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Fundamental frequencies and temperature

1. 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.)

2. 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)

3. 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?
 

Hootenanny

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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?
 

Hootenanny

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Gokul43201

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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...
 
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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