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## Homework Statement

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]

## 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.

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