Q of tuning fork; resonant and natural frequency

[Agent M27]

Homework Statement

A musicians tuning fork rings at A above middle C, 440Hz. A sound meter level meter indicates the sound intensity decreases by a factor of 5 in 4 seconds. What is Q of the tuning fork?

Homework Equations

Q=[itex]\frac{\omega}{2\beta}[/itex]

[itex]\omega[/itex]=[itex]\sqrt{\omega_{0}^{2}-2\beta^{2}}[/itex]

The Attempt at a Solution

I understand how to solve this problem, I had the right answer if I chose the resonant frequency correctly to be 440Hz. Instead I understood the problem to be giving me the natural frequency since after a certain amount of time it equilibrated there, so to speak. Using this idea I solved for the resonant frequency and I got an answer on the order of 1000, which agrees with what my professor said in class, "the Q of a tuning fork is roughly 1000". However he took 440Hz as the resonant frequency and obtained a Q of roughly 7000. His answer was "because that is its resonant frequency" I kid you not, so I guess my understanding of the difference between the two is incorrect. So why is 440Hz the resonant frequency in this problem?

 

[rude man]

The tuning fork frequency does not change with time. If it starts at 440 it "ends" at 440. (I put "ends" in quot. marks since you are going after a linear solution which theoretically has no "end").

You need the expression for the time decay of the acoustic pressure. That decay is a function of Q.