图 5 所示的两个固定点 $A$、$B$ 位于光滑水平地面上，相距 $5\,\mathrm{m}$。

质量为 $1.25\,\mathrm{kg}$ 的质点 $P$ 连接于一根原长为 $2\,\mathrm{m}$、弹性模量为 $20\,\mathrm{N}$ 的弹性轻绳的一端，另一端连接于 $A$。

第二根弹性轻绳的原长为 $1.2\,\mathrm{m}$，弹性模量为 $\lambda$ 牛顿，一端连接于 $P$，另一端连接于 $B$。

最初，$P$ 静止平衡于 $O$ 点，其中 $AO=3\,\mathrm{m}$。

(a) 证明 $\lambda=15$。

现在将质点沿地面向 $B$ 发射。

在 $t$ 秒时，$P$ 相对于 $O$ 沿 $OB$ 方向的位移为 $x$ 米。

(b) 证明两根绳均绷紧时，$P$ 做简谐运动，且 $\ddot{x}=-18x$。

$P$ 的初速率为 $10\,\mathrm{m\,s^{-1}}$。

(c) 求绳 $PB$ 变松的瞬间 $P$ 的速率。

一次完整振动中，两根绳均绷紧的时间为 $T$ 秒。

(d) 求 $T$ 的值。

静止时两根绳的张力相等，但原长、伸长和模量均不同，应分别用胡克定律。

Initially $AP=AO=3\,\mathrm{m}$ and $PB=5-3=2\,\mathrm{m}$. Hence the extensions of $AP$ and $PB$ are respectively

By Hooke’s law the tensions are

Horizontal equilibrium requires $T_A=T_B$, so

向 $B$ 移动时，左绳伸长增加、右绳伸长减少。恢复方程只适用于两绳同时绷紧的区间，不能直接把它扩展到整次振动。

When $P$ has displacement $x$ towards $B$, the two lengths and extensions are

While both strings are taut, $-1<x<0.8$. Resolving towards $B$,

Thus

The acceleration is proportional to the displacement from $O$ and directed towards $O$. Hence the motion is SHM, with $\omega=\sqrt{18}\,\mathrm{s^{-1}}$, while both strings remain taut.

初速度在中心给出简谐运动的幅度参数。右绳伸长减至零时 $x=0.8$；这早于此段简谐运动的形式端点。

For the initial SHM segment, let its amplitude parameter be $b$. At $O$, $x=0$ and the speed is $10\,\mathrm{m\,s^{-1}}$, so

The string $PB$ becomes slack when $PB=1.2\,\mathrm{m}$, that is,

Using $v^2=\omega^2(b^2-x^2)$ gives

Therefore

The parameter $b$ describes the solution while both strings are taut; it is not the amplitude of the complete motion after either string becomes slack.

能量法要把两根绳的弹性势能都计入。右绳刚松时弹性势能为零，但左绳的伸长已由 $1$ 增至 $1.8$。

Initially the total kinetic and elastic potential energy is

When $PB$ becomes slack, $x=0.8$, so the extension of $AP$ is $1.8\,\mathrm{m}$ and that of $PB$ is zero. Conservation of energy gives

Thus

Hence

两绳都绷紧的范围为 $-1<x<0.8$。用中心出发的正弦相位分别算中心到两个边界的时间；保守系统往返通过同一区间的速率相同，所以两段各经过两次。

The string $PB$ reaches its natural length at $x=0.8$, and $PA$ reaches its natural length when

Thus both strings are taut only for $-1<x<0.8$. Boundary instants do not change the total duration.

For the initial motion from $O$ towards $B$,

Let $t_R$ be the time from $O$ to the right-hand boundary. The first, acute phase gives

The system has no dissipative forces. On returning to $O$, $P$ therefore has speed $10\,\mathrm{m\,s^{-1}}$, now towards $A$. For this passage, let $s$ be time measured from that crossing. While both strings are taut,

The time $t_L$ from $O$ to the left-hand boundary $x=-1$ is consequently

During one complete oscillation, each of the intervals from $O$ to these boundaries is traversed once in each direction. The times are equal on the outward and return passages by conservation of energy. Hence

The inverse trigonometric values here are in radians. The portions of the oscillation with one string slack are not included in $T$.