In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ____________ rad/s when k = 20 N/m.

Solution

Physics sir · 3 steps

Two free blocks on a spring: replace the springs by one spring, the masses by the reduced mass.

1. Combine the springs. S₁ and S₂ are joined end to end, so they are in series and carry the same tension. Their effective constant is k_(eq) = (k × 4k)/(k + 4k) = 4k/5 = (4 × 20)/5 = 16 N m⁻¹.
2. Use reduced mass. Neither block is fixed, so both move and the centre of mass stays at rest. The two-body motion behaves like one body of reduced mass μ = (m_A m_B)/(m_A + m_B) = (0.2 × 0.8)/(0.2 + 0.8) = 0.16 kg.

   ω = √(k_(eq)/μ) = √(16/0.16) = √100

3. So. ω = 10 rad s⁻¹, so the answer is 10.

The common slip: 4. Using the total mass of 1 kg in place of the reduced mass gives ω = √(16/1) = 4, not 10.

The idea to keep. When both ends of a spring hold free masses, use ω = √(k/μ) with μ = (m₁ m₂)/(m₁ + m₂). Springs joined end to end add like resistors in parallel.

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