Damped Harmonic Oscillator Approximation?

[cj]

For a simple damped oscillator...

\text {Apparently if } \beta \ll \omega_0 } \text { then ...}

\omega_d \approx \omega_0[1-\frac {1}{2}(\beta/\omega_0)^2]}

Given that:

\beta=R_m/2m \text { (where } R_m= \text {mechanical resistance) } \text { and } \omega _d=\sqrt{(\omega _0^2-\beta ^2)}

How/why is this true? My guess is some kind of
series approximation is used -- but I'm not sure...

 

[arildno]

Let's establish the series approximation:
\omega_{d}=\sqrt{\omega_{0}^{2}-\beta^{2}}=\omega_{0}\sqrt{1-(\frac{\beta}{\omega_{0}})^{2}}

Now, let f(\epsilon)=(1+\epsilon)^{m}
When \epsilon\approx0
we have, by Taylor's theorem:
f(\epsilon)\approx{f}(0)+f'(0)\epsilon=1+m\epsilon
Now, recognize:
m=\frac{1}{2},\epsilon=-(\frac{\beta}{\omega_{0}})^{2}
and you've got the formula.

 

[cj]

Thanks very much. I've got to somehow get
more familiar with Taylor series expansions -- they
seem to be the basis of so many solutions.

cj

Let's establish the series approximation:
\omega_{d}=\sqrt{\omega_{0}^{2}-\beta^{2}}=\omega_{0}\sqrt{1-(\frac{\beta}{\omega_{0}})^{2}}

Now, let f(\epsilon)=(1+\epsilon)^{m}
When \epsilon\approx0
we have, by Taylor's theorem:
f(\epsilon)\approx{f}(0)+f'(0)\epsilon=1+m\epsilon
Now, recognize:
m=\frac{1}{2},\epsilon=-(\frac{\beta}{\omega_{0}})^{2}
and you've got the formula.

 

[arildno]

Thanks very much. I've got to somehow get
more familiar with Taylor series expansions -- they
seem to be the basis of so many solutions.

cj

You are absolutely correct in this.
Taylor expansions occur in every branch of physics; for example, they are often used to simplify and approximate difficult non-linear terms occurring in differential equations.
I'm sure you know this one from the swinging pendulum:
We simply assume the angle to be small, and approximate the term:
\sin\theta(t)\approx\theta(t)
This brings, as you know, the pendulum equation into the form of a simple harmonic oscillator.