For a certain organ pipe, the first three resonance frequencies are in the ratio of 1:3:5 respectively. If the frequency of fifth harmonic is 405 Hz and the speed of sound in air is 324 ms⁻¹ the length of the organ pipe is _________ m.

Solution

Physics sir · 4 steps

Odd harmonics only, 1 : 3 : 5, means a closed pipe, so use fₙ = nv/4L with n = 5.

1. Identify the pipe. An open pipe gives all harmonics, 1 : 2 : 3. Only a pipe closed at one end gives odd harmonics alone, 1 : 3 : 5. So this is a closed pipe.
2. Closed pipe frequencies. The fundamental is v/4L, and the nth harmonic is n times that, with n odd. The fifth harmonic has n = 5.

   f₅ = 5v/4L ⇒ L = 5v/4f₅

3. Put in numbers. L = (5 × 324)/(4 × 405) = 1620/1620 m.
4. So. The length of the organ pipe is 1 m.

The common slip: 2. Using the open-pipe formula f₅ = 5v/2L gives a length of 2 m, not 1 m. The odd-only ratio tells you the pipe is closed, so the denominator is 4L.

The idea to keep. Read the pipe type from the frequency ratio: 1:2:3 is open, 1:3:5 is closed. In a closed pipe the nth harmonic is nv/4L, and only odd n exist.

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