# Beat frequency problem

1. Oct 31, 2014

### toothpaste666

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

A violin string which is 30 cm long is tuned using a 440 Hz reference tone.

A) What is the wavelength fundamental mode of the string?

B) When the string has a tension of 34 N a violinist hears 5.7 beats per second. What are the frequencies at which the string might be vibrating?

C) The tension is increased and the violinist hears 3.8 beats per second. What is the original frequency of the string?
D) What should the tension of the string be so that the fundamental frequency is 440 Hz

2. Relevant equations
beatf = f2-f1
f= v/2l
lambda = 2l

3. The attempt at a solution
A) the fundamental wavelength is twice the length of the string. 2 x 30 cm = 60 cm = .6m
B) the difference between the 440 Hz reference tone and the frequency of the violin is 5.7 so the string is either vibrating at 434.3 or 445.7
C) This is where I am stuck. this is my attempt. the beat frequency is 3.8 and the reference tone is 440 so the frequency is now either 443.8 or 436.2 I completely lost on what to do from here though

2. Oct 31, 2014

### Staff: Mentor

(C) Does the frequency increase or decrease if you increase tension?

3. Oct 31, 2014

### toothpaste666

well the velocity of a wave is v=sqrt(FT/u) where FT is the tension and u is the linear density. also v = lambda * f where lambda is the wavelength and f is the frequencey. therefore
lambda * f = sqrt(FT/u)
f = sqrt(FT/u)/lambda
so if the tension was increased the frequency would increase i think

4. Oct 31, 2014

### Staff: Mentor

Right. Now you can go back to the frequencies calculated in (B) and see if they are both possible with the additional knowlege from (C).

5. Oct 31, 2014

### toothpaste666

ahh it cant be 445.7 in part B because it is higher than both the new frequencies and the frequency should have increased if the tension was increased. so part C is 434.3?

6. Oct 31, 2014

### toothpaste666

so that would mean for part D)
at 34 N of tension:
f*lambda = sqrt(FT/u)
434.3 * .6 = sqrt(34/u)
260.58 = sqrt(34/u)
67902 = 34/u
u= 34/67902 = 5x10^-4 where u is the linear density

if we want f = 440
FT = u (lambda * f)^2 = (5x10^-4) (.6 * 440)^2 = 34.8

is this correct?

7. Nov 1, 2014

### Staff: Mentor

Units are missing.
Apart from that, it looks good.

8. Nov 1, 2014

### toothpaste666

oh sorry. 34.8 N thanks for your help!