Question

Two simple pendulums of length $5m$ and $20m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed $n$ oscillations. The value of $n$ is then:-

• A. $5$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

It is given that,

Length of first pendulum, $l_1=5m$

Length of second pendulum,$l_2=20m$

Time period is given by:

$T_1=2\pi \sqrt{\frac{l_1}{g}}$

$T_1=2\pi \sqrt{\frac{5}{10}}=2\pi \sqrt{\frac{1}{2}}.......\left(1\right)$

$T_2=2\pi \sqrt{\frac{20}{10}}=2\pi \sqrt{2.}......\left(2\right)$

Solving equation (1) and (2)

$T_1=\sqrt{2}\pi$

$T_2=2\sqrt{2}\pi$

$so,$

$T_2=2T_1$

The value of n is 2