Oscillations — Physics STD 11 Science — Question

Two forces acting on the bob are:

1. Weight mg of the bob acting vertically downward.
2. Tension T in string along PS, resolving mg into two components.

1. $\text{mg}\cos\theta$ opposite to tension T
2. $\text{mg}\sin\theta$ directed towards O.

Tension in the string, T $=\text{mg}\cos\theta$

$\therefore$ Restoring force acting on bob is $\text{F}=-\text{mg}\sin\theta-\text{ve}$ sign shows force is directed towards mean position: If $\theta$ is small, then

$\sin\theta=\theta\frac{(\text{arc OP})}{\text{l}}=\frac{\text{x}}{\text{l}}$

$\text{F = -mg}\theta=-\text{mg}\frac{\text{x}}{\text{l}}$