Finding phi constant based on initial position and initial velocity

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[tex]x(t)=Acos(ωt+ϕ)\\ v(t)=-Aωsin(ωt+ϕ)\\ \\ Let\quad n=integer,\quad A=0.5,\quad ω=2.\\ \\ Given\quad initial\quad position\quad (0,\quad 0.25)\quad and\quad initial\quad velocity\quad (0,\quad 1),\quad find\quad ϕ.\\ \\ x(t)=0.5cos(2t+ϕ)\\ 0.25=0.5cos(ϕ)\\ ϕ=±1.0471975512+2πn\\ \\ v(t)=-sin(2t+ϕ)\\ 1=-sin(ϕ)\\ ϕ=-1.5707963268[/tex]

So which one is it? :S

Immediate help is very appreciated!
 
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Solve your position equation for t; utilize x=0. Solve your velocity equation for t; you have two components for the velocity, what is its initial magnitude? After that, you can set both equations equal to one another and solve for phi.
 


Solving for t yields 4 equations - not 2 (2 equations from position and 2 from velocity).
 


You are given (x, y)=(0, 0.25). You have an equation for x. Plug in the value for x, then solve for t. You will have an equation that is a function of phi (1 unknown).

You then have an equation for the velocity given as (vx, vy)=(0, 1). Furthermore, you have an equation for the velocity, which would merely be the magnitude of the force vector. You have two components, and hence can determine the magnitude. You can solve this equation for t, which would also be in terms of phi.

Set these two equations equal to one another and you should be able to solve for phi. 2 equations, 1 unknown.
 


The given is in the form of (t, x) and (t, v).
 


Okay. What if you ignore the values for t, and solve both of the equations for t. Then use the two to find phi?