Phase Difference between Sinusoidal Waves

[AdkinsJr]

Homework Statement

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

The Attempt at a Solution

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

I read a response in this thread too;
https://www.physicsforums.com/showthread.php?t=165354

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.

 

[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

 

[NascentOxygen]

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).

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[/color]
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.

 

[Simon Bridge]

At t=2, we need to solve -5x +16 = (2n+1)Pi

How did you get the minus sign in -5x ?

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

 

[AdkinsJr]

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[/color]
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.

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)?

 

[NascentOxygen]

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

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.