What is the Phase Constant for a Simple Harmonic Oscillator with Given Graphs?

[foundAnAnswer]

Homework Statement

Part (a) of the figure below is a partial graph of the position function x(t) for a simple harmonic oscillator with an angular frequency of 1.35 rad/s; Part (b) of the figure is a partial graph of the corresponding velocity function v(t). The vertical axis scales are set by x_s = 6.5 cm and v_s = 7.0 cm/s. What is the phase constant of the SHM if the position function x(t) is given by the form x = x_mcos(ωt + ϕ)?

Homework Equations

x = x_mcos(ωt + ϕ)
v = -ωx_msin(ωt + ϕ)

The Attempt at a Solution

For the X vs. t graph the line crosses t=0 when x = 2.6. For the V vs. t graph the line crosses t=0 when v=-5.6.

I thought then I could just plug all the number in and find out when they are equal

2.6 = 6.5cos(1.35*0+ϕ)
-5.6 = -8.775sin(1.35*0+ϕ)

I subtracted 2.6 from both sides for the first equation and added 5.6 to both sides for the second. I then set them equal. I used my calculator to attempt to solve them.

Looking at them separately it looks like it should be 1.15 radians for X and .69 radians for V (Roughly).

I can't figure out what I'm doing wrong. Perhaps I shouldn't be reading the graph like I am. or I am simply reading it wrong.

 

[foundAnAnswer]

Found an answer. I don't understand why this is correct, but dividing velocity by position gives the following

v/x = tan^-1(\frac{Velocity @ t=0}{Position @ t=0 TIMES Angular Frequency})

So that leaves


tan^-1(\frac{-5.6}{2.6 * 1.35}) = -1.01

I think then since it is shifted I changed it to positive 1.01.

That answer was taken as correct.

 

[regas99]

One equation represents position and the other represents velocity. Since the units are different, you can never "set them equal."