Question

A rigid rod of mass $m$, with a ball of mass $M$ attached to the free end is restrained to oscillate in a vertical plane as shown in the figure. Find the natural frequency of oscillation.

• A. $f=\frac{1}{2\pi }\sqrt{\frac{3k}{27M+7m}}$
• B. $f=\frac{1}{\pi }\sqrt{\frac{3k}{27m+7m}}$
• C. $f=\frac{1}{2\pi }\sqrt{\frac{k}{27M+7m}}$
• D. $f=\frac{1}{\pi }\sqrt{\frac{k}{27M+7m}}$

Answer 1

The correct option is A $f=\frac{1}{2\pi }\sqrt{\frac{3k}{27M+7m}}$

At equilibrium position, deformation of the spring be $x_0$.


Using principle of moments about point ${}^{\prime }{O}^{\prime }$ we get ,
$k{x}_0\frac{L}{4}=Mg\left(\frac{3L}{4}\right)+mg\frac{L}{4}.....\left(1\right)$
When the rod is rotated clockwise through an angle $\theta$ from equilibrium position, the spring will be compressed further by $x$.
Then, restoring torque about point ${}^{\prime }{O}^{\prime }$
$\tau =-\left[k\right(x+x_0)\frac{L}{4}\cos \theta -Mg\frac{3L}{4}\cos \theta -mg\frac{L}{4}\cos \theta ]$
Using $\left(1\right)$ in the above equation, we get
$\tau =-\left[kx\left(\frac{L}{4}\right)\right]\cos \theta$
For small values of $\theta$, $\cos \theta \approx 1$
$\tau =-\frac{kL}{4}x$
But we know that, $\tau =I\alpha$ and $x=\frac{L}{4}\theta$
$\Rightarrow I\alpha =-\frac{k{L}^2}{16}\theta ........\left(2\right)$
Moment of inertia about point ${}^{\prime }{O}^{\prime }$ is given by
$I=I_{MO}+I_{mO}$
$I=M{\left(\frac{3}{4}L\right)}^2+\frac{m{L}^2}{12}+m{\left(\frac{L}{4}\right)}^2=\frac{9}{16}M{L}^2+\frac{7}{48}m{L}^2$
Substituting this in $\left(2\right)$ we get,
$\alpha =-\frac{3k}{27M+7m}\theta$
Comparing this with equation of Angular SHM $\alpha =-{\omega }^2\theta$ we get,
$\omega =\sqrt{\frac{3k}{27M+7m}}$
Since, $f=\frac{\omega }{2\pi }$
$f=\frac{1}{2\pi }\sqrt{\frac{3k}{27M+7m}}$
Thus, option (a) is the correct answer.

Alternate solution :

Let the suppose the spring is stretched by $x$ units, causing the system to make an angle $\theta$ about point ${}^{\prime }{O}^{\prime }$.
For small values of $\theta$ we can say that, $x=\frac{L}{4}\theta$.
Restoring force in the spring $F=-kx=-k\frac{L}{4}\theta$
Restoring torque about point ${}^{\prime }{O}^{\prime }$ is given by
$\tau =F\times \frac{L}{4}=-k\frac{L^2}{16}\theta$
But we know that,
$\tau =I\alpha$
Moment of inertia about point ${}^{\prime }{O}^{\prime }$ is given by
$I=I_{MO}+I_{mO}$
$I=M{\left(\frac{3}{4}L\right)}^2+\frac{m{L}^2}{12}+m{\left(\frac{L}{4}\right)}^2=\frac{9}{16}M{L}^2+\frac{7}{48}m{L}^2$
Using this we get,
$\alpha =-\frac{3k}{27M+7m}\theta$
Comparing this with $\alpha =-{\omega }^2\theta$ and using $f=\frac{\omega }{2\pi }$ we get,
$f=\frac{1}{2\pi }\sqrt{\frac{3k}{27M+7m}}$
Thus, option (a) is the correct answer.