Two open organ pipes of lengths 60 cm and 90 cm resonate at 6ᵗʰ and 5ᵗʰ harmonics respectively. The difference of frequencies for the given modes is _________ Hz. (Velocity of sound in air =333 m / s)

Solution

Physics sir · 4 steps

Both ends open, so every harmonic is allowed: use fₙ = nv/2L for each pipe and subtract.

1. Open pipe rule. A pipe open at both ends has antinodes at both ends, so its nth harmonic has frequency fₙ = nv/2L, for every whole number n.

   fₙ = (n v)/2L

2. First pipe. With L = 0.60 m and n = 6: f₆ = (6 × 333)/(2 × 0.60) = 1998/1.2 = 1665 Hz.
3. Second pipe. With L = 0.90 m and n = 5: f₅ = (5 × 333)/(2 × 0.90) = 1665/1.8 = 925 Hz.
4. So. The difference is 1665 - 925 = 740 Hz.

The common slip: 370. Using the closed-pipe formula f = nv/4L halves both frequencies and gives 370 Hz. Both pipes here are open, so the denominator is 2L.

The idea to keep. Open pipe: fₙ = nv/2L with all harmonics allowed. Closed pipe: fₙ = nv/4L with only odd n. Check the pipe type first, and convert cm to m before you divide.

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