Question

A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium positions, its time period of oscillation is:

• A. $T=2\pi \sqrt{\frac{L}{g}}$
• B. $T=2\pi \sqrt{\frac{L}{\sqrt{g^2}+\frac{v^4}{R^2}}}$
• C. $T=2\pi \sqrt{\frac{L}{\sqrt{g^2}+\frac{v^2}{R}}}$
• D. $T=2\pi \sqrt{\frac{L}{\sqrt{g^2}-\frac{v^4}{R^2}}}$

Answer 1

Answer: (B)

$T=2\pi \sqrt{\frac{L}{\sqrt{g^2}+\frac{v^4}{R^2}}}$

The period of the pendulum, $T=2\pi \sqrt{\frac{L}{a}}$ where $a=$ $\text{resultant acceleration}=\sqrt{g^2+{\left(\frac{V^2}{R}\right)}^2}[\frac{V^2}{R}=\text{centripital acceleratiop},g=\text{acceleration due to gravity}]$.

$\therefore T=2\pi \sqrt{\frac{L}{g^2+\frac{V^4}{R^2}}}$

Option B is the correct answer.