Question

A simple pendulum of length $l$ having a bob of mass $m$ is suspended in a car that is travelling with a constant speed $v$ around a circular path of radius $R$. If the pendulum undergoes small oscillations about its equilibrium position, the frequency of its oscillation will be

• A. $\frac{1}{2\pi }\sqrt{\frac{g}{l}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{{(g^2+\frac{v^4}{R^2})}^{\frac{1}{2}}}{l}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{g}{R}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{v^2}{Rl}}$

Answer 1

The correct option is B $\frac{1}{2\pi }\sqrt{\frac{{(g^2+\frac{v^4}{R^2})}^{\frac{1}{2}}}{l}}$

Time period of simple pendulum in non-inertial frame is $T=2\pi \sqrt{\frac{l}{g_{eff}}}$
Due to circular motion, centripetal acceleration on the bob is $=\frac{v^2}{R}$ towards the centre.
Hence,
$g_{eff}=\sqrt{g^2+{\left(\frac{v^2}{R}\right)}^2}$

Time period is
$T=2\pi \sqrt{\frac{l}{{(g^2+\frac{v^4}{R^2})}^{\frac{1}{2}}}}$
Frequency of oscillation
$\Rightarrow f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{{(g^2+\frac{v^4}{R^2})}^{\frac{1}{2}}}{l}}$