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

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SUMMARY

The discussion focuses on calculating amplitude (A) and phase (\phi) in a damped oscillator represented by a mass-spring-damper system. The governing equation is \(\ddot{y}+2\delta\dot{y}+w_0^2y=0\), with \(\delta = \frac{3}{5}w_0\). The solution for the displacement is given by \(y(t)=Ae^{-\delta t}\cos(w_d t+\phi)\), where \(w_d=\sqrt{w_0^2+\delta^2}\). The user successfully derived the phase angle as \(\phi=\arctan{\frac{-\delta}{w_d}}\) but struggled with the initial condition \(y(0)=\frac{m}{M}y_0\), which was later clarified.

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Homework Statement


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


Homework Equations



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

The Attempt at a Solution



from this i find [itex]k[/itex] and [itex]b[/itex]. 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:

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

I'm to find [itex]A[/itex] and [itex]\phi[/itex]

my try:

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

however i do not find a substitute for [itex]y(0)[/itex]. The solution says [itex]y(0)=\frac{m}{M}y_0[/itex], 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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