For a particle performing S.H.M., when displacement is 'x', the potential energy and restoring force acting…
For a particle performing S.H.M., when displacement is 'x', the potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between $\mathrm{x}, \mathrm{E}$ and $\mathrm{F}$ is
$\frac{E}{F}+x=0$
$\frac{2 E}{F}+x=0$
$\frac{E}{F}-x=0$
$\frac{2 E}{F}-x=0$
Solution
Displacement $=x, \quad$ P.E. $=E, \quad$ Force $=F$
$F=-k x$
$E=\frac{1}{2} m \omega^{2} x^{2}=\frac{1}{2} k x^{2}$
$2 E=(-) \frac{F}{x} \times x^{2}=(-) F x$
$\frac{2 E}{F}+x=0$