Question

The time period of a simple pendulum inside a stationary lift is $\sqrt{3}\text{s}$. When the lift moves downward with an acceleration $\frac{g}{4}$, find the time period of simple pendulum.

• A. $1\text{s}$
• B. $2\text{s}$
• C. $3\text{s}$
• D. $6\text{s}$

Answer 1

The correct option is B $2\text{s}$

Time period of pendulum when lift is stationary,
$T=2\pi \sqrt{\frac{l}{g}}=\sqrt{3}\text{s}$
For the lift accelerating downwards,
$a=\frac{g}{4}$,
Effective value of acceleration due to gravity is,
$\Rightarrow g^{\prime }=g-a=g-\frac{g}{4}=\frac{3g}{4}$

$\Rightarrow$Time period of pendulum in accelerating lift will be:
$T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}$
$T^{\prime }=2\pi \sqrt{\frac{4l}{3g}}=\sqrt{\frac{4}{3}}\times 2\pi \sqrt{\frac{l}{g}}$
$T^{\prime }=\sqrt{\frac{4}{3}}\times \sqrt{3}=2\text{s}$