Example

Question:

Two identical springs of spring constant \(k\) are attached to a block of mass \(m\) and to fixed supports as shown in Fig. 13.14. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.

Solution:

Let the mass be displaced by a small distance \(x\) to the right side of the equilibrium position. Under this situation:
- The spring on the left side gets elongated by length \(x\)
- The spring on the right side gets compressed by the same length \(x\)

The forces acting on the mass are: \[ F_1 = -kx \quad \text{(force by left spring, trying to pull mass towards mean position)} \] \[ F_2 = -kx \quad \text{(force by right spring, trying to push mass towards mean position)} \]
The net force \(F\) acting on the mass is: \[ F = -2kx \]
Since the force is proportional to displacement and directed towards the mean position, the motion is simple harmonic.

The time period of oscillations is: \[ T = 2\pi \sqrt{\frac{m}{2k}} \]