Simple Harmonic Motion 4 Question 8

8. 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 (2009)

(a) $\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}$

(b) $\frac{1}{2 \pi} \sqrt{\frac{k}{M}}$

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

(d) $\frac{1}{2 \pi} \sqrt{\frac{24 k}{M}}$

Show Answer

Answer:

Correct Answer: 8. (c)

Solution:

  1. $x=\frac{L}{2} \theta$,

Restoring torque $=-(2 k x) \cdot \frac{L}{2}$

$$ \begin{aligned} \alpha & =-\frac{k L(L / 2 \theta)}{I}=-\frac{k L^{2} / 2}{M L^{2} / 12} \cdot \theta=-\frac{6 k}{M} \theta \\ \therefore \quad f & =\frac{1}{2 \pi} \sqrt{\left|\frac{\alpha}{\theta}\right|}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}} \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane