Question

Two simple pendulums whose lengths are $100\text{cm}$ and $121\text{cm}$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again.

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

Answer 1

The correct option is B $10$

Let $T_1$ and $T_2$ be the time period of the two pendulums
$T_1=2\pi \sqrt{\frac{100}{g}}$ and $T_2=2\pi \sqrt{\frac{121}{g}}$
$(T_1<T_2$ because $l_1<l_2)$
$\Rightarrow \frac{T_2}{T_1}=\frac{\sqrt{121}}{\sqrt{100}}$
$\therefore \frac{T_2}{T_1}=\frac{11}{10}$
$\Rightarrow 10T_2=11T_1$
Hence smaller pendulum will complete one oscillation more than longer pendulum.

Let longer pendulum complete $n$ oscillations and shorter length pendulum complete $(n+1)$ oscillations

For oscillating in same phase:
$(n+1)T_1=n{T}_2$
$(n+1)\times 2\pi \sqrt{\frac{100}{g}}=n\times 2\pi \sqrt{\frac{121}{g}}$
$11n=10n+10$
$\therefore n=10$