A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two spri

Q: Why do the torques from both springs add up? Shouldn't they cancel each other out since they are on opposite sides?

Even though the springs are on opposite sides, when the rod rotates clockwise, the top end moves right (stretching the top spring to the left) and the bottom end moves left (stretching the bottom spring to the right). Both forces pull in opposite linear directions, but both create a counter-clockwise torque about the pivot. Since both torques tend to rotate the rod in the same counter-clockwise direction, they add up.

Q: Why did we approximate the displacement of the spring as purely horizontal?

For a very small angle $\theta$, the actual path of the rod's ends is an arc of a circle. The horizontal component of this displacement is $\frac{L}{2}\sin\theta \approx \frac{L}{2}\theta$, and the vertical component is $\frac{L}{2}(1-\cos\theta) \approx 0$. Thus, the vertical movement is negligible, and we can safely assume the springs only stretch horizontally.

Q: Does the length of the rod affect the frequency of oscillation?

No! As we saw in the derivation, the $L^2$ term from the restoring torque (which scales with the square of the lever arm) cancels out completely with the $L^2$ term from the moment of inertia. Therefore, the frequency depends only on the spring constant $k$ and the mass of the rod $M$.

Q: What if the rod was pivoted at one of its ends instead of the center?

If the rod were pivoted at one end, the moment of inertia would change to $I = \frac{1}{3}ML^2$, and the spring attachments and lever arms would be different. This would change the restoring torque and consequently the frequency of oscillation.

Q: Why is there a negative sign in the torque equation $\tau = -\frac{kL^2}{2}\theta$?

The negative sign indicates that the torque is a restoring torque. It acts in the direction opposite to the angular displacement $\theta$. If we displace the rod clockwise (positive direction), the torque acts counter-clockwise (negative direction) to bring it back.

Question

A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane.

The rod is gently pushed through a small angle $	heta$ in one direction and released. The frequency of oscillation is

Options:

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

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

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

(D) $\frac{1}{2\pi} \sqrt{\frac{24k}{M}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform rod of mass $M$ and length $L$ balanced horizontally or vertically in space, pivoted exactly at its center point $O$.

At both ends of the rod, we attach two identical springs of spring constant $k$.

The other ends of these springs are anchored to rigid walls.

In the equilibrium position, the rod is perfectly vertical, and both springs are at their natural, unstretched lengths.

There are no net forces or torques acting on the system.

Now, let's disturb this equilibrium.

Suppose we gently rotate the rod clockwise by a very small angular displacement $	heta$.

What happens to the springs?

- The top end of the rod, let's call it $A$, moves to the right by a small horizontal distance $x$.
- The bottom end of the rod, let's call it $B$, moves to the left by the same horizontal distance $x$.

Since the angle $	heta$ is extremely small, we can approximate the motion of the ends as purely horizontal.

Using simple trigonometry, the horizontal displacement $x$ is related to the angular displacement $	heta$ by:

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

### The Restoring Forces and Torques

As the top end moves right, the top spring is stretched by $x$.

It exerts a restoring force $F_s = kx$ to the left.

Similarly, as the bottom end moves left, the bottom spring is stretched by $x$.

It exerts a restoring force $F_s = kx$ to the right.

Both of these forces act to pull the rod back to its vertical equilibrium position.

Let's calculate the torque exerted by these forces about the pivot $O$:

1. The force at the top end acts at a distance of $\frac{L}{2}$ from the pivot, creating a counter-clockwise (restoring) torque:

$$	au_1 = -F_s \cdot \frac{L}{2} = -k\left(\frac{L}{2}	hetaight) \cdot \frac{L}{2} = -\frac{kL^2}{4}	heta$$

2. The force at the bottom end also acts at a distance of $\frac{L}{2}$ from the pivot, creating a counter-clockwise (restoring) torque in the same direction:

$$	au_2 = -F_s \cdot \frac{L}{2} = -k\left(\frac{L}{2}	hetaight) \cdot \frac{L}{2} = -\frac{kL^2}{4}	heta$$

Since both torques act in the same direction to restore the rod, the total restoring torque $	au$ is the sum of the two:

$$	au = 	au_1 + 	au_2 = -2 \cdot \left(\frac{kL^2}{4}	hetaight) = -\frac{kL^2}{2}	heta$$

### Rotational Dynamics and Moment of Inertia

To find the angular acceleration $\alpha$ of the rod, we use Newton's second law for rotation:

$$	au = I\alpha$$

where $I$ is the moment of inertia of the uniform rod about the pivot at its center.

The standard formula for the moment of inertia of a uniform rod of mass $M$ and length $L$ about its central axis is:

$$I = \frac{1}{12}ML^2$$

Substituting the expressions for $	au$ and $I$ into the rotational equation of motion, we get:

$$-\frac{kL^2}{2}	heta = \left(\frac{1}{12}ML^2ight)\alpha$$

### Solving for the Frequency

Let's isolate the angular acceleration $\alpha$:

$$\alpha = -\frac{\frac{kL^2}{2}}{\frac{1}{12}ML^2}	heta$$

Notice how beautifully the $L^2$ terms cancel out from both the numerator and the denominator!

This tells us that the length of the rod has absolutely no effect on its angular acceleration.

Simplifying the fraction:

$$\alpha = -\left(\frac{12}{2} \cdot \frac{k}{M}ight)	heta = -\left(\frac{6k}{M}ight)	heta$$

This is the classic equation for angular simple harmonic motion:

$$\alpha = -\omega^2	heta$$

By comparing the two equations, we find the angular frequency $\omega$:

$$\omega^2 = \frac{6k}{M} \implies \omega = \sqrt{\frac{6k}{M}}$$

The frequency of oscillation $f$ is related to the angular frequency by $f = \frac{\omega}{2\pi}$:

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

This matches **Option (c)** perfectly!

Select Answer:

Visualized Solution

Visualizing the Equilibrium State

Displacing the Rod by a Small Angle $θ$

Linear Displacement of the Ends

Restoring Forces Exerted by the Springs

Restoring Torque about the Pivot

Total Restoring Torque

Moment of Inertia of a Uniform Rod

Rotational Equation of Motion

Solving for Angular Acceleration $α$

Finding the Angular Frequency $ω$

Frequency of Oscillation

The Way Forward & Conceptual Insights

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Restoring Forces and Torques

Rotational Dynamics and Moment of Inertia

Solving for the Frequency

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