$\mathrm{r}=l \sin \theta$

Forces acting on the bob are.

$(i)$ $\mathrm{T}=$ tension in the string

$(ii)$ $\mathrm{Mg}=$ weight of the bob

Resolving $T$ along the vertical and horizontal

direction, we get

${\mathrm{T} \cos \theta=\mathrm{Mg}}$         $...(1)$

${\mathrm{T} \sin \theta=\mathrm{Mr} \omega^{2}=\mathrm{M}(l \sin \theta) \omega^{2}}$   

or         ${\mathrm{T}=\mathrm{M} l \omega^{2}}$   $...(2)$

Dividing eq. $(2)$ by eq. $(1)$, we get

$\frac{1}{\cos \theta}=\frac{l \omega^{2}}{g}$

$\mathbf{o r}$          $\omega^{2}=\frac{g}{l \cos \theta}$

$\text { time period } \quad \mathrm{t}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{l \cos \theta}{\mathrm{g}}}$