A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions. The springs have spring cons... A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions. The springs have spring constant k, and at equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position. Read more

Steps to Solve

1. Identify the forces and torques acting on the rod

In the equilibrium position, the forces from the two springs produce a torque about the pivot.

• The torque due to the spring at the midpoint (distance $l/2$ from the pivot) is given by: $$ \tau_1 = -k y_1 \cdot \frac{l}{2} $$

• The torque due to the spring at the far end (distance $l$ from the pivot) is: $$ \tau_2 = k y_2 \cdot l $$

Here, $y_1$ and $y_2$ are the displacements of the springs from their equilibrium positions.

2. Set up the equation of motion

Using the principle of rotational dynamics, the net torque ($\tau_{net}$) is equal to the moment of inertia ($I$) times the angular acceleration ($\alpha$):

$$ \tau_{net} = I \alpha $$

For small angles, $y_1 = y_2 = \theta$ (small angular displacement).

The moment of inertia of the rod about the pivot is given by: $$ I = \frac{1}{3} m l^2 $$

Thus, the net torque equation becomes: $$ -\left( k \theta \cdot \frac{l}{2} + k \theta \cdot l \right) = \frac{1}{3} m l^2 \cdot \frac{d^2 \theta}{dt^2} $$

3. Combine the equations together

Simplifying the left side, we have: $$ -k\theta\left( \frac{l}{2} + l \right) = \frac{1}{3} m l^2 \cdot \frac{d^2 \theta}{dt^2} $$

This simplifies to: $$ -\frac{3k l}{2} \theta = \frac{1}{3} m l^2 \cdot \frac{d^2 \theta}{dt^2} $$

4. Write the standard form of simple harmonic motion

Rearranging gives us: $$ \frac{d^2 \theta}{dt^2} + \frac{3k}{2m} \theta = 0 $$

This is the standard form of simple harmonic motion, where: $$ \omega^2 = \frac{3k}{2m} $$

5. Determine the frequency of oscillation

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

Substituting our expression for $\omega$, we get: $$ f = \frac{1}{2\pi} \sqrt{\frac{3k}{2m}} $$