A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s. The maximum force exerted by spring on block is _________ N

Solution

Physics sir · 4 steps

Get the spring constant from the period, then the force is largest at the extreme position.

1. Angular frequency. With π ≈ 3.14, the period gives ω = 2π/T = (2 × 3.14)/3.14 = 2 rad s⁻¹.
2. Spring constant. For a spring-mass system ω² = k/m, so k = mω² = 5 × 2² = 20 N m⁻¹.

   Fₘₐₓ = kA = mω² A

3. Largest stretch. The spring force kx is largest when the stretch is largest, at x = A = 1 m. So Fₘₐₓ = 20 × 1 = 20 N.
4. So. The maximum force exerted by the spring on the block is 20 N.

The common slip: 10. Forgetting to square ω and using F = mω A gives 10 N, not 20 N.

The idea to keep. In SHM the restoring force peaks at the extremes, where it equals mω² A. Turn the period into ω first, and remember ω enters squared.

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