质量为 $m$ 的质点 $P$ 连接于一根原长为 $a$、弹性模量为 $\lambda$ 的弹性轻弹簧的一端，另一端连接于天花板上的固定点 $A$。

点 $B$、$C$、$D$ 满足

• $B$ 位于 $A$ 正下方，$AB=\dfrac12a$；
• $C$ 位于 $A$ 正下方，$AC=a$；
• $D$ 位于 $A$ 正下方，$AD=3a$。

将质点 $P$ 竖直向下拉至 $D$ 后由静止释放。

已知 $P$ 在 $B$ 瞬时静止，

(a) 证明 $\lambda=\dfrac43mg$。

(b) 用 $a$ 和 $g$ 表示 $P$ 到达 $C$ 的瞬间的速率。

$P$ 在 $E$ 点达到最大速率。

(c) 证明 $AE=\dfrac74a$。

(d) 证明 $P$ 绕 $E$ 做简谐运动。

这是可以拉伸和压缩的弹簧。在下端点有伸长 $2a$，上端点有压缩 $a/2$，两端都有弹性势能；两端速率都是零。

At $D$, the extension is $3a-a=2a$. At $B$, the compression is $a-a/2=a/2$. Both are turning points, so conservation of energy from $D$ to $B$ gives

Consequently,

Cancelling $a>0$ and solving,

到达自然长位置 $C$ 时弹性势能为零，但质点从 $D$ 上升了 $2a$。把初始弹性势能分配为重力势能的增加与动能。

From $D$ to $C$, $P$ rises by $2a$, and at $C$ the spring is at its natural length. If its speed there is $v$, energy conservation gives

Hence

向上运动时，速度先增后减，最大速率处合力为零。用弹簧拉力等于重力求该处的伸长，再加上原长。

Let the extension at $E$ be $e$. At maximum speed the acceleration is zero, so the upward spring force equals the weight:

Using $\lambda=4mg/3$,

During the ascent from $D$, the upward acceleration is positive for extension greater than $3a/4$ and negative once the extension is less than $3a/4$. Thus this position gives the maximum speed. Therefore

也可先用能量表示任意位置的速率平方，再对位置求导找最大值。这条路线同时检查了压缩区域仍采用弹簧模型。

Let $x=AP-a$ be the signed extension; $x<0$ denotes compression. From $D$ to this position the rise is $2a-x$. Energy conservation gives

Simplifying,

The coefficient of $x^2$ is negative, and

This lies between the turning-point values $-a/2$ and $2a$ and gives the maximum of $v^2$. Hence

位移必须从平衡点 $E$ 量起。取向下为正，用带符号伸长统一表示弹簧的拉力和压力，最后说明加速度与位移反向且成正比。

Let $y$ be the downward displacement of $P$ from $E$. The signed extension of the spring is

Taking downward as positive, Newton’s second law gives

Thus

The acceleration is proportional to the displacement from $E$ and directed towards $E$. Therefore $P$ moves with SHM about $E$, with $\omega^2=4g/(3a)$. This remains valid during compression because the spring can push as well as pull.