Homework Help: Frequency of a String

1. Jul 6, 2008

crazyog

1. The problem statement, all variables and given/known data
While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
(a) What are the possible frequencies of the string?
(b) When she tightens the string slightly, she hears 3 beats/s What is the frequency of the string now?
(c) By what percentage should the piano tuner now change the tension in the string to bring it into tune?

2. Relevant equations
I'm not sure what equation to use for part b.
freq= n/2L

3. The attempt at a solution
For part (a) I found that the two possible frequencies are5 23 +/- 2
so 535 Hz or 521 Hz

for part (b) I'm not sure how to get an actual value
From what I understand...when the string is tightened its frequency increase since the beat frequency goes up, its becoming more out of tune.. (is this correct?)

(I think I need part (b) to do (c))

2. Jul 6, 2008

alphysicist

Hi crazyog,

I think you do part b the same way you did part a. It's just that since they tell you that she tightens the string, that will tell you whether to you plus or minus in your equation. What do you get?

3. Jul 7, 2008

crazyog

Oh ok so I did Freq of beat = absolute value of (F1 - F2)
so...
3 = (524- F2)
solve for F2 = 526 Hz

Then for c I used the equation 2*pi/T = 2*pi*freq
I solve for T for the freq at 526 and at 523...(is that correect)
and I get an answer like T = .00190 and T = .001912 respectively.

4. Jul 7, 2008

alphysicist

I think you might be misreading the equation; this T is the period not the tension. (Remember that tension has units of force.)

What equation relates the tension in the string to other things?

5. Jul 7, 2008

crazyog

Oh ok so I want to use f = v/wavelength which is also equal to (1/2L)(sqrt(T/mu)

can't I set up a ration like F2/F1 = sqrt(T2/T1)
I saw this in the book but I was not sure how they reduced this
then I et T2 = (523/526)^2T1
T2 =0.989T1

but you want to do 1 - 0.989 (right?) so 0.0114
then reduced by 1.14 percent?

6. Jul 7, 2008

alphysicist

That looks right to me.

They reduced the fraction to that, because once they formed a ratio the variables that were the same for F1 and F2 cancelled.