Question

Two simple pendulums of length $1\, m$ and $16\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed $n$ oscillations. The value of $n$ is

• A. 1/3
• B. 2/3
• C. 1
• D. 4/3

Answer 1

Answer: (D)

$T_{1}=2 \pi \sqrt{\frac{1}{g}}, T_{2}=2 \pi \sqrt{\frac{16}{g}}=4 T_{1}$
at any time $t$, phase of pendulums are:
$\phi_{1}=\omega_{1} t=\frac{2 \pi}{T_{1}} t, \phi_{2}=\omega_{2} t=\frac{2 \pi}{T_{2}} t$
First pendulum is faster. Both will be in same phase again when faster pendulum has completed one oscillation more than slower pendulum
$ \phi_{1}-\phi_{2}=2 \pi $
$\Rightarrow \frac{2 \pi}{T_{1}} t-\frac{2 \pi}{4 T_{1}} t=2 \pi $
$\Rightarrow t=\frac{4 T_{1}}{3}$
No. of oscillations completed by shorter pendulum in time $t$ :
$n=\frac{t}{T_{1}}=\frac{4}{3}$