Physics - SHM Question with Solution | TestHub

Let$T=2\mathrm{\pi }\sqrt{\frac{l}{g}}$

$T=2\mathrm{\pi }\sqrt{\frac{0.1}{10}}=\frac{\mathrm{\pi }}{5} \text{s}$

Time taken by the pendulum in going from$\mathrm{A}$to$\mathrm{B}=\frac{T}{4}$

Time taken by the pendulum in going from$\mathrm{B}$to$\mathrm{C}=\frac{T}{12}$

$\therefore$The time period of the pendulum$=2\left(\frac{T}{4}+\frac{T}{12}\right)=\frac{2T}{3}=\frac{2}{3}.\frac{\mathrm{\pi }}{5}=\frac{2\mathrm{\pi }}{15} \mathrm{s}$