一根原长为 $4a$、弹性模量为 $3mg$ 的弹性轻弹簧的一端连接于固定点 $A$。

质量为 $m$ 的质点连接于弹簧的另一端，竖直悬挂于 $A$ 下方。

将质点竖直向下拉至 $B$ 点，其中 $AB=7a$，然后由静止释放。

忽略空气阻力。

(a) 求质点刚由静止释放的瞬间的加速度。

(b) 通过考虑能量，用 $g$ 和 $a$ 表示质点位于 $A$ 下方距离 $5a$ 处时的速率。

质点首次瞬时静止于 $C$ 点，其中 $AC=ka$，$k$ 为常数。

(c) 通过考虑能量，求 $k$ 的值。

释放位置的伸长为 $7a-4a=3a$。先求弹簧拉力，再与重力作差，明确给出加速度方向。

At release the spring extension is $3a$, so Hooke’s law gives

Taking upwards as positive, Newton’s second law gives

Therefore the initial acceleration is

按题意使用能量。初始和最终位置都有弹性势能；质点向上移动 $2a$，所以重力势能增加。

At $B$ the extension is $3a$. At a distance $5a$ below $A$, the extension is $a$ and the rise from $B$ is $2a$. Conservation of energy gives

Therefore

Taking the positive root,

上端转折点可能越过自然长位置，因此用带符号伸长 $a(k-4)$ 的平方写弹性势能。二次方程有一个根是原释放点，必须选另一个转折点，并检查它对应压缩。

At $C$, the signed extension is $(k-4)a$, and the rise from $B$ is $(7-k)a$. Since the particle is instantaneously at rest at both positions, conservation of energy gives

Cancel $mga$ and multiply by $8$:

Expanding and collecting terms,

Factorising,

The root $k=7$ is the original release point $B$. The next turning point is therefore

Here $AC=11a/3<4a$, so the spring is compressed by $a/3$. The use of the squared compression in its elastic potential energy is consistent with the spring model.