Frequency Ratio of Oscillations for Hinged Rod with Springs

To find the ratio of the frequencies of small angular oscillations $\frac{f_1}{f_2}$, we analyze the restoring torques acting on the rod for both configurations.

Let:

• $m$ be the mass of the uniform rod $OO'$,
• $l$ be the length of the rod,
• $k$ be the spring constant of each spring,
• $\theta$ be a small angular displacement of the rod from its vertical equilibrium position.

The moment of inertia of the uniform rod about the hinge $O$ at its end is: $I = \frac{1}{3} m l^2$

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Case 1: Figure 1

When the rod is rotated clockwise by a small angle $\theta$:

1. Spring at midpoint ($y = \frac{l}{2}$):

   • The linear displacement of the midpoint is $x_1 = \frac{l}{2} \theta$.
   • The spring connected to the left wall is stretched, exerting a leftward restoring force $F_1 = k \left(\frac{l}{2} \theta\right)$.
   • The restoring torque about $O$ is: $\tau_{1a} = F_1 \cdot \frac{l}{2} = k \left(\frac{l}{2} \theta\right) \left(\frac{l}{2}\right) = \frac{1}{4} k l^2 \theta$

2. Spring at top-end $O'$ ($y = l$):

   • The linear displacement of the top-end is $x_2 = l \theta$.
   • The spring connected to the right wall is compressed, exerting a leftward restoring force $F_2 = k (l \theta)$.
   • The restoring torque about $O$ is: $\tau_{1b} = F_2 \cdot l = k (l \theta) \cdot l = k l^2 \theta$

The total restoring torque for Fig. 1 is: $\tau_1 = \tau_{1a} + \tau_{1b} = \frac{1}{4} k l^2 \theta + k l^2 \theta = \frac{5}{4} k l^2 \theta$

Using the rotational equation of motion $I \alpha_1 = -\tau_1$: $\omega_1^2 = \frac{\tau_1 / \theta}{I} = \frac{\frac{5}{4} k l^2}{\frac{1}{3} m l^2} = \frac{15 k}{4 m}$

Hence, the frequency of oscillation is: $f_1 = \frac{1}{2\pi} \sqrt{\frac{15 k}{4 m}}$

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Case 2: Figure 2

In this configuration, both springs are attached at the midpoint ($y = \frac{l}{2}$):

1. Upon displacement $\theta$, the midpoint moves by $x = \frac{l}{2} \theta$.
2. The left spring is stretched by $x$, and the right spring is compressed by $x$. Both springs apply restoring forces in the same direction (towards the left).
3. The restoring force from each spring is $F = k \left(\frac{l}{2} \theta\right)$.

The total restoring torque about $O$ for Fig. 2 is: $\tau_2 = 2 \times \left( k \left(\frac{l}{2} \theta\right) \cdot \frac{l}{2} \right) = \frac{1}{2} k l^2 \theta$

Using the equation of motion $I \alpha_2 = -\tau_2$: $\omega_2^2 = \frac{\tau_2 / \theta}{I} = \frac{\frac{1}{2} k l^2}{\frac{1}{3} m l^2} = \frac{3 k}{2 m}$

Hence, the frequency of oscillation is: $f_2 = \frac{1}{2\pi} \sqrt{\frac{3 k}{2 m}}$

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Ratio of Frequencies $\frac{f_1}{f_2}$

$\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2} = \sqrt{\frac{15 / 4}{3 / 2}} = \sqrt{\frac{15}{4} \times \frac{2}{3}} = \sqrt{\frac{5}{2}}$

Thus, the correct option is C.