# Standing Waves (tension vs. wavelength)

1. Dec 11, 2007

### furiouspoodle

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

I'm trying to decode a graph of Tension vs. $$\lambda^2$$ of standing waves on a string to understand the actual meaning of the slope. I also need to derive the equation for the line from two expressions:

$$v= \sqrt{T/\mu}$$
$$v= f\lambda$$

2. Relevant equations

$$\lambda=2L/n$$
$$\mu = m/L$$

3. The attempt at a solution

I'm starting to think I'm completely missing something in my initial approach.

$$v = \sqrt {T/\mu}$$

$$v^2 = T/\mu$$

$$v = f\lambda$$

$$v^2 = (f\lambda)^2$$

$$T/\mu = f^2\lambda^2$$

$$T = \mu f^2\lambda^2$$

$$T = \frac {\mu f^2 4L^2}{n^2}$$

I feel like it should lead to a clearer solution than this one, but I'm not sure what. I first thought that the slope was simply a surface tension (dynes/cm^2) on the string but I don't see how that relates to the equation or slope I'm trying to find.

I also tried this:

$$f = \frac {v}{\lambda}$$

$$f = \frac {\sqrt {T/\mu}}{\lambda}$$

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

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

which seems closer, but I don't know how to apply it in a usable manner. Any help would be appreciated.