Finding Phase Constant and air-track glider

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Homework Help Overview

The problem involves an air-track glider attached to a spring, oscillating with a specified period. The original poster seeks to determine the phase constant based on the glider's initial position and velocity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the phase constant using the relationship between position, velocity, and angular speed. Various participants suggest adjustments to the calculations and question the interpretation of the results.

Discussion Status

Participants are actively discussing the calculations and interpretations related to the phase constant and the phase at a later time. Some guidance has been offered regarding the use of trigonometric functions and the relationship between angular speed and velocity, but no consensus has been reached on the correct approach for the phase at t=0.5s.

Contextual Notes

There are indications of confusion regarding the signs of the values used in calculations, as well as the distinction between angular speed and linear velocity. The original poster also notes that the question has multiple parts, which may contribute to the complexity of the discussion.

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Homework Statement


An air-track glider attached to a spring oscillates with a period of 2.3 s. At t=0 s the glider is 3.09 cm left of the equilibrium position and moving to the right at 35.07 cm/s.
What is the phase constant? (in degrees)



Homework Equations



v0x= -wAsin(ro), x=Acos(ro)

The Attempt at a Solution


I divided v0x/x=wtan(ro). I solved for angular speed which was 2.73 rad/s. Now I solved for (ro) by taking tan inverse of (v/-wx). I received 73 as an answer and that's wrong. Would you please help me? Thank you.
 
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Since you are to the left, you have to subtract 180. The anwser you came up with is to the right of the equilibrium point.
 
I minus that by 180 so it's 107 but that's still wrong. What do you think is wrong?
 
Did you get

r_{0} = \tan^{-1} \frac{35.07}{3.09 \cdot 2.73}?
 
yeah, I tried that as well which was 76.4751. Then I minus it from 180 so it's 103.525 but that's wrong.
 
Both 76.47 and 76.47 - 180 have the same tangent. You can see by drawing a pic that the one we want is (76.47 - 180).

-103.5 degrees = -1.807 radians.
 
is it because it's negative and the distance is negative?
 
Also, in this case, is angular speed the same as velocity? If I wanted to find the phase at t=0 s would I just need to use wt+ro, w being the velocity?
 
Yes, the cosine of 76.4 is positive, but the initial distance is negative.
 
  • #10
Oh sorry w=2.73. Sorry about that.
 
  • #11
No, angular speed and velocity are different. The angular speed is constant, while the velocity is changing.

v = -Aωsin(ωt + a0)
 
  • #12
ok, so for t=.5s I get -102.115 but that's wrong. All I did is wt+constant(-103.48). But that's wrong for some reason. Does the constant change at that time?
 
  • #13
Write -103.48 in radians.
 
  • #14
-1.80607 rad
 
  • #15
Do you get the right answer now? (I'm not sure what you're trying to calculate BTW, did you get the correct answer for the phase constant?)
 
  • #16
I got the correct answer for the phase constant but I'm trying the phase for t=.5s. Sorry for the confusion.
 
  • #17
Ok, so at t = 0.5, you have ωt + a0 = 2.73(0.5) - 1.806. This is in radians. If you want to convert it to degrees, multiply by 360/2pi.
 
  • #18
The question has several parts and it's asking me for the phase at t=.5s. I used the same formula but it's wrong. Could you please let me know what's wrong? thank you.
 
  • #19
Ok thank you very much for all your help. I greatly appreciate it.
 

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