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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 oscillations.

Solution

Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in 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,
$F=-2 k x$
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 oscillations is,
$T=2 \pi \sqrt{\frac{m}{2 k}}$