Question

An organ pipe is 151cm\; cm long. The speed of sound in air is 343 m/s. Part A: What are the fundamental and first three audible overtones if the pipe is closed at one end? What are the fundamental and first three audible overtones if the pipe is open at both ends? Express awnsers to 3 signiicant figures seperated by commas

Answer 1

If one end is closed, and one open, then you get a pattern of successive harmonics that goes
f₁, 3f₁, 5f₁, 7f₁, 9f₁ - that is, odd multiples of the fundamental, so let's find the fundamental:
With one end fixed, or a closed pipe, at the fundamental, the node is at the closed end (fixed end) and the anti-node is at the open end. Therefore the length of the pipe (L) is equal to only ¹/₄l, orl= 4L = 4(1.51 m) = 6.04 m, and the frequency you can get with v = fl.
The resonant frequency now is: f = v/l= (343 m/s)/(6.04 m) = 56.7 Hz
And the next two can be found now:
f₁ = 56.7 Hz (fundamental frequency)
f₂ = 3f₁ = 3 x 56.7 Hz =170.1 Hz
f₃ = 5f₁ = 5 x 56.7 Hz = 283.5 Hz
f₃ = 7f₁ = 7 x 56.7 Hz = 396.9 Hz

If both ends are open, then you get a pattern of successive harmonics that goes
f₁, 2f₁, 3f₁, 4f₁, 5f₁ - that is, multiples of the fundamental, so let's find the fundamental:
At the fundamental, a both ends open organ pipe has a node in the middle, and two anti-nodes at each end, the length of the pipe (L) is equal to ²/₄l, or  L = l/2, or l= 2L = 2(1.51 m) = 3.02 m, and the frequency you can get with v = fl.
The resonant frequency now is: f = v/l= (343 m/s)/(3.02 m) = 113.57 Hz
And the next two can be found now:
f₁ = 113.57Hz (fundamental frequency)
f₂ = 2f₁ = 2 x 113.57 Hz = 227.14 Hz
f₃ = 3f₁ = 3 x 113.57 Hz = 340.71 Hz
f₃ = 4f₁ = 4 x 113.57 Hz =454.28 Hz