Determining the Phase Constant in Simple Harmonic Motion

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To determine the phase constant in simple harmonic motion, the position function z(t) is expressed as z(t) = A cos(ωt + φ), with given parameters including a period of 2.48 s and an amplitude of 4.24 cm. The particle's initial position at t = 0 is z = -2.0 cm, moving towards z = 0, leading to the need for accurate calculations of velocity and phase constant. The attempt involved using the equations for position and velocity, resulting in a calculation for tan(φ) that yielded -13 degrees, which is inconsistent with the expected phase constant of -2.06 rad. The solution requires careful consideration of both position and velocity to correctly identify the appropriate phase constant from the two possible values.
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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(\omegat + \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(\omegat+\phi)
v = A\omega cos(\omegat+\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? :)
 
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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.
 
Thank you!
 
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