Question
Show that for small oscillations the motion of a simple pendulum is simple harmonic. Derive an expression for its time period. Does it depend on the mass of the bob?

Answer

Expression for time period:

Let $m =$ Mass of the bob $l =$ Length of the simple pendulum $OP = x$ When the bob is displaced to point $P$ through small angle $\theta.$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, $T =\text{mg}\cos\theta$
The force $mg \sin$ tends to bring back the bob to its mean position $O.$
$\therefore$ 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
$\sin\theta=\theta\frac{(\text{arc OP})}{\text{l}}=\frac{\text{x}}{\text{l}}$
$\text{F = -mg}\theta=-\text{mg}\frac{\text{x}}{\text{l}}$
$\text{F}\propto$ displacement $(x)$ and $F$ is directed towards mean position $O$.
In $\text{S.H.M}$., Restoring force $\text{F}=-\text{kx} \ ...(\text{ii})$
Comparing $(i)$ and $(ii)\text{k}=\frac{\text{mg}}{\text{l}}$
Inertia factor $=$ Mass of bob $= m\text{T}=2\pi\sqrt{\frac{\text{Inertia factor}}{\text{Spring factor}}}$
$=2\pi\sqrt{\frac{\text{m}}{\frac{\text{mg}}{\text{l}}}}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
No. $T$ does not depend on the mass of the bob.

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