A wire having a linear mass density 9.0 × 10⁻⁴ kg/m is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is ____________ m.

Solution

Physics sir · 4 steps

Two back-to-back resonances differ by exactly the fundamental, v/2L. Find v first, then L.

1. Wave speed. On a stretched string, v = √(T/μ) = √(900/(9.0 × 10⁻⁴)) = √10⁶ = 1000 m s⁻¹.
2. Gap between harmonics. A string fixed at both ends resonates at fₙ = nv/2L. Two consecutive resonances, n and n+1, differ by v/2L, which is the fundamental frequency.

   fₙ₊₁ - fₙ = v/2L = 550 - 500 = 50 Hz

3. Solve for L. L = v/(2 × 50) = 1000/100 = 10 m. As a check, 500 Hz is the 10th harmonic and 550 Hz is the 11th.
4. So. The length of the wire is 10 m.

The common slip: 20. Taking the gap between harmonics as v/L instead of v/2L gives 20 m, not 10 m.

The idea to keep. For a string fixed at both ends, the difference between any two consecutive resonant frequencies equals the fundamental v/2L. Get v from √(T/μ) and the length follows.

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