WAVES AND OSCILLATIONS BYU 18

Solution

1. Analysis of the Equilibrium State

Let us first analyze the forces acting on the rod when it is in the horizontal equilibrium position.

• Let the extension in the spring at equilibrium be $x_0$.
• The spring force is $F_s = k x_0$.
• The spring exerts a force at distance $d$ from the hinge, at an angle $\theta_0$ with the rod.
• The gravitational force $mg$ acts downwards at the end of the rod (distance $l$).

Taking torque about the hinge $O$:

$$ \tau_{\text{gravity}} = \tau_{\text{spring}} $$ $$ mg l = (k x_0) \times (d \sin \theta_0) \quad \dots \text{(i)} $$

2. Analysis of Small Oscillations

Consider the rod is displaced slightly downwards by a small angle $\theta$ (let us denote the angular displacement as $\theta$ to avoid confusion with $\theta_0$, although typical notation might use $\Delta \theta$ or $\alpha$).

Change in Spring Extension:
When the rod rotates downwards by a small angle $\theta$, the point of attachment on the rod moves downwards by an arc length of $d \theta$. The component of this displacement along the line of the spring contributes to the additional extension.

$$ \Delta x = (d \theta) \sin \theta_0 $$

Restoring Torque:
The total force in the spring is now $F_{\text{total}} = k(x_0 + \Delta x)$. The restoring torque tries to bring the rod back to the horizontal position. The torque equation for the displaced state is:

$$ \tau_{\text{net}} = \tau_{\text{gravity}} – \tau_{\text{spring}}’ $$

The perpendicular distance for the spring force remains approximately $d \sin \theta_0$ for small oscillations. The gravity torque remains approximately $mg l$ (as $\cos \theta \approx 1$).

$$ \tau_{\text{net}} = mg l – [k(x_0 + \Delta x)] (d \sin \theta_0) $$

Substituting $mg l = k x_0 d \sin \theta_0$ from the equilibrium equation (i):

$$ \tau_{\text{net}} = k x_0 d \sin \theta_0 – k(x_0 + d \theta \sin \theta_0) (d \sin \theta_0) $$ $$ \tau_{\text{net}} = – k (d \theta \sin \theta_0) (d \sin \theta_0) $$ $$ \tau_{\text{net}} = – (k d^2 \sin^2 \theta_0) \theta $$

3. Time Period Calculation

The equation of motion for angular simple harmonic motion is $\tau = -C \theta$, where $C$ is the torsional constant. From our derivation:

$$ C = k d^2 \sin^2 \theta_0 $$

The moment of inertia $I$ of the mass $m$ attached to the light rod of length $l$ is:

$$ I = m l^2 $$

The angular frequency $\omega$ is given by:

$$ \omega = \sqrt{\frac{C}{I}} = \sqrt{\frac{k d^2 \sin^2 \theta_0}{m l^2}} $$ $$ \omega = \frac{d \sin \theta_0}{l} \sqrt{\frac{k}{m}} $$

The time period $T$ is:

$$ T = \frac{2\pi}{\omega} $$