From the variation of potential energy in the direction of small oscillation of a simple pendulum, find the effective spring constant for the simple pendulum, where m is mass of the bob, l is length of the simple pendulum. 

When a particle is executing SHM, at mean position

$F=-\frac{dU}{dr}=0$. Also, its potential energy is minimum at equilibrium position. So, $\frac{d^{2}U}{d{r}^2}>0$

$\frac{dU}{dr}=-F\Rightarrow \frac{d^{2}U}{d{r}^2}=-\frac{dF}{dr}$

$\text{ At equilibrium, }dF=-{\left|\frac{d^{2}U}{d{r}^2}\right|}_{r=0}dr$

Comparing this standard SHM equation, i.e.

$\text{ At equilibrium, }dF=-{\left|\frac{d^{2}U}{d{r}^2}\right|}_{r=0}dr$

In general equilibrium happens at $r=r_0\left(\text{ say }\right)$

$\therefore k_{eff}={\left|\frac{d^{2}U}{d{r}^2}\right|}_{r=r_0}$

Also, at equilibrium $F=-\frac{dU}{dr}=0$

Now, coming to the situation given, when the bob is given a small horizontal displacement as shown.

$\begin{array}{l}U=mgy\\ y=l(1-\cos \theta )=2l{\sin }^2\frac{\theta }{2}\\ \therefore U=2mgl{\sin }^2\frac{\theta }{2}\end{array}$

$\text{ As }\theta \text{ is small }\sin \frac{\theta }{2}\simeq \frac{\theta }{2}$

$\begin{array}{l}\mathrm{So} U=2mgl\left(\frac{\theta }{2}\cdot \frac{\theta }{2}\right)\text{ or }U=\frac{1}{2}mgl{\theta }^2\\ \theta \simeq \frac{x}{l}\Rightarrow U=\frac{mg}{2l}{x}^2\\ \frac{dU}{dx}=\frac{mg}{2l}\cdot 2x=\frac{mg}{l}\cdot x\\ \Rightarrow \frac{d^{2}U}{d{x}^2}=\frac{mg}{l}\therefore k_{eff}={\left|\frac{d^{2}U}{d{x}^2}\right|}_{x=0}=\frac{mg}{l}\end{array}$