The time period of simple harmonic motion of mass M in the given figure is π √((α M)/(5 k)), where the value of α is _________.

Solution

Physics sir · 4 steps

Shrink the springs to one equivalent spring step by step, then match your answer to the given form.

1. Top pair in parallel. The two upper springs on the right share both ends, so they stretch by the same amount. Parallel stiffnesses add: k + k = 2k.
2. Then in series. This 2k block hangs above the lower spring k, and the same force passes through both, so they are in series: (2k × k)/(2k + k) = 2k/3.
3. Both branches together. The left spring k and the right branch 2k/3 both hold the same block M, so they are in parallel: k_(eq) = k + 2k/3 = 5k/3.

   T = 2π√(M/k_(eq)) = 2π√(3M/5k) = π√(12M/5k)

4. So. Taking the factor 2 inside the root as 2² = 4 gives π√(12M/5k), so α = 12.

The common slip: 3. Reading α off 2π√(3M/5k) without moving the 2 inside the root gives 3, not 12.

The idea to keep. Springs that share both ends add like k₁ + k₂. Springs that carry the same force combine as (k₁ k₂)/(k₁ + k₂). Always match the final form exactly, including the factor outside the root.

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