$\text{As shown in the figures, a uniform rod}OO\prime \text{of length}\mathrm{ \ell }\text{ is hinged at the point}\mathrm{ O }\text{and held in place vertically between}$

$\text{two walls using two massless springs of same spring constant. The springs are connected at the midpoint and}$

$\text{at the top-end}(O\prime )\text{of the rod, as shown in Fig.1 and the rod is made to oscillate by a small angular displacement.}$

$\text{The frequency of oscillation of the rod is}{f}_1.\text{On the other hand, if both the springs are connected at the midpoint}$

$\text{of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is}{}_{}$

$f_2.\text{Ignoring gravity and assuming motion only in the plane of the diagram, the value of }\frac{{\mathrm{ f}}_1}{f_2}\text{is:}$

The correct answer is:
$\sqrt{\frac{5}{2}}$

Step-by-step Explanation:

Let the uniform rod $O{O}^{\prime }$ have mass $m$ and length $\ell$. The rod is hinged at the bottom point $O$.
The moment of inertia of the uniform rod about the hinge $O$ at its end is:
$I=\frac{1}{3}m{\ell }^2$

Let both springs have the same spring constant $k$. We consider a small angular displacement $\theta$ of the rod from its vertical equilibrium position. Since gravity is ignored, the only restoring forces acting on the rod are due to the springs.

Case 1: Springs connected at the midpoint and top-end (Fig. 1)
In the first configuration (shown in Fig. 1):
- One spring is connected at the midpoint of the rod (at a distance $\frac{\ell }{2}$ from the hinge $O$). For a small angle $\theta$, its deformation is $x=\frac{\ell }{2}\theta$. The restoring torque it exerts about point $O$ is:
${\tau }_1=-k\left(\frac{\ell }{2}\theta \right)\left(\frac{\ell }{2}\right)=-\frac{1}{4}k{\ell }^2\theta$
- The second spring is connected at the top-end $O^{\prime }$ (at a distance $\ell$ from the hinge $O$). Its deformation is $x=\ell \theta$. The restoring torque it exerts about point $O$ is:
${\tau }_2=-k\left(\ell \theta \right)\left(\ell \right)=-k{\ell }^2\theta$

The total restoring torque is the sum of these two torques:
${\tau }_{\text{total, 1}}={\tau }_1+{\tau }_2=-(\frac{1}{4}k{\ell }^2+k{\ell }^2)\theta =-\frac{5}{4}k{\ell }^2\theta$

The equation of motion is $I\alpha ={\tau }_{\text{total, 1}}$, which gives:
$I\frac{d^2\theta }{d{t}^2}+\frac{5}{4}k{\ell }^2\theta =0$

Therefore, the frequency of oscillation $f_1$ is:
$f_1=\frac{1}{2\pi }\sqrt{\frac{\frac{5}{4}k{\ell }^2}{I}}$

Case 2: Both springs connected at the midpoint (Fig. 2)
In the second configuration (shown in Fig. 2):
- Both springs are connected at the midpoint of the rod (at a distance $\frac{\ell }{2}$ from the hinge $O$).
- When the rod rotates by $\theta$, the displacement of the midpoint is $x=\frac{\ell }{2}\theta$, and both springs act together to resist this motion.
The total restoring torque about point $O$ is:
${\tau }_{\text{total, 2}}=-2·k\left(\frac{\ell }{2}\theta \right)\left(\frac{\ell }{2}\right)=-\frac{2}{4}k{\ell }^2\theta =-\frac{1}{2}k{\ell }^2\theta$

The equation of motion in this case is:
$I\frac{d^2\theta }{d{t}^2}+\frac{1}{2}k{\ell }^2\theta =0$

Therefore, the frequency of oscillation $f_2$ is:
$f_2=\frac{1}{2\pi }\sqrt{\frac{\frac{1}{2}k{\ell }^2}{I}}$

Calculating the Ratio:
To find $\frac{f_1}{f_2}$, we divide the expressions for $f_1$ and $f_2$:
$\frac{f_1}{f_2}=\frac{\frac{1}{2\pi }\sqrt{\frac{\frac{5}{4}k{\ell }^2}{I}}}{\frac{1}{2\pi }\sqrt{\frac{\frac{1}{2}k{\ell }^2}{I}}}=\sqrt{\frac{\frac{5}{4}}{\frac{1}{2}}}=\sqrt{\frac{5}{2}}$