An organ pipe is 112 c m long. The speed of sound in air is 343 m / s . What are

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Explanation / Answer

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, or l= 4L = 4(1.12 m) = 4.48 m, and the frequency you can get with v = fl.
The resonant frequency now is: f = v/l= (343 m/s)/(4.48 m) = 75.6 Hz
And the next two can be found now:
f1 = 75.6 Hz (fundamental frequency)
f2 = 3f1 = 3 x 75.6 Hz = 230 Hz
f3 = 5f1 = 5 x 75.6 Hz = 383 Hz
f3 = 7f1 = 7 x 75.6 Hz = 536 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(1.12 m) = 2.24 m, and the frequency you can get with v = fl.
The resonant frequency now is: f = v/l= (343 m/s)/(2.28 m) = 153.1 Hz
And the next two can be found now:
f1 = 153.1 Hz (fundamental frequency)
f2 = 2f1 = 2 x 153.1 Hz = 306 Hz
f3 = 3f1 = 3 x 153.1 Hz = 459 Hz
f3 = 4f1 = 4 x 153.1 Hz = 612 Hz

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