What Determines the Phase Constant in Oscillatory Motion?

[BioCore]

Homework Statement

An air-track glider attached to a spring oscillates with a period of 1.5s. At t = 0s the glider is 5.0 cm left of the equilibrium position and moving to the right at 36.3 cm/s.

What is the phase constant, \Phi₀?

Homework Equations

x = Acos (\varpit + \phi ₀) {1}

V_x = -V_max (\varpit + \phi ₀) {2}

The Attempt at a Solution

I tried to rearrange equation two for the phase constant, but I am missing A (amplitude). I have tried using v_max = \varpi A but that only works with maximum velocity which I do not have. Maybe I am not seeing something clearly. Can anyone help me out please?

 

[alphysicist]

Hi BioCore,

Homework Statement

An air-track glider attached to a spring oscillates with a period of 1.5s. At t = 0s the glider is 5.0 cm left of the equilibrium position and moving to the right at 36.3 cm/s.

What is the phase constant, \Phi₀?

Homework Equations

x = Acos (\varpit + \phi ₀) {1}

V_x = -V_max (\varpit + \phi ₀) {2}

This equation is missing the trig function. Is that just a typo?

The Attempt at a Solution

I tried to rearrange equation two for the phase constant, but I am missing A (amplitude). I have tried using v_max = \varpi A but that only works with maximum velocity which I do not have.

They did not give a numerical value for v_{\rm max}, but it's in the equation you need to work with, so by using that relationship you can eliminate one of the unknowns from your two equations.

You'll find the answer by solving the x and v equations together. Based on what the problem gives you, how can you do that to find \phi?

 

[BioCore]

Yes sorry it was a typo. Ok so when I solved the formula's together I get:

\phi ₀ = tan^-1 (\frac{V_x }{-\varpi x})

Which when I solve for I get -60 degrees or pi/3. Now the answer they got is -2pi/3 or -120 degrees.

I can't seem to explain to myself how to it is -120 degrees although I think I know why they say that.

 

[alphysicist]

Yes sorry it was a typo. Ok so when I solved the formula's together I get:

\phi ₀ = tan^-1 (\frac{V_x }{-\varpi x})

Which when I solve for I get -60 degrees or pi/3. Now the answer they got is -2pi/3 or -120 degrees.

I can't seem to explain to myself how to it is -120 degrees although I think I know why they say that.

When you find an inverse trig function with a calculator, you only get one of the two possible answers. So when you get your answers of 60 degrees (your post said -60 but you meant 60, right?) you'll actually need to determine if the answer is 60 degrees or 180 degrees away from that, which would be the -120 degrees.

There are several ways to find out, but probably the quickest way is to see which one solves the original equations you wrote down (for x and v) with the data you got from the problem.

 

[BioCore]

When you find an inverse trig function with a calculator, you only get one of the two possible answers. So when you get your answers of 60 degrees (your post said -60 but you meant 60, right?) you'll actually need to determine if the answer is 60 degrees or 180 degrees away from that, which would be the -120 degrees.

There are several ways to find out, but probably the quickest way is to see which one solves the original equations you wrote down (for x and v) with the data you got from the problem.

Yeah sorry I meant +60 degrees, I am just very tired right now and I think I will go get some rest. But thanks a lot for the help and understanding the answer. Appreciate it a lot.

 

[alphysicist]

Sure, glad to help!