Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T b /Tₐ = √x, where value of x is ___________. (Round off to the Nearest Integer)

Solution

Physics sir · 3 steps

Two equal springs in series are half as stiff, and the period goes as one over root k.

1. Series spring constant. In series, the same force acts on both springs, so their stretches add. That gives 1/kₛ = 1/k + 1/k, so kₛ = k/2.

   T = 2π√(M/k) ⇒ T_b/Tₐ = √(k/kₛ)

2. Put in numbers. The mass M is the same in both figures, so it cancels. T_b/Tₐ = √(k/(k/2)) = √2.
3. So. Comparing with √x gives x = 2.

The common slip: 0.5. Adding the springs directly as 2k gives T_b/Tₐ = 1/√2, so x = 0.5. That is the parallel rule, not the series rule.

The idea to keep. Springs in series add like resistors in parallel, and springs in parallel add directly. Then use T ∝ 1/√k: a softer spring gives a longer period.

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