Determining the Phase Constant in Simple Harmonic Motion

[joeyjane]

Homework Statement

A particle oscillates in simple harmonic motion with a period of 2.48 s and an
amplitude of 4.24 cm. At t = 0, it is at z = -2.0 cm and it is moving toward
z = 0. We wish to write the position of the particle as a function of time in the following
form, z(t) = A cos($$\omega$$t + $$\phi$$).

And then I know A = 4.24 cm, $$\omega$$ = 2.5335 rad/s, and I need to find the phase constant.

Homework Equations

x = A sin($$\omega$$t+$$\phi$$)
v = A$$\omega$$ cos($$\omega$$t+$$\phi$$)

The Attempt at a Solution

I tried plugging in -2 for x, and 3.419 for v (I attempted to find v by doing (4.24*2)/T, but I'm not sure if I did this right), and 2.5335 for $$\omega$$ and then I divided the two equations I listed above. This gave me (-2/3.419)=2.5335 tan$$\phi$$. After some solving, I got tan^-1(-2.3089), which is -13 degrees, or -.227 rad. My answer is supposed to be -2.06 rad. Any help? :)

 

[vela]

Take your equation for z(t) and solve for ϕ. You should find there are two possible values for ϕ. Use the velocity information to choose between the two.

 

[joeyjane]

Thank you!