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Phase Difference between Sinusoidal Waves

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Wave functions:

    [tex]y_1=2.00sin(20.0x-32.0t)[/tex]
    [tex]y_2=2.00sin(25.0x-40.0t)[/tex]

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

    3. The attempt at a solution

    My understanding is that the phase of [tex]Asin(kx-ωt)[/tex] is [tex]∅=kx-ωt[/tex]

    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 [tex]-\frac{1}{4}(20x-32t)[/tex]. 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 [tex]y_1+y_2=0[/tex] at this point. My attempts to solve that equation fails:

    [tex]y_1=-y_2[/tex]

    [tex]sin(20x-32t)=-sin(25x-40t)[/tex]

    I tried to do something with this property:
    [tex]sin(-u)=-sin(u)[/tex]

    [tex]-sin(20x-32t)=sin(32t-20x)[/tex]
    [tex]sin(20x-32t)=-sin(32t-20x)[/tex]

    I plug that into the [tex]y_1=-y_2[/tex] equation,

    [tex]-sin(32t-20x)=-sin(25x-40t)[/tex]

    So now [tex]32t-20x=25x-40t \pm 2n\pi[/tex]

    [tex]n=1,2,...[/tex]

    This doesn't seem to work after pluging in t=2 and solving for x.
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

    Simon Bridge

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    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
     
  4. Oct 30, 2011 #3

    NascentOxygen

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    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.
     
  5. Oct 30, 2011 #4

    Simon Bridge

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

    Note: the question only wants the solution for x is closest to 0.
     
  6. Oct 30, 2011 #5
    Ok, I see what my problem was. But first I'll explain the phase difference in terms of x and t is,

    [tex]-\frac{1}{4}(20x-32t)[/tex]

    This is the same thing you have, it reduces down to [tex]-5x+16[/tex] 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 [tex]\pm \frac{1}{4}(20x-32t)[/tex] 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 [tex]2\pi[/tex]. It should be [tex]\pi + 2n\pi[/tex] or [tex](2n+1)\pi[/tex] 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 [tex](20x-32t)-(25x-40t)[/tex] instead of [tex](25x-40t)-(20x-32t)[/tex]?
     
  7. Oct 31, 2011 #6

    NascentOxygen

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    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.
     
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