# Homework Help: Wave on a wire

1. Mar 8, 2008

### ~christina~

[SOLVED] Wave on a wire

1. The problem statement, all variables and given/known data
A transverse wave on a taut wire has amplitude of 0.200mm and a frequency of 500Hz. It travels with a speed of 196m/s.

a) write an equation in SI units of the form $$y(x,t)= Asin( \omega t- kx)$$ for this wave

b)The linear mass density of this wire is 4.10g/m Find the Tension in the wire

c) what are the transverse velocity and the acceleration of the wave when x= 19.7m and t= 0.101s

2. Relevant equations
F= -kx
$$v= \omega/ k$$ ==> teacher gave me this equation but I can't find it in the book...is it valid?
$$f= 1/T= \omega/ 2 \pi$$

3. The attempt at a solution

a)
$$f= 1/T= \omega/ 2 \pi$$

$$T= 0.002s$$

$$v= \omega/ k$$

$$2 \pi (500Hz)= \omega$$

$$\omega= 3151.59rad/s$$

$$k= \omega / v$$

$$3141.59rad/s / 196m/s= 16.02$$ => I'm not sure it's suppsosed to be that large

I guess I'd just plug in the numbers but I'm not sure if the way I got the numbers is correct.

b) I don't know how to find this

c) I think I would just differentiate the original equaiton with the numbers included and then just plug in the values given and find the numbers.

I have a question though.

Is the transverse velocity always found through the the differential equation?

Thank you very much

2. Mar 8, 2008

### Kurdt

Staff Emeritus
Part a looks ok. For part b, the speed of a wave on a string with tension $T$ and mass per unit length $\rho$ is:

$$v= \sqrt{\frac{T}{\rho}}$$

For part c you would just differentiate wrt time.

3. Mar 8, 2008

### ~christina~

Oh okay.
Thanks alot Kurdt

4. Mar 9, 2008

### ~christina~

I have a Question about this don't I have to differentiate with respect to BOTH time and displacement?.. the problem said that I had to find the transverse veloicty and acceleration of the wave when x= 19.7m and t= 0.101s so how would I do this?

I know that I differentiate but I'm not sure how it would look if I differentiate with both x and t together...

I do know that if it's just t and theta then it would be

$$y(t)= A cos(\omega*t + \theta)$$

$$v(t)= y'(t)= -\omega A sin(\omega*t + \theta)$$

and

$$a(t)= v'(t)= -\omega^2 A cos(\omega t + \theta)$$

But what would it be with 2 variables??

would it be

[tex] y(x/t)= (8x -8a)((Asin (omega*t- kx)) ? => (for here sinc they start with sin
Thank you Kurdt

Last edited: Mar 9, 2008
5. Mar 9, 2008

### Kurdt

Staff Emeritus
You just differentiate wrt time. You treat the x variable as a constant when you do this. Then once you've differentiated you plug in the numbers.

6. Mar 9, 2008

### ~christina~

alright. but does this apply always when they ask you these questions? do I just differentiate partially and then plug in?

7. Mar 9, 2008

### Kurdt

Staff Emeritus
Yes, that is the only way to find the transverse velocity and acceleration when you're given the wavefunction.

8. Mar 9, 2008

### ~christina~

Thank you Kurdt