A pendulum is oscillating at a frequency of $8 \mathrm{~Hz}$. Suddenly the string of the pendulum is clamped at its midpoint then the new frequency of oscillations is

• A

  $16 \mathrm{~Hz}$
• B

  $13.8 \mathrm{~Hz}$
• Correct answer
  C

  $11.28 \mathrm{~Hz}$
• D

  $5.7 \mathrm{~Hz}$

Solution

$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}}$ $\begin{aligned} & \mathrm{f} \propto \frac{1}{\sqrt{\ell}} \\ & \frac{\mathrm{f}_2}{\mathrm{f}_1}=\sqrt{\frac{\ell_1}{\ell_2}}=\sqrt{\frac{\ell \times 2}{\ell}} \\ & \mathrm{f}_2=\sqrt{2} \mathrm{f}_1=\sqrt{2} \times 8 \\ & =11.28 \mathrm{~Hz}\end{aligned}$

Hence, the correct answer is option: 'C'.