How Do You Calculate Amplitude and Phase in a Damped Oscillator?

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To calculate amplitude (A) and phase (φ) in a damped oscillator, the system is described by the equation of motion for a mass-spring-dampener system. The solution involves the function y(t) = Ae^(-δt)cos(w_d t + φ), where w_d is derived from the natural frequency and damping ratio. The initial condition for velocity, given by \dot{y}(0) = 0, leads to the expression for φ as φ = arctan(-δ/w_d). The confusion arises regarding the initial displacement y(0), which is stated to equal m/M * y_0 in the solution. Ultimately, the user finds the solution after clarifying the initial conditions.
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Homework Statement


A mass-spring-dampener system is applied a force mg and is immediatly removed, setting the system in motion. The system is constantly applied force Mg and is static at y=y_0.
Find a formula for both A and \phi


Homework Equations



\ddot{y}+2\delta\dot{y}+w_0^2y=0
\frac{2\pi}{w_0}=T_0
\delta = \frac{3}{5}w_0
F_f=-b\dot{y}
Mg=ky_0

The Attempt at a Solution



from this i find k and b. No problem, not part of my question.

when the force is applied, the system 'moves' in y direction and is set in motion, given function:

y(t)=Ae^{-\delta t}cos\left(w_d t+\phi\right)
w_d=\sqrt{w_0^2+\delta^2}

I'm to find A and \phi

my try:

I understand \dot{y}(0)=0 gives:
\dot{y}=-\delta Ae^{-\delta t}cos(w_d t+\phi)-w_d A e^{-\delta t}sin(w_d t + \phi)
gives:
\phi=\arctan{\frac{-\delta}{w_d}}

however i do not find a substitute for y(0). The solution says y(0)=\frac{m}{M}y_0, but i don't see the logic in that at all

Sorry if its a bit caotic, this is only part of the assignment. ask and i will provide!
 
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nvm, found the solution
 
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