一根弹性轻弹簧的原长为 $L$，弹性模量为 $4mg$。弹簧的一端连接于水平面上的 $O$ 点，另一端连接于质量为 $\dfrac{3m}2$ 的质点 $P$。如图 7 所示，质点静止平衡于 $E$ 点，弹簧竖直，且 $OE=\dfrac58L$。

随后将质点 $P$ 竖直向下压低距离 $\dfrac12L$，并在 $t=0$ 时由静止释放。

(a) 证明 $P$ 随后绕 $E$ 做周期为 $\pi\sqrt{\dfrac{3L}{2g}}$ 的简谐运动。

(b) 求 $P$ 首次向上运动且以大小为 $\dfrac23g$ 的加速度减速的瞬间的 $t$ 的精确值。

弹簧原本受压，取从 $E$ 向上的位移。压缩量为 $3L/8-x$，可作为有符号压缩量使用；超过原长后弹簧会向下拉，仍满足同一恢复方程。

Take displacement $x$ upwards from $E$. At equilibrium the spring’s compression is

At displacement $x$, its length is $5L/8+x$ and its signed compression is $3L/8-x$. The upward spring force is therefore

With particle mass $3m/2$, resolving upwards gives

Hence

The acceleration is proportional to the displacement from $E$ and directed towards $E$, so the motion is SHM with

The amplitude is $L/2$, because the particle is released from rest at $x=-L/2$. Its minimum height above the surface is $5L/8-L/2=L/8>0$. The spring can exert both a compressive and a tensile force, so the same force equation remains applicable throughout the oscillation.

Its period is

向上运动却在减速，说明加速度向下，即有符号加速度为 $-2g/3$。由恢复方程先求位置，再从最初的下端点选择第一次到达该位置的相位。

The amplitude is $L/2$. At release $x=-L/2$ and $\dot x=0$, so

Moving upwards and decelerating means $\dot x>0$ and $\ddot x=-2g/3$. Using part (a),

Substituting into the displacement equation,

During the first upward passage, $0<\omega t<\pi$, so $\dot x>0$. The first relevant phase is $\omega t=2\pi/3$. Therefore