Question

Two pendulums of length $1.21mand1.0m$ start vibrating. At some instant, the two are in the mean position in same phase. After how many vibrations of the longer pendulum, the two will be in phase ?

• A. $10$
• B. $11$
• C. $20$
• D. $21$

Answer 1

The correct option is C $20$

Since time period depend on the square toot of length.

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{L_1}{L_2}}$

it implies $T_1=1.1T_2$

$\Rightarrow 10T_1=11T_2$

It implies that $10$ oscillations of first pendulum and $10$ of the second they will be in phase.

Hence, the answer is $20.$