A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is in SI unit. The value of is ___________.
Show answer and solution
Answer: 5
Solution
When both springs are attached to rigid supports on either side, displacing the block by x compresses one spring and stretches the other. Both spring forces act in the same restoring direction, so the effective spring constant is keff = k₁ + k₂ = 20 + 20 = 40 N/m.
Time period T = 2π√(m/keff) = 2π√(2/40) = 2π√(1/20) = 2π/(√20) = π/√5.
Comparing with π/√x gives x = 5.
Time period T = 2π√(m/keff) = 2π√(2/40) = 2π√(1/20) = 2π/(√20) = π/√5.
Comparing with π/√x gives x = 5.
[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]