Finding Phase Constant and air-track glider

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An air-track glider oscillates with a period of 2.3 seconds, starting 3.09 cm left of equilibrium and moving right at 35.07 cm/s. The phase constant is being calculated using the equations v0x = -wAsin(φ) and x = Acos(φ), with attempts yielding incorrect results. The discussion highlights confusion over the correct application of angular speed and phase calculations, particularly when determining the phase at t=0 and t=0.5 seconds. Participants clarify that angular speed and velocity are distinct, and emphasize the importance of correctly interpreting the signs of distance and phase. Ultimately, the focus remains on accurately calculating the phase constant and subsequent phases at different times.
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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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