The time period of a simple pendulum inside a stationary lift is 3 second. When the lift moves upwards with an acceleration g /3 , the time period will be ( g = acceleration due to gravity)

Question

The time period of a simple pendulum inside a stationary lift is 3 ​ second. When the lift moves upwards with an acceleration g /3 , the time period will be ( g = acceleration due to gravity)

TestHub · Accepted Answer

The time period of a simple pendulum in a stationary lift is given by:

$T_{1} = 2 π \frac{L}{g} = 3$

When the lift moves upwards with an acceleration $a = \frac{g}{3}$ , the effective acceleration due to gravity becomes:

$g_{e ff} = g + a = g + \frac{g}{3} = \frac{4 g}{3}$

The new time period $T_{2}$ is:

$T_{2} = 2 π \frac{L}{g _{e ff}} = 2 π \frac{L}{\frac{4 g}{3}} = \frac{3}{4} \times 2 π \frac{L}{g}$

Substituting the value of $T_{1}$ :

$T_{2} = \frac{3}{2} \times 3 = \frac{3}{2} = 1.5 s$

Answer: 1.5 s

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion