Question

Two simple pendulums of length $1m$ and $16m$ 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)

$T_1=2\pi \sqrt{\frac{1}{g}},T_2=2\pi \sqrt{\frac{16}{g}}=4T_1$
at any time $t$, phase of pendulums are:
${\varphi }_1={\omega }_{1}t=\frac{2\pi }{T_1}t,{\varphi }_2={\omega }_{2}t=\frac{2\pi }{T_2}t$
First pendulum is faster. Both will be in same phase again when faster pendulum has completed one oscillation more than slower pendulum
${\varphi }_1-{\varphi }_2=2\pi$
$\Rightarrow \frac{2\pi }{T_1}t-\frac{2\pi }{4T_1}t=2\pi$
$\Rightarrow t=\frac{4T_1}{3}$
No. of oscillations completed by shorter pendulum in time $t$ :
$n=\frac{t}{T_1}=\frac{4}{3}$