Question

Two pendulums have time period $T$ and $5T/4$. They start $\text{SHM}$ at the same time from the mean position. What will be the phase difference between them at the moment, when bigger pendulum completes one oscillation?

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${90}^{\circ }$

Bigger pendulum( greater $\left(l\right)$) will have greater time period.
Time period of smaller pendulum $T_1=T$
Time period of bigger pendulum $T_2=\frac{5T}{4}$

They both start at the same time from mean position,
after time $t,$ phase difference between them is:
$\Delta \varphi ={\omega }_{1}t-{\omega }_{2}t=(\frac{2\pi }{T}-\frac{2\pi }{5T/4})\times t$
when bigger pendulum completes oscillation, $t=\frac{5T}{4}$
$\Rightarrow \Delta \varphi =\frac{2\pi }{5T}\times \frac{5T}{4}$
$\therefore \Delta \varphi =\frac{\pi }{2}$
Hence, option $\left(b\right)$ is correct