Question

Two simple pendulums of length $1m$ and $1.21m$ are at their mean position with their velocities in the same direction at some instant. After how many oscillations of bigger pendulum they will again be in same phase?

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

The correct option is B $10$

If $T_1$ be the time period of pendulum 1 $(l=1.21m)$ and $T_2$ that of the second one $(l=1m)$

We can say $\frac{T_1}{T_2}=\sqrt{\left(\frac{L_1}{L_2}\right)or\frac{T_1}{T_2}}=1.1=\frac{11}{10}$

i.e. $10T_1=11T_2$

That means $10$ oscillations of $T_1$ is completed in the same time in which $11$ oscillations of $T_2$ is completed. Hence after that time they again be in the same phase, that is, swinging together.