# Wave speed of a transversing guitar string?

1. Apr 10, 2010

### magician13134

Hi, I'm completely stuck on a homework question and I really don't even know where to start...

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.382 m. The maximum transverse acceleration of a point at the middle of the segment is 9000 m/s2 and the maximum transverse velocity is 3.90 m/s.

Part (a) asked "What is the amplitude of this standing wave?" and I was able to correctly get A=1.69×10−3m

Now part (b) is asking "What is the wave speed for the transverse traveling waves on this string?" and I don't even know what equations to use. If someone could just point me in the right direction, tell me a useful equation, or give me a little hint, that would be great :)

2. Apr 10, 2010

### AtticusFinch

$$\frac{a_{max}}{v_{max}} = \omega$$

$$\frac{v_{max}}{\omega} = A$$

You know that $$v = \lambda f$$ or alternatively $$v = \frac{2\pi f}{\frac{2\pi}{\lambda}} = \frac{\omega}{k}$$

You know omega from a) and you can figure out k easily so that's what you should do for b).

3. Apr 10, 2010

### magician13134

Ok, so I'm trying to work that out, but I still can't seem to get the right answer.

So I know $$\omega = 9000/3.9 = 2308$$ and that $$v = \frac{\omega}{\frac{2\pi}{\lambda}}$$ and... I think $$\lambda = 4L$$, right? So I got 561.2, but apparently that's not right. Did I mess up with $$\lambda = 4L$$ (using 0.382 for L) or is there something else I'm missing?

Oh, and thanks!

Oh shoot, $$\lambda = 4L$$ is only for tubes, isn't it... Shoot. Ok, I'll keep working at it.

4. Apr 11, 2010

### AtticusFinch

Yes it's only for tubes. You need to find what the wavelength would be for a string with nodes at both ends (just draw a wave that only has two nodes).

5. Apr 11, 2010

### magician13134

It was $$\lambda = 2L$$, so the answer was just 281

6. Apr 11, 2010

### AtticusFinch

Yep, good job.