Question

Two pendulums have time periods $T$ and $5T/4$. They start SHM at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase

• A. $5$
• B. $4$
• C. $11$
• D. $9$

Answer 1

The correct option is A $5$

When the pendulum $A$ with time period $\frac{5T}{4}$ completes phase of $2\pi$ in $\frac{5T}{4}$ , the other pendulum $B$ leads by phase $\frac{\pi }{2}$.
So, after four oscillations of pendulum $A$ i.e., after $5T$, pendulum $B$ leads by $2\pi$ i.e., they come in phase.
i.e., after 5 oscillations of small pendulum (time taken $=5T$), they will again come in the same phase.