Explanation

(b) [image:14_image59.emf]
When the bob is displaced to position $P$, through a small angle $\theta$ from the vertical, the various forces acting on the bob at $P$ are
(i)the weight $mg$ of the bob acting vertically downwards
(ii) the tension $T$ in the string acting along $PS$
Resolving $mg$ into two rectangular components, we get
(a)$mg\cos\theta$ acts along $PA$, opposite to tensions, we get
(b)$mg\sin\theta$ acts along $PB$, tangent to the arc $OP$ and directed towards $O$.
If the string neither slackens nor breaks but remains taut, then
$T = mg\cos\theta$
The force $mg\sin\theta$ tends to bring the bob back to its mean position $O$.
$\therefore$ Restoring force acting on the bob is
$F = - mg\sin\theta$