Simple Harmonic Motion - Finding the Phase

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Homework Help Overview

The discussion revolves around determining the phase of a particle in simple harmonic motion (SHM) based on a position-time graph. The original poster attempts to calculate the phase at a specific point on the graph but encounters discrepancies in their reasoning and assumptions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between time and phase, questioning the original poster's assumption regarding the timing of point a. They discuss the increase in phase corresponding to different points on the graph and the implications of the unit circle in understanding phase angles.

Discussion Status

There is an ongoing exploration of the phase calculation, with participants providing insights and corrections to the original poster's assumptions. Some guidance has been offered regarding the relationship between time intervals and phase changes, but no consensus has been reached on a definitive method without the accompanying diagram.

Contextual Notes

Participants note the importance of the graph for determining the correct phase angle and the potential ambiguity in the absence of visual information. The discussion reflects the constraints of interpreting SHM without complete context.

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?
 
Last edited:
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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...
 
Last edited:
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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