Calculating the mass per unit length of a string based on the graph of f vs. 1/L

[buttermellow]

Homework Statement

The frequency of a vibrating string is set to 15, 20, 25, 30, or 35 Hz and the length needed to attain a standing wave (mode 1) is recorded. A graph of frequency versus 1/L is recorded. Calculate the mass per unit length of the string.

The resulting graph has the equation y=9.15x + 1.03

Homework Equations

f= m/2L x (sqrt(T/mu))

m=1
T=tension= 1470 g m/s²
L=1m

The Attempt at a Solution

I assumed it had something to do with the slope, which would be equal to 1/2L x sqrt(T/mu). That doesn't make solving for mu any easier though, so what's the point? If I set 9.15 equal to this, mu comes out to be 1.09 g/m, is this right? Gah, I'm so confused!

 

[kuruman]

Homework Statement

The frequency of a vibrating string is set to 15, 20, 25, 30, or 35 Hz and the length needed to attain a standing wave (mode 1) is recorded. A graph of frequency versus 1/L is recorded. Calculate the mass per unit length of the string.

Are you sure you quoted the question correctly? What does "A graph of frequency versus 1/L is recorded" mean when L = 1 m = constant? What is your independent variable?

 

[buttermellow]

The question is from a lab, and yes, it was worded poorly.

The frequency was set and the length changed until a standing wave at the first mode was attained. After doing that for each frequency, a graph of frequency vs. 1/L was constructed (not recorded, as I said earlier). From this graph (and I assume related equations) we are supposed to find the mass per unit length of the string.

 

[kuruman]

For the first harmonic (fundamental) λ = 2L so that v = f(2L) = sqrt(T/μ) which gives

f = (1/2L)sqrt(T/μ) (there is no extra m multiplying the square root).

Let f = y and x = (1/L). Then

y = (1/2)sqrt(T/μ)*x

This says that if you plot y (a.k.a. f) vs. x (a.k.a. 1/L) you should get a straight line that passes through zero and has a slope equal to (1/2)sqrt(T/μ). So if you know the slope, you can find the linear mass density from

slope = (1/2)sqrt(T/μ)

Try fitting a straight line by constraining the intercept to be zero. If it doesn't work, then you will have to explain what the intercept means physically in terms of what you did in the lab.

Finally, I am not sure what you mean by "T=tension= 1470 g m/s²". To get a result that makes sense, you need to express the tension in Newtons.