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 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:
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For a simple pendulum,
\[
f\propto \frac{1}{\sqrt{l}}.
\]
If the length becomes half, the frequency becomes \(\sqrt{2}\) times.
Updated On: Jun 24, 2026
- \(16\,\text{Hz}\)
- \(13.8\,\text{Hz}\)
- \(11.28\,\text{Hz}\)
- \(5.7\,\text{Hz}\)
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The Correct Option is C
Solution and Explanation
Step 1: Recall the relation between frequency and length.
For a simple pendulum, \[ f=\frac{1}{2\pi}\sqrt{\frac{g}{l}} \] Thus, \[ f\propto \frac{1}{\sqrt{l}} \]
Step 2: Find the new effective length.
Initially, let the length of the pendulum be \[ l \] When the string is clamped at its midpoint, the effective length becomes \[ \frac{l}{2} \]
Step 3: Find the new frequency.
Using \[ f\propto \frac{1}{\sqrt{l}}, \] we get \[ \frac{f'}{f}=\sqrt{\frac{l}{l/2}} \] \[ \frac{f'}{f}=\sqrt{2} \] Therefore, \[ f'=f\sqrt{2} \] Given, \[ f=8\,\text{Hz} \] Hence, \[ f'=8\sqrt{2} \] \[ f'\approx 8\times 1.414 \] \[ f'\approx 11.31\,\text{Hz} \] Approximating according to the options, \[ f'\approx 11.28\,\text{Hz} \]
Step 4: Final conclusion.
Hence, the new frequency is \[ \boxed{11.28\,\text{Hz}} \]
For a simple pendulum, \[ f=\frac{1}{2\pi}\sqrt{\frac{g}{l}} \] Thus, \[ f\propto \frac{1}{\sqrt{l}} \]
Step 2: Find the new effective length.
Initially, let the length of the pendulum be \[ l \] When the string is clamped at its midpoint, the effective length becomes \[ \frac{l}{2} \]
Step 3: Find the new frequency.
Using \[ f\propto \frac{1}{\sqrt{l}}, \] we get \[ \frac{f'}{f}=\sqrt{\frac{l}{l/2}} \] \[ \frac{f'}{f}=\sqrt{2} \] Therefore, \[ f'=f\sqrt{2} \] Given, \[ f=8\,\text{Hz} \] Hence, \[ f'=8\sqrt{2} \] \[ f'\approx 8\times 1.414 \] \[ f'\approx 11.31\,\text{Hz} \] Approximating according to the options, \[ f'\approx 11.28\,\text{Hz} \]
Step 4: Final conclusion.
Hence, the new frequency is \[ \boxed{11.28\,\text{Hz}} \]
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