# Wavelength^2 vs. Tension (graph/conceptual)

1. Apr 17, 2010

### anti404

hi,
in our lab last Thursday we were doing standing waves on a string attached to a pulley and vibrator(at f=120Hz), and we produced loops by creating a tension force in the string.
by calculating the wavelength(2*[distance from node to node]/#of loops), and the tension force(mass added to the end of the string*g), we are supposed to create a plot wavelength^2 vs. T, and then find the slope of the line, and use that to compare our experimental frequency vs. the known frequency.
however, I have no idea what quantity the slope of the line of a v^2/T graph represents. the units would be, uh, m/kg*s, which doesn't really help me, either. =/
any help would be much appreciated, thanks!

2. Apr 18, 2010

### Stonebridge

The speed of the wave on the string is given by the formula
v= √(T/μ)
where T is the tension in the string, and μ is its mass per unit length.
The speed of a wave is also given by v=fλ
If you eliminate v between those two equations you will get a formula that relates the wavelength and the tension.
This should tell you why a graph of wavelength squared against tension could be useful.

3. Apr 18, 2010

### anti404

so fλ=√(T/μ), or λ^2=(T/μ)*1/f?.. does that mean the slope of the line=1/f?
sorry, I still don't really understand this... =/.

4. Apr 18, 2010

### RoyalCat

You've made just a small mistake.

$$\lambda ^2 =\frac{1}{\mu f^2} T$$

Since you're plotting the wavelength squared against the tension, your slope is $$\frac{\Delta (\lambda ^2)}{\Delta T}= \frac{1}{\mu f^2}$$

Since you know the frequency to a certain degree of accuracy, you can find the linear mass density from the slope, and compare it with an independent measurement (Weighing the rope and measuring its length).

Last edited: Apr 18, 2010
5. Apr 18, 2010

### Stonebridge

It gives λ²=(T/μ).(1/f²) You forgot to square the f.

If you compare that with the equation of a straight line y=mx+c [m is the gradient]
then for a graph of λ² against T, the gradient is (1/μ)(1/f²)

6. Apr 18, 2010

### anti404

so basically, m(slope)=1/μf², or f(experimental)=$$\sqrt{1/mu*m}$$
if so(and even if not), awesome, guys! thanks a lot. =]