In figure (A), mass '2 m^' is fixed on mass 'm ' which is attached to two springs of spring constant k. In figure (B), mass 'm' is attached to two springs of spring constant 'k' and '2 k′. If mass 'm' in (A) and in (B) are displaced by distance 'x′ horizontally and then released, then time period T₁ and T₂ corresponding to (A) and (B) respectively follow the relation.

Solution

Physics sir · 4 steps

Springs on both sides of a block act in parallel, and in (A) the spring force moves all 3m together.

1. Springs in parallel. Push the block by x. One spring stretches and the other compresses, and both forces point back towards the mean position. The restoring force is (k₁ + k₂)x, so k_(eff) = k₁ + k₂.
2. Figure (A). The 2m block is fixed on m, so the moving mass is 3m. The two springs give k_(eff) = k + k = 2k. So T₁ = 2π√(3m/2k).
3. Figure (B). Only m moves, and k_(eff) = k + 2k = 3k. So T₂ = 2π√(m/3k).

   T₁/T₂ = √(3m/2k × 3k/m) = √(9/2) = 3/√2

4. So. T₁/T₂ = 3/√2, which is option (a).

The idea to keep. For a block between two springs, the stiffnesses add like parallel springs. Always count every mass that moves with the block, because T = 2π√(m/k_(eff)).

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