这是一份liverpool利物浦大学PHYS101的成功案例

The position of a particle moving in a plane is
$$
\mathbf{r}=r \hat{\mathbf{r}}
$$
As the particle moves, both $r$ and $\hat{\mathbf{r}}$ may change, i.e. they are both implicitly time-dependent. The velocity of the particle is calculated by differentiation of the product $r \hat{\mathbf{r}}$.
$$
\mathbf{v}=\frac{\mathrm{d}}{\mathrm{d} t}(r \hat{\mathbf{r}})=\frac{\mathrm{d} r}{\mathrm{~d} t} \hat{\mathbf{r}}+r \frac{\mathrm{d} \hat{\mathbf{r}}}{\mathrm{d} t} .
$$
Since $\hat{\mathbf{r}}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}$,
$$
\frac{\mathrm{d} \hat{\mathbf{r}}}{\mathrm{d} t}=-\sin \theta \frac{\mathrm{d} \theta}{\mathrm{d} t} \mathbf{i}+\cos \theta \frac{\mathrm{d} \theta}{\mathrm{d} t} \mathbf{j}=\frac{\mathrm{d} \theta}{\mathrm{d} t} \hat{\theta}
$$
where we have used the definition for $\hat{\boldsymbol{\theta}}$. Thus,
$$
\mathbf{v}=\frac{\mathrm{d} r}{\mathrm{~d} t} \hat{\mathbf{r}}+r \frac{\mathrm{d} \theta}{\mathrm{d} t} \hat{\boldsymbol{\theta}}
$$

PHYS101 COURSE NOTES ：

$$
v(t)=\frac{\mathrm{d} x}{\mathrm{~d} t}=-A \omega \sin \omega t
$$
and
$$
a(t)=\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=-A \omega^{2} \cos \omega t .
$$
So we can write
$$
a(t)=-\omega^{2} x
$$
We can connect the force with the motion by simply substituting for $x$ using Hooke’s Law, i.e.
$$
F=\frac{k}{\omega^{2}} a .
$$