A wire of length L and mass per unit length 6.0 × 10 –3 kgm –1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are : 420 Hz and 490 Hz. Then L in meters is :

Solution

Physics sir · 4 steps

Two neighbouring resonances of a string differ by exactly the fundamental, so find v and use the gap.

1. Wave speed. The speed on a stretched string is v = √(T/μ) = √(540/(6.0 × 10⁻³)) = √90000 = 300 m s⁻¹.
2. Consecutive harmonics. A wire fixed at both ends resonates at fₙ = nv/2L. Two consecutive frequencies differ by one step in n, so their gap is the fundamental v/2L.

   fₙ₊₁ - fₙ = v/2L = 490 - 420 = 70 Hz

3. Solve for L. L = v/(2 × 70) = 300/140 ≈ 2.14 m. As a check, 420 = 6 × 70 and 490 = 7 × 70, so these are the 6th and 7th harmonics.
4. So. L ≈ 2.1 m, which is option (b).

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

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