What Are the Frequencies of the Lowest Three Harmonics in a 1.70 m Organ Pipe?

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The discussion focuses on calculating the frequencies of the lowest three harmonics for a 1.70 m organ pipe that is open at one end and closed at the other. The speed of sound is given as 340 m/s. The correct formula for harmonics in this type of pipe is f = m(v/4L), where m represents the harmonic number (1, 3, 5). After applying the formula, the calculated frequencies are 50 Hz, 150 Hz, and 250 Hz, which corresponds to the first three harmonics. This method ensures accurate frequency determination for the specified organ pipe configuration.
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Homework Statement


An organ pipe is 1.70 m long and it is open at one end and closed at the other end. What are the frequencies of the lowest three harmonics produced by this pipe? The speed of sound is 340 m/s. Only one answer is correct.
200 Hz, 400 Hz, 600 Hz
200 Hz, 300 Hz, 400 Hz
200 Hz, 600 Hz, 1000 Hz
50 Hz, 100 Hz, 200 Hz
100 Hz, 200 Hz, 300 Hz
50 Hz, 150 Hz, 250 Hz
100 Hz, 300 Hz, 500 Hz
50 Hz, 100 Hz, 150 Hz


Homework Equations





The Attempt at a Solution


I am not really sure how to calculate these?
 
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Do I use...
F = (V/2L) = 100 Hz
F = (V/2L) * 2 = 200 Hz
F = (V/2L) * 3 = 300 Hz
So my answer would be...
100 Hz, 200 Hz, 300 Hz
 
No Wait if it's open-closed it would be...
F = (V/2L) = 100 Hz
F = (V/2L) * 3 = 300 Hz
F = (V/2L) * 5 = 500 Hz
So my answer would be...
100 Hz, 300 Hz, 500 Hz
 
Oh no wait...lol...thats not the right equation is it? It should be...
F = m * (v/4L), m = 1, 3 5
so I get...
50 Hz, 150 Hz, 250 Hz
is that right?
 
For a pipe that's open at one end and closed at the other, you'll want to use \lambda = 4L for the first harmonic, \lambda = 4L / 3 for the second harmonic, and \lambda = 4L/5 for the thrid harmonic, using f = v / \lambda for the frequency.
 
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