ExamBro
ExamBro
JEE Advanced · Physics · 15. Oscillations

A uniform rod of length \(L\) and mass \(M\) is pivoted at the centre. Its two ends are attached to two springs of equal spring constants \(k\). The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle \(\theta\) in one direction and released. The frequency of oscillation is

  1. A \(\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}\)
  2. B \(\frac{1}{2 \pi} \sqrt{\frac{k}{M}}\)
  3. C \(\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}\)
  4. D \(\frac{1}{2 \pi} \sqrt{\frac{24 k}{M}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}\)

Step-by-step Solution

Detailed explanation


\(\text {Restoring torque }=-(2 k x) \cdot \frac{L}{2} \)
\(\alpha=-\frac{k L(L / 2 \theta)}{I}=-\left[\frac{k L^2 / 2}{M L^2 / 12}\right] \cdot \theta\)
\(=-\left(\frac{6 k}{M}\right) \theta \)
\(\therefore f =\frac{1}{2 \pi} \sqrt{\left|\frac{\alpha}{\theta}\right|}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}\)
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app