Simple Harmonic Motion - Finding the Phase

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jumbogala
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Homework Statement


Here is a position-time graph for a ball on a spring in SHM.

What is the phase of the particle at point a on the graph?

Homework Equations


phase = ωt + ɸ0

The Attempt at a Solution


First I found the initial phase, ɸ0. 0.5A / A = 0.5 = cos(ɸ0). Then ɸ0 = -π/3. It's negative because the particle is moving to the right.

Now for point a, I figured this is at 0.75T, where T is the period.This is kind of hard to see so it could be wrong. But then:

phase = (2π/T)(0.75T) - (π/3) = (7/6)π or 210 degrees. However, the answer is -120 degrees.

I also tried with points b and c but they were wrong too. Can anyone help?
 
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I'm not sure how to figure out the increase in phase directly... does it depend on knowing time still?

Because I'm confused how my assumption that t at a = 0.75T is wrong. The time from the first crest to the first trough is definitely 0.5T, right? And the time from 0 to a trough or crest is 0.25T. So the time from 0 to a should be 0.5(0.25T) = (1/8)T...
 
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x1 = 60
x2 = 0

x2 - x1 = 60

So... 360 degrees corresponds to one revolution (T), then 60/360 is the amount of time?

But that can't be right either, because (2pi / T)(60/360)T - (pi/3) = 0
 
Ah, I figured it out - I think this is right now:

So I need to solve the equation cos((2pi/T)t - (pi/3)) = -0.5

cos(x) = -0.5 when x = 120 and 240, 480 and 600, etc.

So the answer is 240 degrees (same as -120 degrees).

But if the question didn't show the diagram, how would you know which angle to use (120 or 240?)
 
Okay, that makes sense :)

Thanks for your help!