Show that the motion of simple pendulum is simple harmonic and hence derive an equation for its time period. What is seconds pendulum?

Consider a simple pendulum of length “l”, consists a bob of mass “m” is suspended from rigid support “s”. 

The pendulum pulled through a side with small angular displacement "$\mathrm{\theta }$" and released then it begins to oscillation about the mean position “O”. 

At any instant of time, the pendulum at point A, the weight “mg” resolved into two components. "mgcos$\mathrm{\theta }$" balanced by the tension in a string. mgsin$\mathrm{\theta }$ pulled the pendulum to its means position, 

Hence "mgsin$\mathrm{\theta }$" is called restoring force.
$\therefore \mathrm{F}=-\mathrm{mgsin}\mathrm{\theta }$   [-ve sign indicate that opposite direction of applied force ]
$\text{ If " }\mathrm{\theta }"\text{ is very small, then }\sin \mathrm{\theta }\approx \mathrm{\theta }$
$\therefore \mathrm{F}=-\mathrm{mg\theta }$
From Newton’s Second law F = ma
$\therefore \mathrm{ma}=-\mathrm{mg\theta }\left(\text{ or }\right)\mathrm{a}=-\mathrm{g\theta }$

$\text{ From fig. angle }\mathrm{\theta }=\frac{\mathrm{arc}\left(\mathrm{x}\right)}{\mathrm{radius}\left(\mathrm{l}\right)}\left(\text{ or }\right)\mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{l}}$

$\therefore \mathrm{a}=-\mathrm{g}\frac{\mathrm{x}}{\mathrm{l}}\text{ (or) }\mathrm{a}=-\left(\frac{\mathrm{g}}{\mathrm{l}}\mathrm{x}\right) ...\left(1\right)$

$\text{ Here }\mathrm{g}/\mathrm{l}\text{ is constant then }\mathrm{a\alpha }-\mathrm{x}\text{ it gives the }$simple harmonic motion

From (1) & (2)

$+{\mathrm{\omega }}^2\mathrm{x}=+(\mathrm{g}/\mathrm{l})\mathrm{x}\therefore \mathrm{\omega }=\sqrt{\mathrm{g}/1}$

The time period (T) of a pendulum,

$\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{\sqrt{\mathrm{g}/\mathrm{l}}}\therefore \mathrm{T}=2\mathrm{\pi }\sqrt{1/\mathrm{g}}$

Seconds pendulum: A simple pendulum whose time period of oscillation is two seconds is called a seconds pendulum.