Driven, Damped Oscillator; Plot x(t)/A

[oddjobmj]

Homework Statement

The problem is long so I will post the whole thing but ask only for help on part C.

The steady-state motion of a damped oscillator driven by an applied force F0 cos(ωt) is given by xss(t) = A cos(ωt + φ).
Consider the oscillator which is released from rest at t = 0. Its motion must satisfy x(0) = 0 and v(0) = 0. After a very long time, we expect that x(t) = xss(t). To satisfy these conditions we can take as the solution
x(t) = xss(t) + xtr(t),
where xtr(t) is a solution to the equation of motion of the free damped oscillator.

(A) Show that if xss(t) satisfies the equation of motion for the forced damped oscillator, then so does x(t) = xss(t) + xtr(t), where xtr(t) satisfies the equation of motion of the free damped oscillator.

(B) Sketch the resulting motion for the case that the oscillator is driven at resonance (i.e., ω = ω0 where ω02=k/m) with b=mω0.

(C) Choose the arbitrary constants in xtr(t) so that x(t) satisfies the initial conditions, assuming that the oscillator is driven at resonance with b=mω0. Hand in an accurate graph of x(t)/A.

Homework Equations

x(t)/A=sin(w₀t)-\frac{2}{sqrt(3)}sin(\frac{sqrt(3)}{2}w₀t)e^-w_0t/2

A=\frac{F_0}{mw_0^2}

The Attempt at a Solution

I am not quite sure what I am being asked for. How do I pick the constants in x_tr(t)? Does it matter what I use or can I just plug in 1 for all of them and plot the result? Doing that yields a plot that seems to exhibit the characteristics that a driven, damped oscillator would.

Thank you!

 

[vela]

The response of the system when $b=m\omega_0$ is given by
$$x(t) = e^{-\omega_0 t/2}\left[c_1 \cos \left(\frac{\sqrt{3}\omega_0 t}{2}\right) + c_2 \sin \left(\frac{\sqrt{3}\omega_0 t}{2}\right)\right] + \frac{F_0}{m[(\omega_0^2-\omega^2)^2 + \omega_0^2\omega^2]}[(\omega_0^2-\omega^2)\cos \omega t + \omega_0\omega \sin \omega t]$$ where $c_1$ and $c_2$ are arbitrary constants to be determined by the initial conditions. At resonance, we have $\omega=\omega_0$, and the solution simplifies to
$$x(t) = e^{-\omega_0 t/2}\left[c_1 \cos \left(\frac{\sqrt{3}\omega_0 t}{2}\right) + c_2 \sin \left(\frac{\sqrt{3}\omega_0 t}{2}\right)\right] + \frac{F_0}{m\omega_0^2}\sin \omega_0 t.$$ The condition $x(0)=0$ requires that $c_1=0$. Taking the derivative of x(t), we get
$$v(t) = c_2e^{-\omega_0 t/2}\left[-\frac{\omega_0}{2}\sin \left(\frac{\sqrt{3}\omega_0 t}{2}\right) + \frac{\sqrt{3}\omega_0}{2} \cos \left(\frac{\sqrt{3}\omega_0 t}{2}\right)\right] + \frac{F_0}{m\omega_0}\cos \omega_0 t.$$ The condition $v(0)=0$ requires that
$$\frac{\sqrt{3}c_2\omega_0}{2} + \frac{F_0}{m\omega_0} = 0.$$ Solving for $c_2$ gives
$$c_2 = -\frac{2F_0}{\sqrt{3}m\omega_0^2}.$$ The solution is therefore
$$x(t) = \frac{F_0}{m\omega_0^2}\left[\sin \omega_0 t - \frac{2}{\sqrt{3}}\sin\left(\frac{\sqrt{3}}{2}\omega_0 t\right)\right].$$ To plot the function $x(t)/A$ where $A = \frac{F_0}{m\omega_0^2}$, define $T_0 = \frac{2\pi}{\omega_0}$ and plot in units of $T_0$.