Let us consider a body of mass ‘m’ is connected at one end
of a thread of length ‘l’ which passes through a rigid support at point O.
Here,
The point O on the rigid support is called point of
suspension and c.g. of the bob is called point of oscillation.
The distance between point of suspension and oscillation is
called effective length.
At extreme position, let, mg is the wt. of the body and T be
the tension produced in the string. The component mgcos$\theta $ is balanced by
the tension (T).
The component mgsin$\theta $ provides restoring force to
move the bob towards mean position.
∴ f = -mgsin$\theta $ , -ve sign is for restoring force.
or, ma = -mg sin$\theta $
or, a = -g sin$\theta $
for small angle, sin$\theta $$ \approx $$\theta $
a = -g$\theta $ --------------i)
By trigonometry,
$\theta $ = $\frac{{{\rm{arc\: length}}}}{{{\rm{radius}}}}$
= $\frac{{{\rm{AB}}}}{{{\rm{OA}}}}$
$\theta $ = $\frac{{\rm{y}}}{{\rm{l}}}$
----------------ii)
Putting the value of $\theta {\rm{\: }}$in eqn ii) we get,
a = -g $\frac{{\rm{y}}}{{\rm{l}}}$
a = -($\frac{{\rm{g}}}{{\rm{l}}}$)y
---------------iii)
Here, acceleration is directly proportional to displacement
and they are opposite to each other.
Hence, motion of a simple pendulum is SHM
Expression for time period
If a body is in SHM then its acceleration is,
a = -${\omega ^2}$y -----------------iv)
Comparing eqn iii) and iv)
${\omega ^2}$ = $\frac{{\rm{g}}}{{\rm{l}}}$
or, $\omega $ = $\sqrt {\frac{{\rm{g}}}{{\rm{l}}}} $
or, $\frac{{2{\rm{\pi }}}}{{\rm{T}}}$ = $\sqrt {\frac{{\rm{g}}}{{\rm{l}}}}
$ ($\omega $ = $\frac{{2{\rm{\pi }}}}{{\rm{T}}}$ )
or, T = 2${\rm{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} $