# 震动与波浪 Vibrations & Waves PHYS125001

$$A=\frac{F}{2 \pi \sqrt{4 \pi^{2} m^{2}\left(f^{2}-f_{0}^{2}\right)^{2}+b^{2} f^{2}}}$$
Show that the maximum occurs not at $f_{\mathrm{a}}$ but rather at the frequency
$$f_{\text {res }}=\sqrt{f_{o}^{2}-\frac{b^{2}}{8 \pi^{2} m^{2}}}=\sqrt{f_{o}^{2}-\frac{1}{2} \mathrm{FWHM}^{2}}$$

## PHYS125001 COURSE NOTES ：

\begin{aligned} a &=F / m \ &=4 T h / \mu w^{2} . \end{aligned}
The time required to move a distance $b$ under constant acceleration $a$ is found by solving $h=\frac{1}{2} a t^{2}$ to yield
\begin{aligned} t &=\sqrt{2 b / a} \ &=w \sqrt{\frac{\mu}{2 T}} . \end{aligned}
Our final result for the velocity of the pulses is
\begin{aligned} |v| &=w / t \ &=\sqrt{\frac{2 T}{\mu}} . \end{aligned}