$A$ 和 $B$ 是两个固定点，相距 $22a$，其中 $B$ 在 $A$ 的正下方。一根原长为 $4a$、模量为 $5mg$ 的轻弹性绳，一端连接在 $A$，另一端连接质量为 $km$ 的质点 $P$。另一根原长为 $8a$、模量为 $6mg$ 的轻弹性绳，一端连接在 $B$，另一端连接在 $P$。质点 $P$ 位于 $B$ 的正上方。

(a) 证明当系统处于平衡时，

将质点 $P$ 竖直向上拉动，使得 $BP=18a$，随后由静止释放。在其随后的运动中，$P$ 首次瞬时静止时所在位置满足 $BP=8a$。

(b) 求 $k$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要重力加速度的数值时，取 $g=10\,\mathrm{m\,s^{-2}}$。请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

设 $BP=x$。在 $8a<x<18a$ 时两根绳均伸长，上绳向上拉，下绳向下拉。用胡克定律求张力，再令合力为零。

Between the two stated rest positions $BP=8a$ and $BP=18a$, both strings are stretched. The total potential energy has equal values at the two rest positions, so it has a stationary point in the interior. This gives an equilibrium in the two-string range.

With $x=BP$, the extensions are $18a-x$ and $x-8a$. Balancing the upward tension with the downward tension and weight gives

Multiplying by $4a/(mg)$ gives $5(18a-x)=3(x-8a)+4ak$. Expanding gives $90a-5x=3x-24a+4ak$, hence $8x=114a-4ak$. Therefore

两处瞬时静止，动能都为零。出发时只有下绳储存弹性势能，到达 $BP=8a$ 时只有上绳储存弹性势能，重力势能减少 $10akmg$。

At $BP=18a$, the upper string is at its natural length and the lower string is stretched by $10a$. Using elastic potential energy $\lambda e^2/(2l)$, its elastic energy is

At $BP=8a$, the lower string is at its natural length and the upper string is stretched by $10a$. Its elastic energy is $\frac{5mg(10a)^2}{2(4a)}=\frac{125}{2}mga$. Taking $B$ as the gravitational datum, conservation of energy gives

Thus $10k=25$ and $\boxed{k=\frac52}$. Part (a) then gives $BP=13a$, which lies strictly between $8a$ and $18a$.