Question

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

• A. \(1.5\text{ s}\)
• B. \(2\text{ s}\)
• C. \(\sqrt{3}\text{ s}\)
• D. \(3\text{ s}\)

Hint:

If lift moves upward → effective gravity increases → time period decreases.

Answer 1

The Correct Option is A

Concept:
Time period of a simple pendulum: \[ T = 2\pi \sqrt{\frac{l}{g}} \] When lift accelerates upward, effective gravity becomes: \[ g_{\text{eff}} = g + a \] Step 1: Write initial time period. \[ T_1 = 2\pi \sqrt{\frac{l}{g}} = \sqrt{3} \]
Step 2: Find new effective gravity. \[ a = \frac{g}{3} \Rightarrow g_{\text{eff}} = g + \frac{g}{3} = \frac{4g}{3} \]
Step 3: Write new time period. \[ T_2 = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{\frac{4g}{3}}} \] \[ T_2 = 2\pi \sqrt{\frac{3l}{4g}} = \sqrt{\frac{3}{4}} \cdot 2\pi \sqrt{\frac{l}{g}} \] \[ T_2 = \frac{\sqrt{3}}{2} \times T_1 \]
Step 4: Substitute value of \(T_1\). \[ T_2 = \frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2} = 1.5 \text{ s} \]
Step 5: Conclusion. \[ T_2 = 1.5 \text{ s} \]