Simple Harmonic Motion - Finding the Phase

In summary, the particle moves from the left to the right on the graph, and at point a on the graph, the phase is -120 degrees.
  • #1

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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  • #2
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?
  • #3
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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  • #4
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?
  • #5
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
  • #6
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.
  • #7
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?)
  • #8
You wouldn't be able to because without additional information, like the diagram, you don't know in which direction the particle is moving.
  • #9
Okay, that makes sense :)

Thanks for your help!

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where an object oscillates back and forth around an equilibrium point with a constant amplitude and a constant period. This type of motion can be seen in systems such as mass-spring systems and pendulums.

2. How do you find the phase of a simple harmonic motion?

The phase of a simple harmonic motion is the fraction of the period that has elapsed since the object passed through its equilibrium point. To find the phase, you can use the equation θ = 2πt/T, where θ is the phase, t is the time, and T is the period of the motion.

3. What is the relationship between the phase and the amplitude of a simple harmonic motion?

The phase and amplitude of a simple harmonic motion are related by the equation x = A*cos(ωt + θ), where x is the displacement of the object from its equilibrium point, A is the amplitude, ω is the angular frequency, and θ is the phase. This equation shows that the phase determines the starting position of the motion, while the amplitude determines the maximum displacement.

4. How does the mass affect the phase of a simple harmonic motion?

The mass does not affect the phase of a simple harmonic motion. The phase is only dependent on the time and the period of the motion. However, the mass does affect the amplitude of the motion, as a larger mass will result in a smaller amplitude.

5. Can the phase of a simple harmonic motion be negative?

Yes, the phase of a simple harmonic motion can be negative. This indicates that the object is starting its motion on the opposite side of the equilibrium point compared to its initial position. The phase can have any value between 0 and 2π, representing a full period of the motion.

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