Question

As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint 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 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:

• A. \(2\)
• B. \(\sqrt{2}\)
• C. \(\sqrt{\frac{5}{2}}\)
• D. \(\sqrt{\frac{2}{5}}\)

Hint:

For oscillations of a rod with spring torques, calculate the restoring torque by considering spring displacements, then use \(\omega = \sqrt{\frac{\text{restoring torque constant}}{\text{moment of inertia}}}\) to find frequencies.

Answer 1

The Correct Option is C

Step 1: Setup and parameters
The rod \(OO'\) has length \(l\) and is hinged at \(O\). Two springs of the same spring constant \(k\) are attached either at the midpoint and top-end (Fig. 1) or both at the midpoint (Fig. 2).
Step 2: Moment of inertia
The rod rotates about \(O\), so its moment of inertia is \[ I = \frac{1}{3} m l^2 \]
Step 3: Torque by springs in Fig. 1
- One spring is attached at midpoint \(l/2\), the other at top-end \(l\). - If the rod rotates by a small angle \(\theta\), the displacement at midpoint is \(\frac{l}{2} \theta\) and at top-end is \(l \theta\). - Force by each spring: \(F = k \times \text{displacement}\). So the restoring torque is \[ \tau_1 = -k \left(\frac{l}{2}\right)^2 \theta - k (l)^2 \theta = -k \theta \left(\frac{l^2}{4} + l^2\right) = -k \theta \frac{5l^2}{4} \]
Step 4: Torque by springs in Fig. 2
- Both springs attached at midpoint \(l/2\), each stretched by \(\frac{l}{2}\theta\). - Total restoring torque: \[ \tau_2 = -2 \times k \left(\frac{l}{2}\right)^2 \theta = -2 k \theta \frac{l^2}{4} = -\frac{k l^2}{2} \theta \] Step 5: Angular frequency
Using \(\tau = I \alpha = I \frac{d^2 \theta}{dt^2}\), and \(\tau = -K_{\text{eff}} \theta\), the angular frequency for small oscillations is \[ \omega = \sqrt{\frac{K_{\text{eff}}}{I}} \] For Fig. 1: \[ \omega_1 = \sqrt{\frac{\frac{5}{4} k l^2}{\frac{1}{3} m l^2}} = \sqrt{\frac{5}{4} k \times \frac{3}{m}} = \sqrt{\frac{15 k}{4 m}} \] For Fig. 2: \[ \omega_2 = \sqrt{\frac{\frac{k l^2}{2}}{\frac{1}{3} m l^2}} = \sqrt{\frac{k}{2} \times \frac{3}{m}} = \sqrt{\frac{3 k}{2 m}} \] Step 6: Ratio of frequencies
Since \(f = \frac{\omega}{2\pi}\), frequency ratio equals angular frequency ratio: \[ \frac{f_1}{f_2} = \frac{\omega_1}{\omega_2} = \sqrt{\frac{15 k / 4 m}{3 k / 2 m}} = \sqrt{\frac{15/4}{3/2}} = \sqrt{\frac{15}{4} \times \frac{2}{3}} = \sqrt{\frac{30}{12}} = \sqrt{\frac{5}{2}} \]