Question

A man measures the period of a simple pendulum inside a stationary lift and finds it to be T second. if the lift accelerates upward with an acceleration g/4, then the period of the pendulum will be

• A. $T$
• B. $\frac{T}{4}$
• C. $\frac{2T}{\sqrt{5}}$
• D. $2T\sqrt{5}$

Answer 1

Answer: (C)

$\frac{2T}{\sqrt{5}}$

The time period of pendulum in a stationary lift

$T=2\pi \sqrt{\frac{l}{g}}$

When the lift moves up with an acceleration $A$, the time period will be:

$T^{\prime }=2\pi \sqrt{\frac{1}{(g+a)}}$

So, the ratio of time period of the pendulum will be:

$\Rightarrow \frac{T^{\prime }}{T}=\sqrt{\frac{g}{(g+a)}}$

$=\sqrt{\frac{g}{(g+\frac{g}{4})}}=\sqrt{\frac{4}{5}}$

$\Rightarrow T^{\prime }=\frac{2T}{\sqrt{5}}$