As shown in the figures, a uniform rod OO' of length l is hinged at the point O and held in place ve

Q: Why do we use $x = r\theta$ for the spring extension? Isn't the path curved?

Yes, the path is technically an arc of a circle. However, for very small angular displacements (which is a standard assumption for SHM), the arc length is practically indistinguishable from the horizontal straight-line distance. Therefore, we can safely approximate the horizontal extension or compression of the spring as $r\theta$.

Q: Why is the moment of inertia $\frac{Ml^2}{3}$ and not $\frac{Ml^2}{12}$?

The moment of inertia depends entirely on the axis of rotation. The formula $\frac{Ml^2}{12}$ is for a rod rotating about an axis passing through its center of mass. In this problem, the rod is hinged at its bottom end $O$. Using the parallel axis theorem, the moment of inertia about the end is $\frac{Ml^2}{12} + M\left(\frac{l}{2}\right)^2 = \frac{Ml^2}{3}$.

Q: In the first figure, one spring compresses and the other stretches. Don't their forces cancel out?

No, they actually work together! Imagine tilting the rod to the right. The left spring gets compressed and pushes the rod back to the left. The right spring gets stretched and pulls the rod back to the left. Both forces create a torque in the same counter-clockwise direction, so their torques add up.

Q: Why does the ratio of $f_1/f_2$ equal $\omega_1/\omega_2$? What happened to the $2\pi$?

The relationship between linear frequency $f$ and angular frequency $\omega$ is $f = \frac{\omega}{2\pi}$. When you take the ratio of the two frequencies, you get $\frac{f_1}{f_2} = \frac{\omega_1 / 2\pi}{\omega_2 / 2\pi}$. The $2\pi$ terms in the numerator and denominator perfectly cancel each other out.

Q: What would happen if the problem didn't say 'ignoring gravity'?

If gravity were included, the weight of the rod ($Mg$) acting downwards through its center of mass would create an additional torque. Since gravity tries to pull the rod further down away from the vertical, it acts as a destabilizing torque, opposing the restoring torque of the springs. This would decrease the net restoring torque, effectively lowering the frequency of oscillation.

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 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:

Options:

(A) 2

(B) $\sqrt{2}$

(C) $\sqrt{\frac{5}{2}}$

(D) $\sqrt{\frac{2}{5}}$

Sigma Percentile · Accepted Answer

This problem is a beautiful exploration of angular simple harmonic motion (SHM) and rotational dynamics. We are presented with two identical uniform rods, both hinged at their bottom ends and held in a vertical equilibrium by massless springs. The only difference between the two setups is the placement of these springs. Our mission is to determine the ratio of their oscillation frequencies when given a small angular nudge.

### The Master Equation for Angular SHM

Before we analyze the specific setups, let's recall the fundamental principle of angular SHM. When a rigid body is slightly displaced from its equilibrium position by an angle $	heta$, it experiences a restoring torque $	au$ that tries to bring it back. If this torque is directly proportional to the angular displacement, the body executes simple harmonic motion governed by the equation:

$$	au = -C	heta$$

Here, $C$ is the torsional constant of the system. The angular frequency $\omega$ of the oscillation is then given by:

$$\omega = \sqrt{\frac{C}{I}}$$

where $I$ is the moment of inertia of the body about the axis of rotation. Since both rods are identical and hinged at one end, their moment of inertia is $I = \frac{Ml^2}{3}$. Our task now is to find the torsional constant $C$ for both configurations.

### Analyzing the First Configuration

In the first figure, the left spring is attached at the midpoint (distance $\frac{l}{2}$ from the hinge), and the right spring is attached at the top end (distance $l$ from the hinge). When the rod is tilted by a small angle $	heta$, the points on the rod move horizontally. Using the small-angle approximation ($x = r	heta$), the left spring compresses by $x_1 = \frac{l}{2}	heta$, and the right spring stretches by $x_2 = l	heta$.

According to Hooke's Law, the forces exerted by the springs are $F_1 = k\left(\frac{l}{2}	hetaight)$ and $F_2 = k(l	heta)$. Both of these forces create a restoring torque about the hinge. The net torque is the sum of the individual torques (Force $	imes$ perpendicular distance):

$$	au_{net} = F_1\left(\frac{l}{2}ight) + F_2(l)$$

Substituting the forces, we get:

$$	au_{net} = k\left(\frac{l}{2}	hetaight)\left(\frac{l}{2}ight) + k(l	heta)(l) = \frac{kl^2}{4}	heta + kl^2	heta = \frac{5}{4}kl^2	heta$$

This gives us the torsional constant $C_1 = \frac{5}{4}kl^2$. Plugging this into our frequency formula:

$$\omega_1 = \sqrt{\frac{\frac{5}{4}kl^2}{\frac{Ml^2}{3}}} = \sqrt{\frac{15k}{4M}}$$

### Analyzing the Second Configuration

Now, let's look at the second figure. This setup is highly symmetric. Both springs are attached exactly at the midpoint of the rod. When tilted by an angle $	heta$, the midpoint moves by $x = \frac{l}{2}	heta$. Both springs undergo this exact same deformation, so each exerts a force $F = k\left(\frac{l}{2}	hetaight)$.

Since both forces act at a distance of $\frac{l}{2}$ from the hinge, the net restoring torque is simply twice the torque of a single spring:

$$	au_{net} = 2 	imes F 	imes \left(\frac{l}{2}ight) = 2 	imes k\left(\frac{l}{2}	hetaight)\left(\frac{l}{2}ight) = \frac{1}{2}kl^2	heta$$

This gives us the torsional constant $C_2 = \frac{1}{2}kl^2$. The angular frequency for the second rod is:

$$\omega_2 = \sqrt{\frac{\frac{1}{2}kl^2}{\frac{Ml^2}{3}}} = \sqrt{\frac{3k}{2M}}$$

### The Final Calculation

The question asks for the ratio of the linear frequencies, $\frac{f_1}{f_2}$. Since linear frequency is directly proportional to angular frequency ($f = \frac{\omega}{2\pi}$), the ratio $\frac{f_1}{f_2}$ is identical to $\frac{\omega_1}{\omega_2}$.

Let's divide the two expressions we derived:

$$\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2} = \frac{\sqrt{\frac{15k}{4M}}}{\sqrt{\frac{3k}{2M}}}$$

The $k$ and $M$ terms cancel out completely, leaving us with pure numbers:

$$\frac{f_1}{f_2} = \sqrt{\frac{15}{4} 	imes \frac{2}{3}} = \sqrt{\frac{5}{2}}$$

**The correct option is (C).** This problem elegantly demonstrates how the spatial distribution of restoring forces drastically alters the dynamic response of a rotational system.

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Visualized Solution

$V i s u a l A n c h or$

$A n g u l a r S H M L o g i c$

$A na l y z in g F i g 1 S e t u p$

$F or ces in F i g 1$

$R es t or in g T or q u e in F i g 1$

$F r e q u e n cy o f F i g 1$

$A na l y z in g F i g 2 S e t u p & T or q u e$

$F r e q u e n cy o f F i g 2$

$F ina l R a t i o$

$T h e W a y F or w a r d$

The Sigma Insight: Angular SHM

The Master Equation for Angular SHM

Analyzing the First Configuration

Analyzing the Second Configuration

The Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Anchor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular SHM Logic

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Fig 1 Setup

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Restoring Torque in Fig 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frequency of Fig 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Fig 2 Setup & Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frequency of Fig 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward