This harmonic oscillator is driven and damped, with the form:
$$\ddot{x} + \lambda \dot{x} + \omega_0^2 x = A \cos(\omega_d t)$$
Now, I have used the ansatz (guess): $x(t) = B \cos(\omega_d t + \phi)$, and have written B in the form:
$$B = \frac{A} {\sqrt{(\omega_o^2-\omega_d^2)^2+\lambda\omega_d^2}}$$
Next, I am required to "approximate B using the Lorentzian form"
$$B = \frac{C}{(\omega_d – \Omega)^2+\biggl(\frac{\Gamma}{2}\biggr)^2}$$
However, this is where I am stuck. I know that because it says "approximate" I will somehow have to drop out terms from my first expression to B, but I don't know where to start. How can I write B in this form?
EDIT: I have found a wikipedia article on resonance which show a form very similiar to what I seek, however, I can't seem to find a derivation http://en.wikipedia.org/wiki/Resonance
Best Answer
Since the maximum is the most important point of the curve, I suggest matching the derivatives 0-2 of the two curves at $\omega_0$. This is equivalent to doing a Taylor/power expansion on both functions and matching the first three coefficients. Since there are three constants, we can match three criteria (=equations).
First derivative: set $\Omega=\omega_0$, and by differentiating both curves you can show that the first derivative of both curves at $\omega_0$ is zero.
Zeroth derivative: By setting the two curves equal and solving for $C$, you find that $C=\frac{A (\frac{\Gamma}{2})^2}{\sqrt{\lambda}\omega_d}$.
Second derivative: We still haven't set $\Gamma$. The necessary equation comes from setting the second derivative at $\omega_0$ equal. This gives $\frac{\Gamma}{2}=\sqrt{\frac{\lambda}{2}}$.
The plot (all parameters in the original resonance curve are 2; blue is original, red is Lorentzian) looks pretty good to me: