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 of the equilibrium position as shown in figure. Under this situation, the spring on the left gets elongated by a length equal tox and that on the right 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 is, tryping to pull the mass towards the mean position)
`F_(2) = - kx` ( force exerted by the spring on the right is, tryping to push the mass towards the means position) .
The net force, F acting on teh mass is then given by `F= - 2 kx`.
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 SHM. The time period of oscialltions is ,
`T= 2pisqrt((m)/(2k))`