Question

Time period of a simple pendulum measured inside a stationary lift is T. If the lift starts moving upwards with an acceleration of $\frac{g}{3}$, what will be its time period:

• A. $\frac{T}{3}$
• B. $3T$
• C. $\frac{\sqrt{3}}{2T}$
• D. $\sqrt{\frac{3}{2}}T$

Answer 1

The correct option is C

$\frac{\sqrt{3}}{2T}$

Time period of simple pendulum
$T=2\pi \sqrt{\frac{l}{g_{eff}}}$
$T=2\pi \sqrt{\frac{l}{g}}$ When lift stationary
When lift starts moving upwards with an acceleration $\frac{g}{3}$
$T=2\pi \sqrt{\frac{l}{g+a}}a=\frac{g}{3}{T}^{\prime }=2\pi \sqrt{\frac{l}{g+\frac{g}{3}}}=2\pi \sqrt{\frac{3l}{4g}}{T}^{\prime }=\frac{\sqrt{3}}{2}T$