As per given figures, two springs of spring constants k and 2 k are connected to mass m. If the period of oscillation in figure (a) is 3 s, then the period of oscillation in figure (b) will be √x s. The value of x is ___________.

Solution

Physics sir · 4 steps

Find the effective spring constant in each figure, then let T ∝ 1/√k do the rest.

1. Figure (a), series. One spring hangs below the other, so both carry the same force. Their stiffnesses combine as 1/kₛ = 1/k + 1/2k, which gives kₛ = 2k/3.
2. Figure (b), parallel. Both springs hold the block side by side and stretch by the same amount. Their stiffnesses add directly: kₚ = k + 2k = 3k.

   T_b/Tₐ = √(kₛ/kₚ) = √((2k/3)/3k) = √(2/9) = √2/3

3. Use the ratio. Since T = 2π√(m/k) and the mass is the same, T ∝ 1/√k. So T_b = 3 × √2/3 = √2 s.
4. So. T_b = √x s with x equal to 2.

The common slip: 40.5. Treating the period as growing with stiffness, T ∝ √k, gives T_b = 9/√2 s and x = 40.5. A stiffer system oscillates faster, so T_b must be smaller than 3 s.

The idea to keep. With the same mass, T ∝ 1/√k, so take the ratio of the two effective spring constants. Series makes the system softer, parallel makes it stiffer.

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