Two light identical springs of spring constant k are attached horizontally at the two ends of an uni

Q: Why do we assume small deflections?

For large angles, the end of the rod moves in an arc, and the spring force is no longer perfectly perpendicular to the rod. Small angle approximations ($\sin\theta \approx \theta$ and $\cos\theta \approx 1$) ensure the motion remains simple harmonic.

Q: Why do both springs contribute to the restoring torque in the same direction?

When the rod rotates, one spring is compressed and pushes back, while the other is stretched and pulls back. Both of these forces create a torque that tries to rotate the rod back to its original equilibrium position.

Q: What is the moment of inertia of the rod?

Since the rod is pivoted at its center, its moment of inertia is $I = \frac{ml^2}{12}$. If it were pivoted at one end, it would be $\frac{ml^2}{3}$.

Q: Why does the length $l$ cancel out in the final expression?

The restoring torque depends on $l^2$ (since $x \propto l$ and torque arm $\propto l$), and the moment of inertia also depends on $l^2$. Thus, the length of the rod doesn't affect the final frequency of oscillation.

Q: What if the springs were attached at different distances from the pivot?

You would need to calculate the torque for each spring separately using their specific distances from the pivot, $r_1$ and $r_2$. The total torque would be $(kr_1^2 + kr_2^2)\theta$.

Question

Two light identical springs of spring constant $k$ are attached horizontally at the two ends of an uniform horizontal rod AB of length $l$ and mass $m$. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure.

The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is

Options:

(A) $\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$

(B) $\frac{1}{2\pi}\sqrt{\frac{3k}{m}}$

(C) $\frac{1}{2\pi}\sqrt{\frac{6k}{m}}$

(D) $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform rod of mass $m$ and length $l$, perfectly balanced and pivoted exactly at its center, $O$. Attached to its ends, $A$ and $B$, are two identical springs, each possessing a spring constant $k$. The rod is free to rotate in the horizontal plane. This is a classic example of a torsional oscillator, where linear spring forces translate into rotational restoring torques.

### The Physics of Small Deflections

When we gently push the rod, displacing it by a small angle $	heta$, the magic of Simple Harmonic Motion (SHM) begins. As the rod rotates, the ends $A$ and $B$ sweep through a small arc. For very small angles, this arc length can be approximated as a straight line distance, $x$.

Since the pivot is at the center, the distance from the pivot to either end is $\frac{l}{2}$. Therefore, the linear displacement of each end is given by the arc length formula:

$$x = \frac{l}{2}	heta$$

### Restoring Forces and Torque

As the ends move, the springs are either stretched or compressed. According to Hooke's Law, each spring exerts a restoring force $F = kx$. Substituting our expression for $x$, we get:

$$F = k\left(\frac{l}{2}	hetaight)$$

This force acts perpendicularly to the rod (again, assuming a small angle $	heta$). The torque $	au_1$ produced by one spring about the pivot $O$ is the force multiplied by the perpendicular distance $\frac{l}{2}$:

$$	au_1 = F 	imes \frac{l}{2} = k\left(\frac{l}{2}	hetaight) 	imes \frac{l}{2}$$

Crucially, when the rod rotates, one spring pushes while the other pulls, but both actions create a torque in the *same* rotational direction—opposing the initial displacement. Thus, the total restoring torque $	au$ is simply twice the torque of a single spring:

$$	au = 2 	imes \left[k\left(\frac{l}{2}	hetaight) 	imes \frac{l}{2}ight] = \frac{kl^2}{2}	heta$$

In the standard angular SHM equation, $	au = C	heta$, where $C$ is the torsional constant. Comparing the two, we find:

$$C = \frac{kl^2}{2}$$

### The Master Equation

To find the time period of oscillation, we need the moment of inertia $I$ of the rod. For a uniform rod rotating about an axis through its center, the moment of inertia is a standard, well-known result:

$$I = \frac{ml^2}{12}$$

The time period $T$ for an angular harmonic oscillator is given by:

$$T = 2\pi\sqrt{\frac{I}{C}}$$

Let's substitute our values for $I$ and $C$ into this master equation:

$$T = 2\pi\sqrt{\frac{\frac{ml^2}{12}}{\frac{kl^2}{2}}}$$

Notice the beautiful symmetry here: the $l^2$ terms in the numerator and denominator perfectly cancel each other out! This tells us a profound physical truth—the frequency of this specific oscillator is completely independent of the rod's length. Simplifying the fraction $\frac{2}{12}$ to $\frac{1}{6}$, we arrive at the final time period:

$$T = 2\pi\sqrt{\frac{m}{6k}}$$

### Final Calculation

The problem asks for the frequency of oscillation, $f$, which is simply the reciprocal of the time period ($f = \frac{1}{T}$):

**$$f = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}$$**

This elegant result perfectly matches option (c).

Select Answer:

Visualized Solution

$A na l y z in g t h e S e t u p$

$C o n d i t i o n f or S H M$

$C a l c u l a t in g R es t or in g T or q u e$

$T o t a l R es t or in g T or q u e$

$M o m e n t o f I n er t ia$

$T im e P er i o d$

$F r e q u e n cy o f O sc i l l a t i o n$

The Sigma Insight: Angular SHM

Analyzing the Setup

The Physics of Small Deflections

Restoring Forces and Torque

The Master Equation

Final Calculation

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