How to Solve for Amplitude and Phase Constant in an Oscillation Problem?

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The discussion focuses on solving for the phase constant and amplitude of an oscillating air-track glider attached to a spring. The glider's displacement is 4.6 cm left of equilibrium, with a velocity of 33.4 cm/s, and the period of oscillation is 1.50 seconds. The equations for displacement and velocity are provided, but the user struggles with incorporating the angular frequency and determining the phase constant correctly. It is suggested that the user needs to solve for amplitude first and then use that value to find the phase constant. The importance of correctly accounting for time in the equations is emphasized to avoid confusion in calculations.
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I'm not quite sure on what I did wrong. Can anyone please help me with this?


Homework Statement



An air-track glider attached to a spring oscillates with a period of 1.50sec . At the glider is 4.60cm left of the equilibrium position and moving to the right at 33.4cm/s.
What is the phase constant , if the equation of the oscillator is taken to be ? Give an answer in the range -pi < Q < pi.

Homework Equations


phase constant = Q
Amplitude = A
angular frequency = w
x(t) = Acos(wt+Q)
v(t) = -Awsin(wt + Q)
w = (2pi)/period


The Attempt at a Solution


4.6 = Acos(Q)
33.4 = -Awsint(Q)

Q = atan(.334/(.046*(2pi/1.5))
 
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It appears that this is close to the right answer, but I'm missing some "additive constant". I probably posted this in the wrong section just by judging by the number of responses on this. If someone would kindly help me, I'd really appreciate it. :frown:
 
I think you might have missed a constant in your calculation there, but is there a way that you can figure out, t, at these instant where you're given x and v.

The term w*t seems to have disappeared, from the argument of both trig functions, not sure you can do this, or at least can't call it Q unless you're at t=0 or some integral number of cycles after that. In otherwords, you neet to solve for A and Q both if I'm not mistaken. You can get A from the approacch you were using once you know the angle. See it that leads anywhere
 
The term w*t is no longer present b/c it was a function of time, and at the instance when t =0 or time = 0sec, w * t or (2pi/period) *t became 0. I then solved for A (Amplitude) from the first equation then substituted that into the second eqaution. If I'm not mistaken, isn't that the way to solve for two variables using two equations?
 
Last edited:
irvine752 said:
The term w*t is no longer present b/c it was a function of time, and at the instance when t =0 or time = 0sec, w * t or (2pi/period) *t became 0. I then solved for A (Amplitude) from the first equation then substituted that into the second eqaution. If I'm not mistaken, isn't that the way to solve for two variables using two equations?

The displacement is 4.6 cm and the velocity 33.4 cm/s at some time t different from 0. As denverdoc mentioned, you can write those equations but with a value different then Q and solve for A.
Once you know A, you can solve for Q.
 
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