Question

A pendulum is oscillating at a frequency of \( 8 \, \text{Hz} \). Suddenly the string of the pendulum is clamped at its midpoint. Then the new frequency of oscillations is

• A. 16 Hz
• B. 13.8 Hz
• C. 11.28 Hz
• D. 5.7 Hz

Hint:

Clamping the pendulum at its midpoint effectively halves the length, increasing frequency by a factor of \( \sqrt{2} \).

Answer 1

The Correct Option is C

The frequency \( f \) of a simple pendulum is given by: \[ f = \frac{1}{2\pi} \sqrt{\frac{g}{l}} \] When the pendulum is clamped at its midpoint, the new length becomes \( l' = \frac{l}{2} \). The new frequency is: \[ \begin{align} f' = \frac{1}{2\pi} \sqrt{\frac{g}{l/2}} = \frac{1}{2\pi} \sqrt{\frac{2g}{l}} = \sqrt{2} \cdot f \] Given \( f = 8 \, \text{Hz} \), so: \[ \begin{align} f' = \sqrt{2} \cdot 8 \approx 1.414 \cdot 8 = 11.31 \approx 11.28 \, \text{Hz} \]