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

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Homework Help Overview

The discussion revolves around calculating the frequencies of the lowest three harmonics produced by 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.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various formulas for calculating harmonic frequencies, questioning the correct application of equations related to open-closed pipes. There are attempts to derive frequencies based on the length of the pipe and the speed of sound.

Discussion Status

Multiple approaches to the problem have been presented, with participants suggesting different sets of frequencies based on their calculations. Some guidance on the correct formula for an open-closed pipe has been offered, but no consensus has been reached on the final answer.

Contextual Notes

Participants are navigating through potential misunderstandings of the harmonic equations and the specific conditions of the organ pipe. There is a focus on ensuring the correct harmonic series is applied for the given setup.

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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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