Question

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest, its period is 'T'. With what acceleration 'a' should the lift be accelerated upward in order to reduce the period to $T/2$?

• A. $2g$
• B. $3g$
• C. $4g$
• D. $g$

Hint:

Upward acceleration increases effective gravity, decreasing the time period.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Accelerating a lift changes the effective gravity \( g_{\text{eff}} \) acting on the pendulum.

Step 2: Key Formula or Approach:
Time period \( T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} \).
For upward acceleration \( a \), \( g_{\text{eff}} = g + a \).

Step 3: Detailed Explanation:
Initially, at rest: \( T = 2\pi \sqrt{\frac{l}{g}} \) \dots (i)
When moving up with acceleration \( a \): \( T' = \frac{T}{2} = 2\pi \sqrt{\frac{l}{g+a}} \) \dots (ii)
Dividing (i) by (ii):
\[ \frac{T}{T/2} = \sqrt{\frac{g+a}{g}} \]
\[ 2 = \sqrt{\frac{g+a}{g}} \]
Squaring both sides:
\[ 4 = \frac{g+a}{g} \implies 4g = g + a \]
\[ a = 3g \]

Step 4: Final Answer:
The acceleration should be \( 3 \text{ g} \).