Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $AB$ of length $l$ and mass $m.$ The rod is pivoted at its center 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

• A

  $\frac{1}{2\pi }\sqrt{\frac{3k}{m}}$
• B

  $\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
• Correct answer
  C

  $\frac{1}{2\pi }\sqrt{\frac{6\mathrm{k}}{m}}$
• D

  $\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$

Solution

Torque on the rod about $O,$



$\tau =kx.\frac{l}{2}\times 2$

$\mathrm{I }\alpha =k.\frac{l}{2}.\theta .\frac{l}{2}\times 2$

$\mathrm{I }\alpha =k\frac{l^2}{2}.\theta$

$\therefore \mathrm{ }{\omega }^2=\frac{k\frac{l^2}{2}}{M\frac{l^2}{12}}=\frac{6k}{M}$

$f=\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$

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