A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________ Hz. (Round off to the nearest integer)

Solution

Physics sir · 3 steps

Fix the speed of sound in the new gas first; the pipe then hands you the difference as v/2L.

1. Speed in the gas. The speed of sound is v = √(γ P/ρ). With the same pressure and the same γ, v ∝ 1/√ρ, so doubling the density divides the speed by √2: v = 300/√2 = 212.1 m s⁻¹.
2. Open pipe harmonics. A pipe open at both ends supports every harmonic, fₙ = nv/2L. Going from n = 1 to n = 2 adds exactly one step of v/2L.

   f₂ - f₁ = v/2L = 212.1/(2 × 1) = 106.1 Hz

3. So. The fundamental is about 106.1 Hz and the second harmonic about 212.1 Hz, so the difference rounds to 106 Hz.

The common slip: 150. Using 300 m s⁻¹ directly and ignoring the denser gas gives 150 Hz, not 106.

The idea to keep. In an open pipe, neighbouring harmonics are always v/2L apart. When the medium changes, correct v first: at the same pressure, v ∝ 1/√ρ.

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