Waves and optics pertaining to sinusoidal waves

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1. Sep 29, 2014

jweie29nh

1. The problem statement, all variables and given/known data
https://us-mg5.mail.yahoo.com/neo/launch?.rand=768lpb97vo87e#4636076903 The picture displayed shows three graphs and in each problem the objective is to find the phase constant. I'm having an issue attempting any form of strategy to solve these problems. Any help would be greatly appreciated thanks.

2. Relevant equations
For the first graph D(x=0,t=0)=asin(kx-wt+phase constant)
1=2sin(0-0+phase constant)
phase constant = 30 degrees

For the second graph
1=2sin(0-0+phase constant)
phase constant = 30 degrees

For the third graph
1=2sin((2pi*4)/4 - (2pi)(T/2)/4+phase constant)

Don't have an idea how to solve for the third graph

3. The attempt at a solution
1. For the first graph I answered 30 degrees
2. For the second graph I answered 30 degrees

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2. Sep 29, 2014

BvU

Your answers would imply that the two pictures both represent $D = 2 \sin(kx-\omega t + \pi/6)$ and are therefore exactly the same.
Clearly, they are not. Perhaps the equation $\sin\phi = 1/2$ has another solution ?

3. Sep 29, 2014

jweie29nh

I agree there are two different solutions however I'm not sure which problem has the solution pi/6 and why it would be considered an answer over the other problem.

4. Sep 29, 2014

BvU

Did you read the 'Note' ? (I have to ask, since in your 1 and 2 answers you do have the same $\phi_0$)

5. Sep 29, 2014

BvU

From another angle :): In the first picture, is D increasing or decreasing when you take a small step in time ?

6. Sep 29, 2014

jweie29nh

Yes I did read your note, and I just read your second one now. It seems much more clearer and I have a better understanding. So the first graph has a phase constant of pi/6 since d is increasing while in the second graph d is decreasing.

7. Sep 29, 2014

jweie29nh

So in that case the second graph would indicate the answer would be 5pi/6

8. Sep 29, 2014

BvU

:)I didn't mean my note, but the note under the exercise!

Then: if you make a little time step in the first picture, does D really increase ?

Because, if I fill in x = 0, I get $D = 2 \sin(-\omega t + \pi/6)$ and what is the time derivative $d\;D\over dt$ at time zero for that ?

9. Sep 30, 2014

jweie29nh

The derivative of time for that equation would be D=-2wcos(−ωt+π/6) but what does this equation have to do with the problem?

I suppose time doesn't increase or decrease since it is following a pattern.

10. Sep 30, 2014

BvU

Time is running. The picture is a snapshot at t=0. There is even a big fat arrow showing which way the wave is going.
So what is this derivative at t=0 ? Is it positive or negative ? This derivative has everything to do with the problem: to determine the phase angle, you have to know if the D is increasing or decreasing.

Look at it this way: If you are sitting on your surfboard at x=0 on t=0 and this huge wave moves as is indicated by the arrow, are you going up or down ?

11. Sep 30, 2014

jweie29nh

-pi/6 is the derivative at t=0, so d is decreasing. It is going down

12. Oct 1, 2014

BvU

And in the second picture, same procedure ?

13. Oct 1, 2014

jweie29nh

I believe the same method is applicable but your guidance is widely accepted

14. Oct 1, 2014

BvU

:) The idea is that you are now in a good position to handle the second picture by yourself ! And yes, the approach is the same.
You see that $D(0,0) = 2 \sin\phi_0 = 1$ which has two solutions (*). Which one is the right one, you decide by looking at the time derivative of D at x=0, t=0.

(*) Actually, it has infinitely many solutions, but only two are in the range $[0,2\pi]$