Simple Harmonic Motion - Finding the Phase

AI Thread Summary
The discussion revolves around determining the phase of a particle in simple harmonic motion (SHM) using a position-time graph. The initial phase was calculated as -π/3, but confusion arose regarding the timing of point a, initially assumed to be at 0.75T. After further analysis, it was clarified that the time from the start to the first crest is not 0.25T, leading to the conclusion that the correct phase at point a is 240 degrees, equivalent to -120 degrees. The importance of visual aids, like diagrams, was emphasized for accurately determining the direction of motion and corresponding phase angles. Understanding the relationship between phase and time in SHM is crucial for solving such problems effectively.
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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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Your assumption that ta=0.75 T is wrong. What's the increase in phase going from the very left to the first maximum? Similarly, what's the increase in phase from the first minimum to point a?
 
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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The time from 0 to the first crest is not 0.25 T. Think about the unit circle. What angles gives you cos x1=0.5 and cos x2=1? What's the difference x2-x1? How much time does that increase correspond to?
 
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
 
That wasn't meant to give you the answer directly. It was to get you to see that the time from t=0 (-60 degrees) to get to the crest (0 degrees) isn't 0.25 T as you were assuming.
 
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?)
 
You wouldn't be able to because without additional information, like the diagram, you don't know in which direction the particle is moving.
 
Okay, that makes sense :)

Thanks for your help!
 
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