So my professor was discussing the case of a mass suspended from a vertical massless spring in some viscous liquid.
He arrives at the equation of motion which was :
x: + [itex]\frac{b}{m}[/itex]x. + [itex]\frac{k}{m}[/itex]x = 0
x: is the second derivative of displacement wrt time. similarly x. is the first derivative.
He then defined b/m = gamma k/m= w^2
He then used the trial solution x=[itex]Ae^{t\tau}[/itex] formed an auxillary equation and solved it to get :
[itex]\frac{-\gamma}{2}[/itex]±[itex]\sqrt{\frac{\gamma^2}{4}-w^2}[/itex]
He then examined the discriminant of the above equation to formulate the general solution for light damping.
I understand in light damping w^2 < (gamma^2)/4
But how does he arrive at the following general solution :
[itex]Ae^\frac{t\gamma}{2}cos(wt + \phi)[/itex]
Where did the e^t*gamma/2 come from ? Why is there no sine function even though we have an imaginary root case ? Why is there a phi in there ?
He arrives at the equation of motion which was :
x: + [itex]\frac{b}{m}[/itex]x. + [itex]\frac{k}{m}[/itex]x = 0
x: is the second derivative of displacement wrt time. similarly x. is the first derivative.
He then defined b/m = gamma k/m= w^2
He then used the trial solution x=[itex]Ae^{t\tau}[/itex] formed an auxillary equation and solved it to get :
[itex]\frac{-\gamma}{2}[/itex]±[itex]\sqrt{\frac{\gamma^2}{4}-w^2}[/itex]
He then examined the discriminant of the above equation to formulate the general solution for light damping.
I understand in light damping w^2 < (gamma^2)/4
But how does he arrive at the following general solution :
[itex]Ae^\frac{t\gamma}{2}cos(wt + \phi)[/itex]
Where did the e^t*gamma/2 come from ? Why is there no sine function even though we have an imaginary root case ? Why is there a phi in there ?
