# Homework Help: Wavefunction problems

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1. Aug 18, 2016

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

I found a couple of assignements for a physics course I will take later this year- so I started looking into them a bit in advance. It concerns wavefunctions. I'm a bit rusty on my trigonometric identities So I would love if someone could try to help me solve these two questions:

1) $g(x±vt)$ and $h(x±vt)$ are differentiable twice. Show that $y(x,t)= Ag(x±vt) + Bh(x±vt)$ is a solution to the wave equation where $A$ and $B$ are constants and $v$ is the speed of the wave.

2) Which of these functions $(y(x,t))$ is/are a solution(s) to the wave equation. If so- which direction does the wave propogate?

a) $y = A(x^2-v^2t^2+2vtx)$
b) $y = A(x^2+v^2t^2+2vtx)$
c) $y= Acos(kx-ωt)$
d) $y = Acos(kx-ωt)- Acos(kx+ωt)$

2. Relevant equations
The wave equation looks like this: $\frac{{\partial ^2 y}}{{\partial x^2 }} = \frac{1}{{v^2 }}\frac{{\partial ^2 y}}{{\partial t^2 }}$ If the functions in $2$ satisfy the wave equation then both sides should be equal.

I know if I get something that has the form (kx-ωt) in the function it will propogate in the positive x-direction. And (kx+ωt) then it will propogate in the negative x-direction.

3. The attempt at a solution

The only one I know for certain to be a solution to the wave equation is c, and that would travel in the positive x-direction. I'm pretty sure that that's not the only one, so how would one go about solving these kinds of questions? I'm hoping to get a good grade when this course is due so I want to get started soon since my trigonometric knowledge is a bit rusty. The first question is confusing for me. Useful tips and/or links for reading/videos/lectures are welcome!

2. Aug 18, 2016

### Ray Vickson

For each of the four $y(x,t)$ forms you are given, substitute that form into the wave equation to see if it "works".

If you are rusty on the trig functions, now would be a good time to go to the internet and search for articles, tutorials, etc., that cover the needed material. Alternatively, get an actual book and look at the relevant parts.

3. Aug 18, 2016

For the first two in $2$ I don't see how they would work out. The second two I can solve the wave equation with (using that omega=2*pi*f, v=lambda*f and k=2*pi/lambda). But for (d) I can't figure out the direction of the wave. How do tackle that?

And regarding the first question- should I look into "the general solution" of a wave function?

4. Aug 18, 2016

### Ray Vickson

In the case of the first two, have you actually substituted the $y$ into the wave equation to check it? This is not just a matter of "not seeing how it would work out"; it is a matter of going through all the details.

5. Aug 18, 2016

I messed up with the chainrule on that one obviously. I get $a$ to be false [2A=(-2)A] but $b$ to be correct. So $b$ is traveling in the negative x-direction.

I also took the second partial derivatives in $1$ and the partial derivative with respect to $x$ is $Ag''(x-vt)+Bh''(x-vt)$ and the partial derivative with respect to $t$ to be $v^2(Ag''(x-vt)+Bh''(x-vt))$ which substituted into the wave equation makes equality and thus is a solution.

$(STUFF)=1/v^2*v^2*(SAME STUFF)=1$

6. Aug 18, 2016

Can someone please help me figure out how to tell the direction of the wave in $2d$?

7. Aug 18, 2016

### Ray Vickson

How do you know it is moving at all?

8. Aug 18, 2016

That's the thing. When I think of what comes to mind intuitively is that it might cancel out completely or just become a permanent standing wave (bad expression maybe). I know that cos(-theta)=cos(theta) but then there's the negative sign in front of it all. Does this flip it about the y-axis to cancel it out? Is my intuition way wrong?

9. Aug 18, 2016

### Ray Vickson

You have it exactly right; it is a standing wave---and that is the official term!

The only distinction between
$$y_1(x,t) = \sin(kx - \omega t) + \sin(kx + \omega t)$$
and
$$y_2(x,t) = \sin(kx - \omega t) - \sin(kx + \omega t)$$
is a phase difference:

$$\begin{array}{rcr} y_1(x,t) &=& 2 \sin(kx) \cos(\omega t) \\ y_2(x,t) &=& -2 \cos(kx) \sin(\omega t) \end{array}$$