# Phase Difference between Sinusoidal Waves

1. Oct 30, 2011

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

Wave functions:

$$y_1=2.00sin(20.0x-32.0t)$$
$$y_2=2.00sin(25.0x-40.0t)$$

The units are centimeters and seconds. What is the positive x value closest to the origin for which the two phases differ by $$\pm\pi$$ at $$t=2.00 s$$

3. The attempt at a solution

My understanding is that the phase of $$Asin(kx-ωt)$$ is $$∅=kx-ωt$$

The phase in y2 is 5/4 the phase in y1 so the difference between the two is $$-\frac{1}{4}(20x-32t)$$. Setting this equal to pi and t=2.00 doesn't work (I don't get .0584 cm, the correct answer).

I'm also aware that $$y_1+y_2=0$$ at this point. My attempts to solve that equation fails:

$$y_1=-y_2$$

$$sin(20x-32t)=-sin(25x-40t)$$

I tried to do something with this property:
$$sin(-u)=-sin(u)$$

$$-sin(20x-32t)=sin(32t-20x)$$
$$sin(20x-32t)=-sin(32t-20x)$$

I plug that into the $$y_1=-y_2$$ equation,

$$-sin(32t-20x)=-sin(25x-40t)$$

So now $$32t-20x=25x-40t \pm 2n\pi$$

$$n=1,2,...$$

This doesn't seem to work after pluging in t=2 and solving for x.

Last edited: Oct 30, 2011
2. Oct 30, 2011

### Simon Bridge

The phase of the wave is the angle inside the sine function - you got that part OK:

P=kx-wt

Since the two waves have different frequencies, the phase-difference will change with time, which is why you are given a specific time to consider, t=2s.

P1=20x-64
P2=25x-80

So the phase difference is P2-P1

3. Oct 30, 2011

### Staff: Mentor

Your expression for phase difference isn't quite right.

Actually, there are an infinite number of solutions. So you should set the phase difference to (2n+1)Pi, where n is any +ve or -ve integer.

At t=2, we need to solve -5x +16 = (2n+1)Pi
Try a couple of values of n, to see what the smallest x value turns out to be (in truth, that closest to 0 is what the question is asking).

The fact that you posted the correct answer was very helpful, it allowed me to be confident that I was showing you the correct way. Otherwise, I would not have answered.

4. Oct 30, 2011

### Simon Bridge

Note: the question only wants the solution for x is closest to 0.

5. Oct 30, 2011

Ok, I see what my problem was. But first I'll explain the phase difference in terms of x and t is,

$$-\frac{1}{4}(20x-32t)$$

This is the same thing you have, it reduces down to $$-5x+16$$ after setting t=2. I wrote it the other way because I was trying to play with the equations to see which phase had a greater value in radians than the other by writing them with the same coefficients infront of the x and t variables. The phase of y2 is 5/4 the phase of y1. If you subtract, you get either $$\pm \frac{1}{4}(20x-32t)$$ depending on which phase you place in front. It looks like the negative one works, but I'm not comprehending why it works at the moment.

I was setting this equal to integral mulitples of $$2\pi$$. It should be $$\pi + 2n\pi$$ or $$(2n+1)\pi$$ as you say. It works for n=2, giving x=.0584 cm.

Looking back though, I still don't see how to decide which phase to subtract. Why take $$(20x-32t)-(25x-40t)$$ instead of $$(25x-40t)-(20x-32t)$$?

6. Oct 31, 2011

### Staff: Mentor

I suppose it makes no difference, so flip a coin. But I think I tried both, just to be sure the best x value I got was the closest possible to 0.