粒子的加速度 $a$ m/s$^2$ 与它到固定点的距离 $x$ 米的平方成反比。
粒子的距离减少到 $0.5x$。 求现在的加速度与原加速度的比值。 答：... : ...

1. Express the inverse square relationship using a constant of proportionality.

Since $a$ is inversely proportional to the square of the distance $x$,
$a \propto \frac{1}{x^2}$
so
$a=\frac{k}{x^2}$
where $k$ is a constant.

2. Find the new acceleration when the distance is reduced to $0.5x$.

Let the new acceleration be $A$.

When the distance is $0.5x$,
$A=\frac{k}{(0.5x)^2}$
$A=\frac{k}{0.25x^2}=\frac{4k}{x^2}$

3. Compare the new acceleration with the original acceleration as a ratio.

Original acceleration:
$a=\frac{k}{x^2}$

New acceleration:
$A=\frac{4k}{x^2}=4a$

Therefore,
$A:a=4a:a=4:1$