Question

If an elevator is retarding downwards by $\frac{g}{3}$, then the period of oscillation of a simple pendulum in the elevator is

• A. $2\pi \sqrt{\frac{3l}{4g}}$
• B. $2\pi \sqrt{\frac{3l}{2g}}$
• C. $\pi \sqrt{\frac{3l}{4g}}$
• D. $2\pi \sqrt{\frac{l}{g}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{3l}{4g}}$

When it is retarding downwards by $\frac{g}{3}$, the effective acceleration is
$g^{\prime }=g+\frac{g}{3}=\frac{4g}{3}$

Thus, $T=2\pi \sqrt{\frac{l}{4g/3}}=2\pi \sqrt{\frac{3l}{4g}}$