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: Formula
Time period $T = 2\pi\sqrt{\frac{l}{g_{eff}}}$.

Step 2: Analysis
- At rest, $g_{eff} = g$.
- Moving up with acceleration $a$, $g_{eff} = g + a$.
Since $T \propto \frac{1}{\sqrt{g_{eff}}}$, we have $\frac{T_1}{T_2} = \sqrt{\frac{g+a}{g}}$.

Step 3: Calculation
$\frac{T}{T/2} = \sqrt{\frac{g+a}{g}} \implies 2 = \sqrt{\frac{g+a}{g}}$.
Squaring both sides: $4 = \frac{g+a}{g} \implies 4g = g + a \implies a = 3g$.

Step 4: Conclusion
Hence, the lift must accelerate upward at $3g$. Final Answer: (B)