A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with fundamental frequency of 600 Hz under same tension _______ cm.

Solution

Physics sir · 4 steps

Same tension, same wire, so the wave speed is fixed: frequency and length go inversely.

1. Fundamental mode. In the fundamental mode the wire holds half a wavelength, so λ = 2L. With v = fλ, the frequency is f = v/2L.

   f = 1/2L√(T/μ)

2. Speed stays fixed. The tension T and the mass per unit length μ do not change, so v = √(T/μ) is the same in both cases. That leaves fL = constant.
3. Put in numbers. 400 × 90 = 600 × L₂, so L₂ = 36000/600 = 60 cm.
4. So. The resonating length for 600 Hz is 60 cm. A higher note needs a shorter length.

The common slip: 135. Taking frequency as directly proportional to length gives 90 × 600/400 = 135 cm. The length must get shorter, not longer, when the frequency goes up.

The idea to keep. For a stretched string, f ∝ 1/L when tension and wire are unchanged. A higher frequency always means a shorter vibrating length, so check that your answer got smaller.

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