这是一份leeds利兹大学PHYS121001作业代写的成功案例

$$
\begin{aligned}
&F_{x} / m=-\frac{v^{2}}{r} \cos \theta \text {, so } \
&F_{x}=-m \frac{v^{2}}{r} \cos \theta .
\end{aligned}
$$
Since our goal is an equation involving the period, it is natural to eliminate the variable $v=$ circumference $/ T=2 \pi r / T$, giving
$$
F_{x}=-\frac{4 \pi^{2} m r}{T^{2}} \cos \theta .
$$
The quantity $r \cos \theta$ is the same as $x$, so we have
$$
F_{x}=-\frac{4 \pi^{2} m}{T^{2}} x
$$

PHYS121001 COURSE NOTES ：

$$
\frac{F_{r}}{F_{t}}=\frac{2 \pi m}{b f}\left(f^{2}-f_{r e s}^{2}\right)
$$
This is the ratio of the wasted force to the useful force, and we see that it becomes zero when the system is driven at resonance.
The amplitude of the vibrations can be found by attacking the equation $|F|=b v=2 \pi b A f$, which gives
$$
A=\frac{\left|F_{\mathrm{t}}\right|}{2 \pi b f} .
$$
However, we wish to know the amplitude in terms of $|\boldsymbol{F}|$, not $\left|F_{\mathrm{t}}\right|$. From now on, let’s drop the cumbersome magnitude symbols. With the Pythagorean theorem, it is easily proven that
$$
F_{t}=\frac{F}{\sqrt{1+\left(\frac{F_{t}}{F_{t}}\right)^{2}}},
$$