# Q of tuning fork; resonant and natural frequency

1. Apr 4, 2012

### Agent M27

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

Q=$\frac{\omega}{2\beta}$

$\omega$=$\sqrt{\omega_{0}^{2}-2\beta^{2}}$

3. 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?

2. Apr 4, 2012

### 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.