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)

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

Answer 2

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