Question

Two simple pendulums whose lengths are $100\, cm$ and $121\, cm$ 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)

10

Answer 2

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, $l_{1}=100\, cm,\, l_{2}=121\, cm$
$\therefore \frac{T_{2}}{T_{1}}=\sqrt{\frac{121}{100}}=\frac{11}{10}$.
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
$n \times 11=(n+1) \times 10$
$\Rightarrow n=10$