# Homework Help: Wave functions

1. Sep 18, 2010

### dirk_mec1

1. The problem statement, all variables and given/known data
http://img685.imageshack.us/img685/5585/63862334.png [Broken]

2. Relevant equations
-

3. The attempt at a solution
$$y_1(0,t)=y_2(0,t) \longrightarrow 1+\frac{B}{A}e^{2i \omega t} = \frac{C}{A}$$

$$y_1_x(0,t)=y_2_x(0,t) \longrightarrow 1+\frac{B}{A}e^{2i \omega t} =\frac{k_2}{k_1} \frac{C}{A}$$

but I can not find an explicit expression for C/A and B/A expressed in v1 and v2 I can only find that k2/k1 =1. What am I doing wrong?

Last edited by a moderator: May 4, 2017
2. Sep 18, 2010

### vela

Staff Emeritus
If k1=k2, the strings are identical, and you get no reflection.

Your wave function in region 1 looks a bit off. Typically, the second term is B e-i(k1+ωt). That still describes a wave moving to the left, but it lets the time dependence cancel out of the equations. Plus it'll result in equations you can actually solve.

3. Sep 18, 2010

### dirk_mec1

You left out the x-dependence did you did this on purpose? Furthermore, the problem lies in the fact of the presence of the plus sign at $$\omega t$$

But your conclusion is that there's a mistake in the question?

4. Sep 19, 2010

### vela

Staff Emeritus
Sorry, that was a typo. There should be an x in there.

It appears there's a mistake in the question. You either get k1=k2, which means there should be no reflection, or C=0, which means there's no transmission.

5. Sep 19, 2010

### dirk_mec1

I've send an e-mail to the instructor. Thanks for your help so far.

6. Sep 20, 2010

### dirk_mec1

The instructor repied back there isn't any mistake in the exercise, it is possible! But I don't know how!

7. Sep 20, 2010

### vela

Staff Emeritus
Well, if you take the problem as given, the only consistent solution I see is B/A=0 and C/A=1, which in turn requires k1=k2. I've asked others to take a look at this thread. Perhaps they'll spot something we're both overlooking.

8. Sep 20, 2010

### dirk_mec1

The instructor gave the answer if you take the real part of both equations you'll get a set of 2 equations which indeed lead to the right answer.

9. Sep 20, 2010

### kuruman

Here is my second opinion, for whatever it's worth. If C and A are constants, then the above expression says they are not because they are functions of time. The problem as stated implies that you may match boundary conditions at a specific time, but not at all times. To fix that, the correct form for the second term in y1(x,t) ought to be B*Exp[i(-kx-ωt)] which is still a wave traveling to the left. However, in this form, if you match boundary conditions at one time, you match them at all times because the e-iωt drop out.

10. Sep 20, 2010

### vela

Staff Emeritus
It may lead to the [STRIKE]right[/STRIKE] intended answer, but the method isn't correct in my opinion. You can't just throw out the imaginary part of the equations because it's inconvenient.

Last edited: Sep 20, 2010
11. Sep 20, 2010

### gabbagabbahey

There has to be a typo. If $y_1=Ae^{i(k_1x-\omega t)}+Be^{i(k_1x+\omega t)}$ and $y_2=Ce^{i(k_2x-\omega t)}$ then the only way the two function can be equal at x=0 at all times (It wouldn't make sense for them to only be equal at one time unless the two strings were disconnected) is if $Ae^{-i\omega t}+Be^{i\omega t}=Ce^{-i\omega t}$ which tell you immediately B must be zero, and hence there is no reflected wave, and C=A.

If on the other hand $y_1=Ae^{i(k_1x-\omega t)}+Be^{-i(k_1x+\omega t)}$ , you don't have that problem, and there is a sensible solution.

12. Sep 20, 2010

### ehild

I think the same. Both the travelling and the reflected waves must have the same frequency, and omega is not the same as (-omega), although cos(wt) = cos(-wt).

ehild

13. Sep 21, 2010

### dirk_mec1

He told us that the real part has a physical meaning therefore only the real part of the functions have to be considered. But I don't get what you mean by the typo. The imaginary part of the function is irrelevant, isn't it?

14. Sep 21, 2010

### dirk_mec1

You forgot the second equations which is the derative.

15. Sep 21, 2010

### gabbagabbahey

That only makes things worse. The only way to also satisfy that condition is if $k_1=k_2$.....which says that the wave must travel with the same speed in both strings!

Considering only the real part will give you the correct answer since $\cos(-\omega t)=\cos(\omega t)$, however it makes no sense to consider waves as the real part of complex wavefunctions if the imaginary parts of the complex wavefunctions you choose don't satisfy the same equations as the real parts. Doing so means you have to work exclusivly with the real parts, which defeats the entire purpose of using complex exponentials (they are often easier to work with than sines and cosines) in the first place.

If, on the other hand, you choose your complex waverfunction so that both imaginary and real parts can satisfy the boundary conditions, you are free to do all the calculations with the complex wavefunctions and then simply take the real part at the very end to get your physical result.

16. Sep 21, 2010

### dirk_mec1

That sounds plausible. But what is the meaning of the complex part of the wave function? If I want to draw a picture I can only do so if I take the real part, agreed?