Question

Two simple pendulums of lengths 1 m and 25 m, respectively, are both given small displacement in the same direction at the same instant. If they are in phase after the shorter pendulum has completed n oscillations, then n is equal to

• A. $\frac{4}{3}$
• B. $\frac{5}{4}$
• C. $\frac{7}{5}$
• D. $\frac{8}{7}$

Answer 1

The correct option is B $\frac{5}{4}$

$T=2\pi \sqrt{\frac{L}{g}}$
$\therefore T\alpha \sqrt{L}$ , as time period decreases when the length of pendulum decreases, the time period of shorter pendulum $\left(T_s\right)$ is smaller then that of longer pendulum $\left(T_l\right)$. That means shorter pendulum performs more oscillations in a given time.
It is given that after n oscillations of shorter pendulum, both are again in phase. So, by this time longer pendulum must have made (n-1) oscillations.
$\therefore n{T}_S=(n-1)T_l$
$\Rightarrow n\sqrt{1}=(n-1)\sqrt{25}$
$n=5n-5$
$\Rightarrow 4n=5$
$\Rightarrow n=\frac{5}{4}$
So, the two pendulums shall be in the same phase for the first time when the shorter pendulum has completed $\frac{5}{4}$ oscillation.