2. (Compulsory 4) Given vectors
$$
\boldsymbol{a}=(\sqrt{3} \sin \omega x, 1), \boldsymbol{b}=(\cos \omega x, 0)(\omega>0) \text {, }
$$

and the function $f(x)=\boldsymbol{b} \cdot(\boldsymbol{a}-k \boldsymbol{b})$ is a periodic function with the smallest positive period $\frac{\pi}{2}$. If the maximum value of the function $f(x)$ is $\frac{1}{2}$, does there exist a real number $t$ such that the graph of the function $f(x)$ can be obtained by translating the graph of the function $g(x)=t a \cdot b$? If so, find the real number $t$ and write a translation vector $\boldsymbol{m}$; if not, explain the reason.
2. Notice that
$$
\begin{array}{l}
f(x)=\boldsymbol{b} \cdot(\boldsymbol{a}-k \boldsymbol{b}) \\
=\sqrt{3} \sin \omega x \cdot \cos \omega x-k \cos ^{2} \omega x \\
=\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1}{2} k \cos 2 \omega x-\frac{1}{2} k \\
=\frac{1}{2} \sqrt{k^{2}+3} \sin (2 \omega x+\theta)-\frac{1}{2} k .
\end{array}
$$

Since the smallest positive period of the function $f(x)$ is $\frac{\pi}{2}$, we have
$$
T=\frac{2 \pi}{2 \omega}=\frac{\pi}{2} \Rightarrow \omega=2 \text {. }
$$

Thus, $f(x)=\frac{1}{2} \sqrt{k^{2}+3} \sin (4 x+\theta)-\frac{1}{2} k$.
Also, $f_{\text {max }}=-\frac{1}{2} k+\frac{1}{2} \sqrt{k^{2}+3}=\frac{1}{2}$, solving for $k$ gives $k=1$.

Therefore, $f(x)=\sin \left(4 x-\frac{\pi}{6}\right)-\frac{1}{2}$.
And $g(x)=t a \cdot b$
$$
=\sqrt{3} t \sin \omega x \cdot \cos \omega x=\frac{\sqrt{3}}{2} t \sin 4 x,
$$

Thus, when $t=\frac{2 \sqrt{3}}{3}$, the graph of the function $g(x)$, when translated by the vector $m=\left(\frac{\pi}{24},-\frac{1}{2}\right)$, coincides with the graph of the function $f(x)$.