Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. 14.14. 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.

Explanation

As shown in figure 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 a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then,

`F_(1) = -kx`, force excerted by the spring on the left side, trying to pull the the mass towards the mean position.
`F_(2) = -kx` (force exerted by the spring on the right side, trying to push the mass towards the mean position.)
Since restoring force displaced the mass towards the equilibrium position
The net force on the mass `F= F_(1)+ F_(2)`
`= -kx - kx = -2kx`, here 2k is constant
`therefore F propto -x`.
Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position therefore, the motion executed by the mass is simple harmonic.
The time period of oscillation is `T= 2pi sqrt((m)/(2k))`.