Question

A person measures the time period of a simple pendulum inside a stationary lift and finds it to be $T$. If the lift starts accelerating upwards with an acceleration $g/3$, the time period of the pendulum will be

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

Answer 1

Answer: (C)

Time period of the simple pendulum in the lift
$T=2 \pi \sqrt{\frac{l}{g a}}$
Time period when lift starts accelerating upwards with acceleration $g / 3$
$T=2 \pi \sqrt{\frac{l}{g+\frac{g}{3}}} $
$T=2 \pi \times \sqrt{3} \sqrt{\frac{l}{4 g}} $
or $T=2 p \sqrt{\frac{l}{g}} \times \frac{\sqrt{3}}{2}$
or $T=\frac{\sqrt{3}}{2} T$