# Homework Help: Wave Interference concept help

1. Feb 3, 2012

### QuarkCharmer

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

2. Relevant equations
$$y_{3} = y_{1}+y_{2}$$
$$f(x,t) = y_{m}sin(kx-ωt+\phi)$$

3. The attempt at a solution
So, I'm going back over a proof done in my physics course, where the sum of two waves equals the resulting wave. The proof was on constructive and destructive interference. I'm having trouble figuring out what was done on a certain step. My notes are verbatim from the lecture.

If some wave Y2 differs from Y1 only by a phase shift,
$$y_{1} = Y_{m}sin(kx-ωt)$$
$$y_{2} = Y_{m}sin(kx-ωt+\phi)$$

The resultant wave Y3 is said to be the sum of Y1 and Y2. Now, in this proof the professor applied a trigonometric identity to convert the resultant wave into one that showed it's oscillatory and amplitory components. It reads:

$$Y_{3} = Y_{m}sin(kx-ωt) + Y_{m}sin(kx-ωt+\phi)$$
$$Y_{3} = Y_{m}(sin(kx-ωt) + Y_{m}sin(kx-ωt+\phi))$$

Now using the identity:
$$sin(A) + sin(B) = 2sin(\frac{A+B}{2})cos(\frac{A-B}{2})$$

$$Y_{3} = Y_{m}2sin(\frac{kx+ωt+kx-ωt+\phi}{2})cos(\frac{kx+ωt-kx+ωt-\phi}{2})$$

Which then simplifies down to:
$$Y_{3} = 2Y_{m}sin(\frac{2kx+\phi}{2})cos(\frac{2ωt-\phi}{2})$$

The point was to show that two waves traveling along the same medium differing by a phase difference produce some resulting wave, but I have no idea how he simplified the equation down. Since both waves are traveling in the same direction, the phase of both of them should be (kx-ωt+phi), which shows the wave moving in the positive x-direction (It just so happens that the first wave Y1 has a phi of 0 for this example). But, in the step he simplified down to, the algebra just doesn't work out for me. It looks as though he chose one of the waves to be moving in the positive x-dir and the other in the negative x-dir. If that's the case then sure, it simplifies down. Is there some magic algebra step that I am missing here? Why did he reverse the direction of one of the waves so that the (A+B) from when the trig identity is applied allowed for it all to cancel out nicely?

Thanks,
QC

2. Feb 3, 2012

### vela

Staff Emeritus
Looks like he mysteriously changed the direction of one of the waves.

Perhaps he had moved on to the subject of standing waves and you didn't catch that.

3. Feb 3, 2012

### QuarkCharmer

Well, we were working on standing waves, so I guess Y2 could be the wave reflecting back, but in this example he CLEARLY defined the waves as traveling in the same direction, citing an example in communication, wave jamming and so on. I have the whole thing on livescribe.

So, if I did the same derivation, allowing both waves to travel in the same positive x-direction. I should still get the same result right? Something of the form:
$$Acos()sin()$$
Where the sine represents the oscillation, and the cos modifies the amplitude correct?

4. Feb 3, 2012

### vela

Staff Emeritus
Yes, you'll get a similar but qualitatively different result.

5. Feb 3, 2012

### QuarkCharmer

Thanks vela,

I'll look into both results in more depth and I guess ask him what the heck was going on in that proof. I just didn't want to go in there and be like "Hey is this a mistake or what", and it turn out that it was simply my misunderstanding of the subject.

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