Forced Oscillations and Resonance — Question 4

Question 4

The exact half-power bandwidth has a geometric center

For fixed-amplitude force driving, $\overline P/\overline P_{\max}=b^2/[(k/\Omega-m\Omega)^2+b^2]$. Let $\Omega_-$ and $\Omega_+$ be the positive frequencies at half the maximum power.

1. Derive both half-power frequencies, their difference and their product.

2. For $m=1\,\mathrm{kg}$, $k=25\,\mathrm{N/m}$ and $b=2\,\mathrm{kg/s}$, find the frequencies. Is the natural frequency exactly their arithmetic midpoint?

Question 4 - Solution

The exact half-power bandwidth has a geometric center

(a) Derivation and calculation

$$\overline P/\overline P_{\max}=\tfrac12\iff(k/\Omega-m\Omega)^2=b^2.$$$$m\Omega_-^2+b\Omega_--k=0,\quad m\Omega_+^2-b\Omega_+-k=0.$$$$\boxed{\Omega_{\pm}=\frac{\sqrt{b^2+4mk}\pm b}{2m}}.$$

(b) Result and interpretation

$$\boxed{\Omega_+-\Omega_-=b/m},\qquad\boxed{\Omega_+\Omega_-=k/m=\omega_0^2}.$$$$m=1,\ k=25,\ b=2:\quad\Omega_-=\sqrt{26}-1=4.09902,\quad\Omega_+=\sqrt{26}+1=6.09902.$$$$\sqrt{\Omega_-\Omega_+}=5.00,\qquad(\Omega_-+\Omega_+)/2=\sqrt{26}=5.09902\,\mathrm{rad/s}.$$The natural frequency is exactly the geometric mean, not the arithmetic midpoint.

Physical check

The half-power width is exact; symmetry about $\omega_0$ is only a narrow-band approximation.