Question 3

SCQMEDIUM

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 same spring constant $k$ . 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)

(B)

$\sqrt{2}$

(C)

$\sqrt{\frac{5}{2}}$

(D)

$\sqrt{\frac{2}{5}}$

Detailed Solution

Let the small angular displacement be $\theta$ . The moment of inertia of the rod about the hinge $O$ is $I = \frac{1}{3}ml^2$ .

Case 1: One spring is at the midpoint ( $l/2$ ) and one at the top ( $l$ ). The restoring torque is $\tau_1 = - [k(\frac{l}{2}\theta) \cdot \frac{l}{2} + k(l\theta) \cdot l] = -k\theta(\frac{l^2}{4} + l^2) = -\frac{5}{4}kl^2\theta$ . The angular frequency is $\omega_1 = \sqrt{\frac{|\tau_1 / \theta|}{I}} = \sqrt{\frac{5kl^2/4}{ml^2/3}} = \sqrt{\frac{15k}{4m}}$ .

Case 2: Both springs are at the midpoint ( $l/2$ ). The restoring torque is $\tau_2 = - [k(\frac{l}{2}\theta) \cdot \frac{l}{2} + k(\frac{l}{2}\theta) \cdot \frac{l}{2}] = -\frac{1}{2}kl^2\theta$ . The angular frequency is $\omega_2 = \sqrt{\frac{|\tau_2 / \theta|}{I}} = \sqrt{\frac{kl^2/2}{ml^2/3}} = \sqrt{\frac{3k}{2m}}$ .

The ratio of frequencies is: $\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2} = \sqrt{\frac{15k/4m}{3k/2m}} = \sqrt{\frac{15}{4} \cdot \frac{2}{3}} = \sqrt{\frac{5}{2}}$

Question 3

Detailed Solution

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