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

(N/A) Let the mass be displaced by a small distance $x$ to the right side of the equilibrium position,as shown in the figure. 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:
$F_{1} = -k x$ (force exerted by the spring on the left side,trying to pull the mass towards the mean position)
$F_{2} = -k x$ (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:
$F = F_{1} + F_{2} = -k x - k x = -2 k x$
Since the net force $F$ is proportional to the displacement $x$ and is directed towards the mean position $(F \propto -x)$,the motion executed by the mass is simple harmonic.
Comparing this with the standard equation of simple harmonic motion $F = -K_{eff} x$,we get the effective spring constant $K_{eff} = 2 k$.
The time period of oscillations is given by:
$T = 2 \pi \sqrt{\frac{m}{K_{eff}}} = 2 \pi \sqrt{\frac{m}{2 k}}$