Q. A man measures the period of a simple pendulum inside a stationary lift ad finds it to be T second. If the lift accelerates upwards with an acceleration g/4, then the period of pendulum will be
A) $2T\sqrt{5}$
B) T
C) $\frac{2T}{\sqrt{5}}$
D) $\frac{T}{4}$
Explanation
(c) $T = 2\pi\sqrt{\frac{l}{g}}$. When lift is accelerated upwards with acceleration$a( = g/4)$. Then effective acceleration due to gravity inside the lift
$g_{1} = g + a = g + \frac{g}{4} = \frac{5g}{4}$
$\therefore\ \ T_{1} = 2\pi\sqrt{\frac{l}{5g/4}} = 2\pi\sqrt{\frac{l}{g}} \times \frac{2}{\sqrt{5}} = \frac{2T}{\sqrt{5}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.