这是一份leeds利兹大学PHYS125001作业代写的成功案例

$$
A=\frac{F}{2 \pi \sqrt{4 \pi^{2} m^{2}\left(f^{2}-f_{0}^{2}\right)^{2}+b^{2} f^{2}}}
$$
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
$$
\begin{aligned}
|v| &=w / t \
&=\sqrt{\frac{2 T}{\mu}} .
\end{aligned}
$$