Question

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.

Answer 1

Let the mass m be displaced by a small distance x to the right as shown in Figure. Due to it, the spring on the left hand side gets stretched by length x and the spring on the right hand side gets compressed by length x. The forces acting on the mass due to springs are

F₁ = – kx towards left hand side 

F₂ = – kx towards left hand side

Therefore, total restoring force on mass m is

F = F₁ + F₂ = – kx + (– kx) = –2 kx ...(i)

Here –ve sign shows that force F is directed towards the equilibrium position O and F x. Therefore, if the mass m is left free, it will execute linear SHM.

Comparing (i) with the relation, F = – kx, we have Spring factor, K = 2 k

Here, interia factor = mass of the block = m