Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is :

Solution

Physics sir · 3 steps

Push the block a little and both springs pull it back the same way, so their stiffnesses add.

1. Displace the block. Move the block a small distance x to the right. The left spring stretches by x and pulls the block left with force 2kx. The right spring compresses by x and pushes the block left with force 2kx.
2. Total restoring force. Both forces point back towards equilibrium, so F = -(2k + 2k)x = -4kx. The springs act in parallel, with k_(eff) = 4k.

   T = 2π√(m/k_(eff)) = 2π√(m/4k) = π√(m/k)

3. So. The time period is π√(m/k), which is option (b).

The idea to keep. Decide series or parallel by what happens on a displacement, not by how the drawing looks. If every spring changes length by the same x, their spring constants add.

Answer from the NTA final answer key. Solution by PrepWiser.