A closed organ pipe of length L and an open organ pipe contain gases of densities ρ₁ and ρ₂ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is x/3L√((ρ ₁)/(ρ ₂)) where x is ___________. (Round off to the Nearest Integer)

Solution

Physics sir · 3 steps

Write each pipe's first overtone, set them equal, and let v = √(B/ρ) handle the gases.

1. Speed in each gas. Sound speed in a gas is v = √(B/ρ), where B is the bulk modulus, the inverse of compressibility. Equal compressibility means equal B, so v₂/v₁ = √(ρ₁/ρ₂).
2. First overtones. A closed pipe has only odd harmonics, so its first overtone is the 3rd harmonic, 3v₁/4L. An open pipe has all harmonics, so its first overtone is the 2nd harmonic, 2v₂/2L' = v₂/L'.

   3v₁/4L = v₂/L' ⇒ L' = 4/3L v₂/v₁ = 4/3L√(ρ₁/ρ₂)

3. So. Compare with x/3L√(ρ₁/ρ₂): x = 4.

The common slip: 6. Taking the closed pipe's first overtone as the 2nd harmonic, 2v₁/4L, gives 6. A closed pipe has no even harmonics.

The idea to keep. Closed pipe: only odd harmonics, so the first overtone is the 3rd. Open pipe: all harmonics, so the first overtone is the 2nd. Get that right before you equate frequencies.

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