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.

Answer 1

Let \(T_1\) and \(T_2\) be the time period of the two pendulums
\(T_1=2\pi \sqrt{\dfrac{100}{g}}\) and \(T_2=2\pi\sqrt{\dfrac{121}{g}}\)
\((T_1<T_2\) because \(l_1<l_2)\)
\(\Rightarrow \dfrac{T_2}{T_1}=\dfrac{\sqrt{121}}{\sqrt{100}}\)
\(\therefore \dfrac{T_2}{T_1}=\dfrac{11}{10} \)
\(\Rightarrow 10T_2=11T_1\)
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:
\((n+1)T_1=nT_2\)
\((n+1)\times 2\pi\sqrt{\dfrac{100}{g}}=n\times 2\pi\sqrt{\dfrac{121}{g}}\)
\(11n=10n+10\)
\(\therefore n=10\)