Can someone explain to me what my professor did?

[Warlic]

Homework Statement

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Two things I don't understand; how did he get that omega is sqrt(k/m-(b/2m)^2)
And second; why is it that x(t) = e^(-bt/2m)* cos (omega*t+phi)
shouldnt it rather be; x(t) = c1*cos(ωt) + c2*sin(ωt)

Homework Equations

The Attempt at a Solution

 

[Warlic]

I figured out why omega is what it is, but still don't understand how he got the equation for x(t)

 

[gneill]

I think he's assuming that the system is underdamped, so will be a damped sinusoid. The general solution can be molded into that shape:

$x(t) = c_1 cos(ωt) + c_2 sin(ωt)$
$~~~= \sqrt{c_1^2 + c_2^2}\left( \frac{c_1}{\sqrt{c_1^2 + c_2^2}} cos(ωt) + \frac{c_2}{\sqrt{c_1^2 + c_2^2}} sin(ωt) \right)$
$~~~= \sqrt{c_1^2 + c_2^2}\left(cos(\phi) cos(ωt) + sin(\phi) sin(ωt) \right)$
$~~~= \sqrt{c_1^2 + c_2^2} cos(ωt - \phi)$ where: $~~~~\phi = tan^{-1}\left( \frac{c_2}{c_1} \right)$

You can fudge the sign of the angle $\phi$ by negating it and interpreting it appropriately.

 

[Herman Trivilino]

Aren't c₁ and c₂ constants? If so, I see a sinusoid, not a damped sinusoid.

 

[gneill]

Aren't c₁ and c₂ constants? If so, I see a sinusoid, not a damped sinusoid.

Ah, right. I left out the damping term in the general solution So:
$x(t) = e^{-\alpha t}(c_1 cos(ωt) + c_2 sin(ωt))$
Go from there. The roots of the auxiliary equation will be complex conjugates of the form $\alpha ± \omega$, where $\omega$ can be further broken down as $\omega = \sqrt{\alpha^2 - \omega_o^2}$

 

[Warlic]

Ah, right. I left out the damping term in the general solution So:
$x(t) = e^{-\alpha t}(c_1 cos(ωt) + c_2 sin(ωt))$
Go from there. The roots of the auxiliary equation will be complex conjugates of the form $\alpha ± \omega$, where $\omega$ can be further broken down as $\omega = \sqrt{\alpha^2 - \omega_o^2}$

This is exactly the point where I don't know where to go from :P. Where does the c2sin(ωt) part go?

 

[gneill]

This is exactly the point where I don't know where to go from :P. Where does the c2sin(ωt) part go?

See post #3. The sin and cos terms can be amalgamated into a single cos (or sin) term with a constant phase shift.

 

[Herman Trivilino]

Look at the undamped case:

$x(t)=c_1 \cos (\omega t)+c_2 \sin (\omega t)$

and

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

are equivalent. Note that each contains two constants of integration.