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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

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I gess that you could start expanding the square root in a Taylor series. – Ana Jan 3 '13 at 18:06
...and which term is smaller? $(\omega_0^2 - \omega_d^2)^2$ or $\lambda \omega_d^2$? – Vibert Jan 3 '13 at 20:51
@Vibert I am not given any information in the question that would allow me to deduce which is smaller. In fact, I have provided all the information that was given to me above. – Mel Jan 3 '13 at 21:28
If you take the limit $\omega_d \rightarrow \infty$ you find $A=C$. Then $\omega_d=0$ and $\omega_d=\omega_0$ give two other equations which should be able to determine $\Omega$ and $\Gamma$. Not sure this will work properly though. Probably a better bet is to Taylor expand both forms around the maximum point and match terms. – Michael Brown Jan 4 '13 at 2:46
how did you guess x(t)=Bcos(ωdt+ϕ)? – user22564 Mar 29 '13 at 19:48

1 Answer 1

up vote 3 down vote accepted

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:

enter image description here

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Thanks. One thing though, how come your blue curve doesn't (appear) to have two peaks? When I plotted both curves, the original curve had two peaks, at $\omega_d = \omega_0$ and $\omega_d = -\omega_0$ – Mel Jan 20 '13 at 13:47
You're absolutely right - I forgot a square root. I redid the calculation, the $C$ hasn't changed, but the $\Gamma$ did. – Rafael Reiter Jan 20 '13 at 14:22

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