In an experiment with sonometer when a mass of 180 g is attached to the string, it vibrates with fundamental frequency of 30 Hz. When a mass m is attached, the string vibrates with fundamental frequency of 50 Hz. The value of m is ___________ g.

Solution

Physics sir · 3 steps

Same string, same length: frequency goes as the square root of tension, so square the frequency ratio.

1. Fundamental frequency. For a string of length L and mass per unit length μ, f = 1/2L√(T/μ). Here L and μ do not change, so f ∝ √T.
2. Tension from mass. The hanging mass sets the tension, T = mg. So f ∝ √m, which means m ∝ f².

   m/(180 g) = (50/30)² = 25/9

3. So. m = 180 × 25/9 = 20 × 25 = 500 g, so the answer is 500.

The common slip: 300. Taking frequency proportional to tension instead of its square root gives 180 × 5/3 = 300 g, not 500.

The idea to keep. On a sonometer with fixed length and wire, f ∝ √T. To scale the frequency by some factor, scale the hanging mass by that factor squared.

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