A 40.0 cm long organ pipe is filled with air at 0$^{\circ}$C and is open at both ends.The velocity of sound in air at 0$^{\circ}$C is 331 m/s.Another 40.0 cm long organ pipe is filled with air at 20$^{\circ}$C and is open at both ends.What is the beat frequency between the fundamental modes of the two organ pipes?

Question

Quizplus · Accepted Answer

The frequency of the fundamental mode of an open-closed pipe is given by:

f1 = (2n - 1) v/4L

where n = 1, 2, 3, ..., v is the velocity of sound, and L is the length of the pipe.

For the first pipe,

f1₁ = (2n - 1) v/4L₁ = (2n - 1) (331 m/s)/(4×0.40 m) = 207 n Hz (where n is a positive integer)

For the second pipe,

f1₂ = (2n - 1) v/4L₂ = (2n - 1) (343 m/s)/(4×0.40 m) = 214 n Hz (where n is a positive integer)

The beat frequency between two frequencies is given by Δf = |f1 - f2|. 
Taking n = 1, we have

Δf = |207 - 214| = 7 Hz

Taking n = 2, we have

Δf = |415 - 428| = 13 Hz

Taking n = 3, we have

Δf = |623 - 642| = 19 Hz

Thus, the beat frequency between the fundamental modes of the two pipes is 7 Hz.

A 400 Cm Long Organ Pipe Is Filled with Air at at 0