $F = -kx$,where $k$ is the force constant. The potential energy is given by $E = \frac{1}{2}kx^2$. From the force equation,we have $k = -\frac{F}{x}$. Substituting this value of $k$ into the potential energy equation: $E = \frac{1}{2} \left(-\frac{F}{x}\right) x^2$ $E = -\frac{1}{2} Fx$ Multiplying by $2$: $2E = -Fx$ Rearranging the terms: $2E + Fx = 0$ Dividing by $F$: $\frac{2E}{F} + x = 0$.

For 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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