质量为 $m$ 的质点 $P$ 连接于一根原长为 $l$ 的弹性轻绳的一端，另一端连接于天花板上的固定点。质点 $P$ 悬挂平衡于天花板下方距离 $D$ 处。

现在将 $P$ 竖直向下拉至距天花板 $3l$ 处，再由静止释放。

已知 $P$ 恰好在到达天花板前瞬时静止，

(a) 证明 $D=\dfrac{5l}3$。

(b) 证明弹性绳被拉伸时，$P$ 做周期为 $2\pi\sqrt{\dfrac{2l}{3g}}$ 的简谐运动。

(c) 用 $g$ 和 $l$ 表示从释放 $P$ 的瞬间到弹性绳首次变松的瞬间所用的精确时间。

先用释放点到天花板前静止点的能量求模量。最终绳已松弛，弹性势能为零；再用静止平衡时的张力求 $D$。

Let the modulus of elasticity be $\lambda$. At the release point the string length is $3l$, so its extension is $2l$. At the uppermost point just below the ceiling, the string is slack and the particle is instantaneously at rest.

The particle rises through $3l$. Conservation of energy between these two rest positions gives

Hence

At the equilibrium position, the extension is $D-l$ and the tension equals $mg$. Thus

It follows that

从平衡位置向下量位移，伸长为平衡伸长加位移。推导恢复方程时必须保留重力与张力，并明确这只适用于绳被拉伸的阶段。

Let $x$ be displacement vertically downwards from equilibrium. While the string is stretched, its extension is

Resolving downwards and using $\lambda=3mg/2$,

Therefore

The acceleration is proportional to the displacement from equilibrium and directed towards equilibrium. Hence the stretched-string motion is SHM with

Its SHM period is

This equation applies for $x>-2l/3$; once the string becomes slack, the particle is in free motion under gravity rather than following this SHM equation.

从最低端点静止释放，用余弦位移。松绳点位于平衡位置上方，因此目标位移是负的；首次相位应在第二象限。

At release, the distance below the ceiling is $3l$, so the displacement below equilibrium is

Thus, until the string first becomes slack,

The string first becomes slack when its length falls to $l$. The displacement from equilibrium then is

Consequently

The particle is moving upwards for the first time, with the phase increasing from $0$. Hence the first relevant phase is $\omega t=2\pi/3$, giving