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

Answer 1

Answer: (B)

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

Case I: If lift is a stationary
Case II: If list is accelerated rewarded

$T-mg=mg$
$T=m(g+\frac{g}{3})$
$T_{2}m+\frac{g}{3}$
$T=2\pi \sqrt{\frac{l}{g}}$
$T^1=2\pi \sqrt{\frac{l}{4\frac{g}{3}}}$
$\therefore T^1=\frac{\sqrt{3}}{2}T$