htbruin

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For a wave in which one end is fixed and the other is open, the standing wave frequency is fn=nf1. N= 1, 3, 5, 7, ect.

How many times greater is the 5th harmonic frequency than the fundamental frequency?

The answer is 5 times. I thought since "n" for the 5th frequency for a one end closed is 9 (1, 3, 5, 7, 9) then it should be 9 times greater.

Can someone explain this? thanks.
 
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BerkReviewTeach

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For a wave in which one end is fixed and the other is open, the standing wave frequency is fn=nf1. N= 1, 3, 5, 7, ect.

How many times greater is the 5th harmonic frequency than the fundamental frequency?

The answer is 5 times. I thought since "n" for the 5th frequency for a one end closed is 11 (1, 3, 5, 7, 9) then it should be 9 times greater.

Can someone explain this? thanks.
I think you might be crossing up harmonics and overtones. f1 is the first harmonic and other harmonics are found by mutliplying the harmonic number by the fundamental frequency, f1. Take the equation at face value.
 

traitorman

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i believe you are also confusing the equations.

for strings (whether it is open on one end or closed at both ends) and for pipes (open at both ends) the frequency is:

f = nv/2L where n = 1,2,3,4,5,6,7, etc

For a pipe that is closed on one end: f = nv/4L where n = 1,3,5,7, etc

If you use the f = nv/2L in your question, it works out.
 

htbruin

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i could understand if it was for a pipe closed at both ends, but shouldn't the fifth harmonic number be 9 for a pipe that is closed at one end though?

e.g. if the first harmonic frequency of a pipe closed at one end is 1hz then is the 2nd harmonic frequency 2hz? If it is,then why does n=1,3,5,7, ect? what's the point?

@traitorman,
i think for the string where one is tied down and the other end is not, the standing wave frequency is f=n(v/4L), not f=n(v/2L)
 

BerkReviewTeach

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i could understand if it was for a pipe closed at both ends, but shouldn't the fifth harmonic number be 9 for a pipe that is closed at one end though?

e.g. if the first harmonic frequency of a pipe closed at one end is 1hz then is the 2nd harmonic frequency 2hz? If it is,then why does n=1,3,5,7, ect? what's the point?
A pipe used to amplify selected sound waves must have at least one open end, which means that the term closed pipe refers to just one end of the pipe being closed. If the pipe were closed at both ends, a wave could not resonate out of it. To amplify sound, you need at least one end to be open, so that the wave can leave without losing a substantial amount of the amplitude to reflection.

I believe that the fifth allowed harmonic is f9, but as far as the question is worded, it sounds like they are referring to all harmonics, allowed or disallowed. It's an issue of convention.

If the fundamental harmonic is 1 Hz, then f2 would be 2Hz. However, if you draw that wave, you'll see that the reflected wave undoes the incident wave, so rather than being amplified it is canced out by destructiive interference. This happens to all even numbered harmonics, so we don't hear them. The term overtone describes the waves we hear, where the first overtone is the first harmonic after the fundamental that we can hear (the one that is allowed).

@traitorman,
i think for the string where one is tied down and the other end is not, the standing wave frequency is f=n(v/4L), not f=n(v/2L)
You are correct. For the string tied at one end, the restriction is a node at the tied end and an antinode at the untied end, so it should fit the same formula as the closed pipe (node and antinode at the repsective ends).
 

traitorman

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wow im a moron. thanks for clearing that up.

I'm also interested in what the answer would be then.

So, according to your explanation, if you say the #th harmonic, n = # regardless of the equation.

And if you say the #th overtone, then you consider whether the equation follows n=1,2,3,4,5 etc or n=1,3,5,7, etc?

Would this distinction be true for all similar questions?