Question

Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerates with \(\frac{g}{6}\) vertically upwards then the time period will be (Where g = acceleration due to gravity)

• A. \(\sqrt{\frac{6}{5}}T\)
• B. \(\sqrt{\frac{5}{6}}T\)
• C. \(\sqrt{\frac{6}{7}}T\)
• D. \(\sqrt{\frac{7}{6}}T\)

Answer 1

The Correct Option is C

To solve the problem of finding the time period of a simple pendulum in a lift accelerating upwards, we must first understand the formula for the time period of a simple pendulum.

The time period \( T \) of a simple pendulum in a stationary system is given by:

T = 2\pi \sqrt{\frac{L}{g}}

where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

When the lift accelerates upwards, the effective acceleration becomes \( g' = g + a \), where \( a \) is the acceleration of the lift. Here, the lift is accelerating upwards with \( a = \frac{g}{6} \).

Thus, the effective acceleration \( g' = g + \frac{g}{6} = \frac{6g}{6} + \frac{g}{6} = \frac{7g}{6} \).

The new time period \( T' \) is calculated using the modified acceleration:

T' = 2\pi \sqrt{\frac{L}{g'}}

Substitute \( g' = \frac{7g}{6} \) into the equation:

T' = 2\pi \sqrt{\frac{L}{\frac{7g}{6}}} = 2\pi \sqrt{\frac{6L}{7g}} = \sqrt{\frac{6}{7}} \times 2\pi \sqrt{\frac{L}{g}} = \sqrt{\frac{6}{7}} \times T

Thus, the new time period of the pendulum when the lift accelerates upwards with \( \frac{g}{6} \) is:

\(\sqrt{\frac{6}{7}}T\)

This matches the correct answer from the provided options, \(\sqrt{\frac{6}{7}}T\).