Phase of the following Waves

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Homework Statement



See figure attached for problem statement.

Homework Equations





The Attempt at a Solution



How does one go about solving a problem like this? I don't see how they get actual values from just the photo.

Can someone explain?
 

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Answers and Replies

  • #2
collinsmark
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How does one go about solving a problem like this? I don't see how they get actual values from just the photo.

Can someone explain?
Look at the y-intercept. Each wave crosses the y-axis exactly halfway between the origin and the wave's maximum amplitude. What sort of "angle" causes the sine function to be half its maximum amplitude? :wink:
 
  • #3
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Look at the y-intercept. Each wave crosses the y-axis exactly halfway between the origin and the wave's maximum amplitude. What sort of "angle" causes the sine function to be half its maximum amplitude? :wink:

I know the angle you're talking about is 30 degrees.

When I look at it, it looks like the first wave has a phase of -45 degrees and the second a phase of -135 degrees, but those options aren't present...

Because at 45 degrees it would be half way to reaching its maximum value at 90 degrees.
 
  • #4
SammyS
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I know the angle you're talking about is 30 degrees.

When I look at it, it looks like the first wave has a phase of -45 degrees and the second a phase of -135 degrees, but those options aren't present...

Because at 45 degrees it would be half way to reaching its maximum value at 90 degrees.
Look at the distance (on the x-axis) from the x-intercept to the x value of the maximum. Then compare that and where the y-intercept is (x=0). Looks to me like x=0 is about 1/3 the distance from the x-intercept to x of the max. 45° is halfway from the x-intercept to where Sine has it's max. The value of sine(45°) = 1/√(2) ≈ 0.707. The y-intercept does not look (to me) like it's close to being 70% of the max value.

If collinsmark is talking about 30°, then he knows what he's talking about! Where else is sin(θ) = 1/2 ?
 

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