Two particles A and B of equal masses are suspended from two massless springs of spring constants K₁ and K₂ respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

Solution

Physics sir · 3 steps

Maximum speed is amplitude times omega, so equal speeds mean the stiffer spring gets the smaller amplitude.

1. Maximum speed. In SHM the particle is fastest at the mean position, where vₘₐₓ = Aω. For a mass on a spring, ω = √(K/m).
2. Set them equal. Both masses are m, so equal maximum speeds give A₁√(K₁/m) = A₂√(K₂/m). The m cancels from both sides.

   A₁/A₂ = ω₂/ω₁ = √(K₂/K₁)

3. So. The ratio of the amplitudes of A and B is √(K₂/K₁), which is option (d).

The idea to keep. Whenever vₘₐₓ is fixed, amplitude goes as 1/ω. For springs that means A ∝ 1/√K, so a stiffer spring oscillates through a smaller distance.

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