Question

A uniform rod of length l and mass M is provided at the centre. Its two ends are attached to two springs of equal spring constant k. The springs are fixed to rigid support as shown in figure and the 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

• A. $\frac{1}{2\pi }\sqrt{\frac{2k}{6M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{k}{M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{24k}{M}}$

Answer 1

The correct option is C $\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$

If the rod is rotated through an angle $\theta$, extension in one spring = compression in the other spring,

$i.e.,x=l\theta /2$

Therefore, force acting on each of the ends of the rod,

$F=kx=k\left(l\theta /2\right)$

Restoring torque on the rod,

$T=-Fl=-k\left(\theta /2\right)l=-k{l}^2\left(\theta /2\right)$

As $T=I\alpha =\left(M{l}^2/12\right)\alpha$

or $\alpha =-\frac{6k}{M}\theta ,\alpha \propto theta$.

Thus, the motion of the rod is simple harmonic with

${\omega }^2=\frac{6k}{M}or\omega =\sqrt{\frac{6k}{M}}orv=\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$