Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :

Solution

Physics sir · 3 steps

Maximum speed is Aω, so equal top speeds mean the amplitudes go opposite to the ω values.

1. Vertical is fine. Hanging the mass vertically only shifts the equilibrium point. Each mass still does SHM with ω = √(k/m), and its top speed is vₘₐₓ = Aω.
2. Find each omega. For the first, ω₁ = √(2k/50). For the second, ω₂ = √(9k/100). The mass unit cancels in the ratio, so grams are fine here.

   A₁ω₁ = A₂ω₂ ⇒ A₁/A₂ = ω₂/ω₁ = √((9k/100)/(2k/50)) = √(9/4) = 3/2

3. So. The amplitudes are in the ratio A₁ : A₂ = 3 : 2, which is option (b).

The idea to keep. Equal maximum speeds means Aω is the same for both, so the slower oscillator (smaller ω) swings through the larger amplitude.

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