JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is ω₁. When the mass block is 2 kg the angular frequency is ω₂. The ratio ω₂ / ω₁ is
- (a) 1 / √2
- (b) 1 / 2
- (c) 2
- (d) √2
Answer and solution
Answer (a) 1 / √2
Solution
Physics sir · 3 steps
Same spring, heavier block: ω goes as 1/√m, so doubling the mass divides it by √2.
- Restoring force. The surface is frictionless, so the only horizontal force on the block is the spring force F = -kx. That gives ma = -kx, which is SHM.
ω = √(k/m)
- Take the ratio. The spring is the same in both cases, so k cancels: ω₂/ω₁ = √(m₁/m₂) = √(1/2).
- So. ω₂/ω₁ = 1/√2, which is option (a).
The idea to keep. For a block on a spring, ω = √(k/m). With the same spring, a heavier block oscillates more slowly, and ω changes as the square root of the mass ratio, not the ratio itself.
Answer from the NTA final answer key. Solution by PrepWiser.