Fundamental frequencies and temperature

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Homework Help Overview

The discussion revolves around determining the temperature at which a specific fundamental frequency can be produced by an organ pipe that is closed at one end. The problem involves understanding the relationship between frequency, temperature, and the speed of sound in air.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between frequency and temperature, with some attempting to derive a formula based on the speed of sound. There is a focus on how changes in temperature affect the frequency of sound produced by the organ pipe.

Discussion Status

Participants are engaged in clarifying the assumptions behind the problem, particularly regarding the relationship between the fundamental frequency and temperature. There is an ongoing debate about the correctness of certain assumptions and calculations, with some guidance being offered to reconsider the approach taken.

Contextual Notes

There is mention of ignoring the change in length of the pipe due to temperature variations, which may affect the overall understanding of the problem. Additionally, the relationship between the fundamental frequency and the harmonics is being questioned, indicating potential misunderstandings in the setup.

Fallen Seraph
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Homework Statement


The frequency of the note f_4 is f_F.
If an organ pipe is open at one end and closed at the other, what length must it have for its fundamental mode to produce this note at a temperature of T?

ans= v/(4*f_F), where v is the speed of sound in air.

Now the part which troubles me:

At what air temperature will the frequency be f? (Ignore the change in length of the pipe due to the temperature change.)

Homework Equations



(fundamental frequency)=v/4L where L is the length of a closed pipe.
and

v=k*sqrt(T) where k is a constant (suitable for this situation, nothing else is known/varying)

The Attempt at a Solution


My initial reaction was to say that since the wavelength in the closed pipe must remain the same (4*L) the frequency varies linearly with speed of sound. So for frequency to the divided by four, so must speed, for speed to be divided by four temperature must be divided by 4^2=16, so my answer was T/16. But I'm being told that this is wrong, and that the answer includes the variable "f".

Any help?
 
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It is asking for a temperature. I'm not sure what you've done so far but you should start from here;

[tex]F_{0}=\frac{k\sqrt{T}}{4L}[/tex]
 
Hmmmm... but surely that equation will just return the answer t/16?
 
Fallen Seraph said:
Hmmmm... but surely that equation will just return the answer t/16?
Reread the question. You want to get something of the form of T = ...
 
Fallen Seraph said:
But I'm being told that this is wrong, and that the answer includes the variable "f".
I think you're either misunderstanding this part of the question, or not paying close enough attention with the calculation.

Let me restate the question in a form that might help: If the fundamantal frequency is f_F at temperature T, then at what temperature is the fundamental frequency = f?
 
I'm afraid I'm really missing something here. I really cannot see what is wrong with this as a solution:

[tex]F(x)=\frac{k\sqrt{x}}{4L}[/tex] Where x is the temperature, F is the fundamental frequency.

For f= [tex]f_F[/tex] we have x = T.

so [tex]F(T)=\frac{k\sqrt{T}}{4L} = f_F[/tex]

we want to find y, where F(y)=f. We know that [tex]\frac{f_F}{4} = f[/tex]

so F(y)=F(T)= [tex]\frac{k\sqrt{y}}{4L} = \frac{k\sqrt{T}}{16L}[/tex]

Which eventually gives y = T/16. I know it's a bit long winded for what it's actually doing, but I can't see what's wrong with it...
 
Fallen Seraph said:
We know that [tex]\frac{f_F}{4} = f[/tex]
And how do we know this?
 
well I assumed that if f_F was f_4, then it's the fourth harmonic of f, so f_4=4f. Is this an incorrect assumption?
 
gah. so it is. that's a bit frustrating... ah well. thanks a lot for the help and time.
 

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