Archived The Lorentzian

mbigras

1. Homework Statement
For a lightly damped harmonic oscillator and driving frequencies close to the natural frequency $\omega \approx \omega_{0}$, show that the power absorbed is approximately proportional to
$$\frac{\gamma^{2}/4}{\left(\omega_{0}-\omega\right)^{2}+\gamma^{2}/4}$$
where $\gamma$ is the damping constant. This is the so called Lorentzian function.

2. Homework Equations
$$\text{Average power absorbed} = P_{avg} = \frac{F_{0}^{2} \omega_{0}}{2k Q} \frac{1}{\left(\frac{\omega_{0}}{\omega}-\frac{\omega}{\omega_{0}}\right)^{2}+\frac{1}{Q^{2}}} \\ \omega_{0} = \sqrt{\frac{k}{m}}\\ m = \frac{b}{\gamma}\\ \text{where b is the damping constant and m is the mass}\\ \Delta \omega = \frac{\gamma}{2}$$

3. The Attempt at a Solution
The course of action that I took goes like:
1.Find $k$ and $Q$ in terms of $\omega_{0}$ and $\gamma$.
2. Chug through and do some algebra (and it is here that its very possible that a mistake was made, but I'll put my result not all the steps).
3. Expand a function about $w_{0}$ and make approximations so that $\Delta \omega$ is small.
(4) See the above equation fall out. This is the stage that I'm stuck at.

$$k = b \frac{\omega_{0}^{2}}{\gamma}\\ Q = \frac{\omega_{0}}{\gamma}\\ 2 \Delta \omega = \gamma \\ P_{avg} = \text{plug in and do lots of algebra...}\\ P_{avg} = \frac{\frac{\omega^{2}\gamma^{2}}{(\omega+\omega_{0})^{2}}}{(\omega_{0}-\omega)^{2}+\frac{\omega^{2}\gamma^{2}}{(\omega+\omega_{0})^{2}}}$$
Then taylor expanding $f(\omega) = \frac{\omega^{2}}{(\omega+\omega_{0})^{2}}$ about $\omega_{0}$....

Am I on the right try here? I'd like that taylor expansion to equal $\frac{1}{4}$ because then the equation would match the one described in the question but I'm trying it by hand and with mathematica and I'm not seeing them match.

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nrqed

Homework Helper
Gold Member
1. Homework Statement
For a lightly damped harmonic oscillator and driving frequencies close to the natural frequency $\omega \approx \omega_{0}$, show that the power absorbed is approximately proportional to
$$\frac{\gamma^{2}/4}{\left(\omega_{0}-\omega\right)^{2}+\gamma^{2}/4}$$
where $\gamma$ is the damping constant. This is the so called Lorentzian function.

2. Homework Equations
$$\text{Average power absorbed} = P_{avg} = \frac{F_{0}^{2} \omega_{0}}{2k Q} \frac{1}{\left(\frac{\omega_{0}}{\omega}-\frac{\omega}{\omega_{0}}\right)^{2}+\frac{1}{Q^{2}}} \\ \omega_{0} = \sqrt{\frac{k}{m}}\\ m = \frac{b}{\gamma}\\ \text{where b is the damping constant and m is the mass}\\ \Delta \omega = \frac{\gamma}{2}$$

3. The Attempt at a Solution
The course of action that I took goes like:
1.Find $k$ and $Q$ in terms of $\omega_{0}$ and $\gamma$.
2. Chug through and do some algebra (and it is here that its very possible that a mistake was made, but I'll put my result not all the steps).
3. Expand a function about $w_{0}$ and make approximations so that $\Delta \omega$ is small.
(4) See the above equation fall out. This is the stage that I'm stuck at.

$$k = b \frac{\omega_{0}^{2}}{\gamma}\\ Q = \frac{\omega_{0}}{\gamma}\\ 2 \Delta \omega = \gamma \\ P_{avg} = \text{plug in and do lots of algebra...}\\ P_{avg} = \frac{\frac{\omega^{2}\gamma^{2}}{(\omega+\omega_{0})^{2}}}{(\omega_{0}-\omega)^{2}+\frac{\omega^{2}\gamma^{2}}{(\omega+\omega_{0})^{2}}}$$
Then taylor expanding $f(\omega) = \frac{\omega^{2}}{(\omega+\omega_{0})^{2}}$ about $\omega_{0}$....

Am I on the right try here? I'd like that taylor expansion to equal $\frac{1}{4}$ because then the equation would match the one described in the question but I'm trying it by hand and with mathematica and I'm not seeing them match.
That was very good and basically done. The key point is that we Taylor expand around $\omega \approx \omega_0$, in which case
$\frac{\omega^{2}}{(\omega+\omega_{0})^{2}} \approx 1/4$ and the answer is obtained.

ehild

Homework Helper
$$P_{avg} = \frac{\frac{\omega^{2}\gamma^{2}}{(\omega+\omega_{0})^{2}}}{(\omega_{0}-\omega)^{2}+\frac{\omega^{2}\gamma^{2}}{(\omega+\omega_{0})^{2}}}$$
Multiply both the numerator and the denominator by (ω+ω0)2.
$$P_{avg} = \frac{\omega^{2}\gamma^{2}}{(\omega_{0}-\omega)^{2}(\omega+\omega_{0})^{2}+\omega^{2}\gamma^{2}}$$
When ω is near to ω0 you can replace ω+ω0 by 2ω, and the formula for Pavg can be simplified by ω2.
$$P_{avg} = \frac{\gamma^{2}}{4(\omega_{0}-\omega)^2+\gamma^{2}}$$

"The Lorentzian"

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