The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is _____________ cm.

Solution

Physics sir · 3 steps

Write both frequencies in terms of v and L, set them equal, and the speed of sound cancels out.

1. Open pipe overtone. An open pipe has every harmonic, fₙ = nv/2Lₒ. The first overtone is the second harmonic, so f = 2v/2Lₒ = v/Lₒ.
2. Closed pipe fundamental. A closed pipe fits a quarter wavelength in its fundamental mode, so f = v/4L_c.

   v/Lₒ = v/4L_c ⇒ Lₒ = 4L_c

3. So. Lₒ = 4 × 20 cm = 80 cm, so the answer is 80.

The common slip: 40. Using the open pipe's fundamental v/2Lₒ instead of its first overtone gives Lₒ = 2L_c = 40, not 80.

The idea to keep. Open pipe: fₙ = nv/2L with all n. Closed pipe: f = (2n-1)v/4L with odd harmonics only. The first overtone is the next allowed harmonic after the fundamental.

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