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MHT CET · Physics · Oscillations

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

  1. A 1.5 s
  2. B 2 s
  3. C \(\sqrt{3} \mathrm{~s}\)
  4. D 3 s
Verified Solution

Answer & Solution

Correct Answer

(A) 1.5 s

Step-by-step Solution

Detailed explanation

\( T_1 = 2\pi\sqrt{\frac{L}{g}} = \sqrt{3} \) \( g' = g + \frac{g}{3} = \frac{4g}{3} \)