A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency f₁. The frequency of oscillations if a mass 9 m is suspended from the same spring is f₂. The value of f₁/f₂ i ________.

Solution

Physics sir · 3 steps

Same spring, so k stays fixed: frequency goes as one over the square root of the mass.

1. Frequency formula. For a mass on a spring, the frequency depends only on the spring constant k and the mass. Gravity shifts the equilibrium position but does not change the frequency.

   f = 1/2π√(k/m)

2. Take the ratio. The spring is the same, so k and 1/2π cancel. That leaves f₁/f₂ = √(m₂/m₁) = √(9m/m) = √9.
3. So. f₁/f₂ = 3. The heavier mass oscillates three times more slowly, so the answer is 3.

The common slip: 9. Forgetting the square root and taking f inversely proportional to m gives 9, not 3.

The idea to keep. For a spring-mass system, f ∝ 1/√m and T ∝ √m. Multiply the mass by n and the frequency is divided by √n, not by n.

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