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

Step 1: Understanding the Concept:
The time period of a simple pendulum depends on the effective acceleration due to gravity in its frame of reference.
When the frame of reference (the lift) accelerates, a pseudo force acts on the pendulum bob, changing the effective gravity.
Step 2: Key Formula or Approach:
The formula for the time period of a simple pendulum is \( T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} \).
Step 3: Detailed Explanation:
When the lift is stationary, the effective gravity is simply \( g \).
The initial time period is given as \( T_1 = 2\pi\sqrt{\frac{L}{g}} = \sqrt{3}\text{ s} \).
When the lift moves upwards with an acceleration \( a = g/3 \), a downward pseudo force acts on the mass.
Therefore, the effective acceleration due to gravity increases to \( g_{\text{eff}} = g + a \).
Substitute the value of \( a \): \( g_{\text{eff}} = g + \frac{g}{3} = \frac{4g}{3} \).
The new time period \( T_2 \) is:
\[ T_2 = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\frac{L}{4g/3}} = 2\pi\sqrt{\frac{3L}{4g}} \] We can factor out the constant terms to relate it to \( T_1 \):
\[ T_2 = \sqrt{\frac{3}{4}} \times \left(2\pi\sqrt{\frac{L}{g}}\right) \] Substituting \( 2\pi\sqrt{\frac{L}{g}} = \sqrt{3} \):
\[ T_2 = \frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2} = 1.5\text{ s} \] Step 4: Final Answer:
The time period of the pendulum when the lift moves upwards is \( 1.5\text{ s} \).