Displacement of Transverse Waves HELP

[grapejellypie]

Displacement of Transverse Waves HELP!

Homework Statement

At time t=0, the displacement of a transverse wave pulse is described by y=2/(x^(4) +1), with both x and y in cm. Write an expression for the wavefunction as a function of position x and time t if it is propagating in the positive x direction at 3.0 cm/s

Homework Equations

I'm not sure if this has to do with partial derivatives...and I don't quite understand partial derivatives.

The Attempt at a Solution

I know that v= 3.0 cm/s...

 

[gabbagabbahey]

Well your given a stationary wave pulse $f(x,0)=y(x)=\frac{2}{x^4+1}$ and you want to find f(x,t) that satisfies the wave equation $f_{xx}(x,t)-\frac{1}{v^2}f_{tt}(x,t)=0$. You probably know that any function of the form $f(x \pm vt)$ will satisfy the wave equation, and that if you want just the solution that travels forward at speed v, you choose the negative sign (i.e. $f(x - vt)$). You are given y(x), so what is y(x-vt)?

 

[grapejellypie]

what do you mean by fxx? is that the second derivative of x?

 

[gabbagabbahey]

Yes,

$$f_{xx}(x,t)=\frac{\partial ^2 f(x,t)}{\partial x^2}$$

 

[grapejellypie]

would y(x-vt) be [2/(x^(4) +1)] - 3.0 cm?

 

[gabbagabbahey]

No, just substitute x-vt everywhere you see an x.

 

[grapejellypie]

thank you so much for all of your help! i really, really appreciate it!

I'm sorry, but I have another question:

why do you set f{xx}(x,t)-1/v^(2) * f{tt}(x,t) = 0?

 

[gabbagabbahey]

You mean $f_{xx}(x,t)-\frac{1}{v^2}f_{tt}(x,t)=0$?

That's the one-dimensional wave equation; have you not seen it before?

Would it help if I wrote it like this:
$$\frac{\partial ^2 f(x,t)}{\partial x^2}-\frac{1}{v^2} \frac{\partial ^2 f(x,t)}{\partial t^2}=0$$

 

[grapejellypie]

i've seen the two second derivatives equal to each other, but i never thought of manipulating the equation to move the variables to one side.

thank you again for all of your help. i really appreciate it!

so just to double-check...the answer would be y(x,t) =2/[(x-3t)^(4)+1] ?

 

[gabbagabbahey]

Yes, you can check your answer yourself too by seeing what happens at t=0, you should get y(x) back. Also, you can take the partial second derivatives and verify that $f_{xx}(x,t)-\frac{1}{v^2}f_{tt}(x,t)=0$. You also know that the pulse should be traveling at 3cm/s to the right; which means that since the pulse is centered at x=0 for t=0, you should have a pulse that is centered at x=3cm for t=1s. These are checks that you should do to convince yourself that you have the correct answer.