Question

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

• A. 11
• B. 10
• C. 21
• D. 20

Answer 1

Answer: (B)

The time period of a simple penduium is
$T=2\pi \sqrt{\frac{l}{g}}$
where, $I$ is length.
$\therefore \frac{T_1}{T_2}=\sqrt{\frac{l_1}{l_2}}$
Given, $l_1=100cm,l_2=121cm$
$\therefore \frac{T_2}{T_1}=\sqrt{\frac{121}{100}}=\frac{11}{10}$.
In a given time shorter penduium will make one oscillation more than the longer.
Let after $n$ oscillations, the pendulums are in same phase, then
$n\times 11=(n+1)\times 10$
$\Rightarrow n=10$