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

Explanation

Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig. 14.15. 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 exerted by the spring on the left side, trying to pull 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) The net force, F, acting on the mass is then given by,
`F = -2kx`
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 oscillations is,
`T = 2pi sqrt((m)/(2k))`