一根原长为 $a$、弹性模量为 $20mg$ 的轻弹簧竖直放置在水平面上，下端固定在该平面上。质量为 $m$ 的质点连接在弹簧上端。

将质点竖直向下压，直至弹簧长度为 $\frac35a$，随后将该系统从静止释放。

求弹簧在随后运动中的最大伸长量。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

初始位置与最高位置的速率均为零。以弹簧的伸长为带符号的变量，把初始压缩量和重力势能的增加量都代入能量方程。

Write $x$ for the extension measured from the natural length, so the initial value is $-2a/5$. The particle accelerates upwards on release because the spring exerts an upward force of $8mg$. At the next turning point, conservation of energy gives

Cancelling $mg$ and multiplying by $5a$ gives

Hence

The negative root is the initial compression. The maximum extension is $3a/10$.

也可直接用质点离地的高度表示两种势能。先求最高点的高度，再减去弹簧自然长度。

Let $h$ be the length of the spring at a turning point, which is also the particle’s height above the plane. Taking gravitational potential energy to be zero at the plane, conservation of energy gives

The left side is $\frac{11}{5}mga$. Multiplying the resulting equation by $5a/(mg)$ gives $11a^2=5ah+50(h-a)^2$. Expanding and collecting terms gives

The root $h=3a/5$ corresponds to release. The upper turning point is $h=13a/10$, giving a maximum extension of $3a/10$.