一根弹性轻弹簧的一端连接于固定点 $A$，另一端连接于质量为 $M$ 的质点。

弹簧的原长为 $3a$，弹性模量为 $\dfrac{18Mg}5$。

将质点静止保持在 $B$ 点，$B$ 位于 $A$ 正下方距离 $5a$ 处。

将质点由静止释放。

假设空气阻力可忽略。

利用机械能守恒原理

(a) 用 $a$ 和 $g$ 表示质点位于 $A$ 正下方距离 $3a$ 处的瞬间的速率；

(b) 用 $a$ 表示质点首次瞬时静止时位于 $A$ 下方的距离。

按题目要求用机械能守恒。初始弹簧伸长为 $2a$，到原长时弹性势能为零，同时质点已上升 $2a$，重力势能增加。

At release, the extension is $5a-3a=2a$. The initial elastic potential energy is

When the particle is $3a$ below $A$, the spring is at its natural length and stores no elastic energy. The particle has risen through $2a$. Conservation of mechanical energy gives

Consequently

在达到原长时仍有向上速度，所以会继续上升并压缩弹簧。设压缩量为正，再从释放点写到下一静止点的能量；最后将压缩量换成到 $A$ 的距离。

The speed in part (a) is non-zero, so the particle continues upwards beyond the natural-length position. Let the compression at the next turning point be $e>0$. Its distance below $A$ is then $3a-e$, and its rise from the release point is $2a+e$.

By conservation of mechanical energy from release to this rest position,

Dividing by $Mg$ and multiplying by $5a$ gives

so

The positive compression is $e=a/3$. The root $e=-2a$ represents the original extended release position and is not the next turning point.

Therefore the required distance below $A$ is