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**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?