A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is π/√x in SI unit. The value of x is ____________.

Solution

Physics sir · 4 steps

Displace the block and both springs pull it back together, so their spring constants add.

1. Both springs act. Push the block a distance x to the right. The left spring stretches by x and pulls it left. The right spring compresses by x and pushes it left. Each gives a force kx in the same direction.
2. Effective constant. The total restoring force is 2kx, so the springs behave as if they were in parallel: k_(eff) = 2k = 2 × 20 = 40 N m⁻¹.

   T = 2π√(m/k_(eff)) = 2π√(2/40) = 2π/√20

3. Match the form. Take the 2 inside the root: 2π/√20 = π/√(20/4) = π/√5.
4. So. Comparing with π/√x gives x = 5.

The common slip: 1.25. Treating the two springs as being in series gives k_(eff) = 10 N m⁻¹ and T = π/√1.25, so x = 1.25, not 5.

The idea to keep. A block between two springs fixed to walls feels both springs pulling the same way, so k_(eff) = k₁ + k₂, even though the springs look like they are in a line.

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