方企勤 第三章 一元函数积分学 第3.4题

3.4.10 (1) 设第 $n$ 次击打后,桩被打进地下 ${x}_{n}$ ,第 $n$ 次击打时,汽锤所做的功为 ${w}_{n}\left( {n = 1,2,\cdots }\right)$ . 依题设,当桩被打进地下的深度为 $x$ 时,土层对桩的阻力为 ${kx}$ ,所以

$$ {W}_{1} = {\int }_{0}^{{x}_{1}}{kx}\mathrm{\;d}x = \frac{1}{2}k{x}_{1}^{2} = \frac{1}{2}k{a}^{2}, $$

$$ {W}_{2} = {\int }_{{x}_{1}}^{{x}_{2}}{kx}\mathrm{\;d}x = \frac{1}{2}k\left( {{x}_{2}^{2} - {x}_{1}^{2}}\right) = \frac{1}{2}k\left( {{x}_{2}^{2} - {a}^{2}}\right) . $$

又 ${W}_{2} = r{W}_{1} \Rightarrow {x}_{2}^{2} - {a}^{2} = r{a}^{2}$ ,即 ${x}_{2}^{2} = \left( {1 + r}\right) {a}^{2}$ . 进一步,因此有

$$ {W}_{3} = {\int }_{{x}_{2}}^{{x}_{3}}{kx}\mathrm{\;d}x = \frac{1}{2}k\left( {{x}_{3}^{2} - {x}_{2}^{2}}\right) = \frac{1}{2}k\left\lbrack {{x}_{3}^{2} - \left( {1 + r}\right) {a}^{2}}\right\rbrack . $$

再由 ${W}_{3} = r{W}_{2} \Rightarrow {r}^{2}{W}_{1} \Rightarrow {x}_{3}^{2} - \left( {1 + r}\right) {a}^{2} = {r}^{2}{a}^{2} \Rightarrow {x}_{3} = \sqrt{1 + r + {r}^{2}}a$ . 即汽锤击打桩 3 次后,可将桩打进地下 $\sqrt{1 + r + {r}^{2}}a\mathrm{\;m}$ .

(2)用数学归纳法,设 ${x}_{n} = \sqrt{1 + r + \cdots + {r}^{n - 1}}a$ ,则

$$ {W}_{n + 1} = {\int }_{{x}_{n}}^{{x}_{n + 1}}{kx}\mathrm{\;d}x = \frac{1}{2}k\left( {{x}_{n + 1}^{2} - {x}_{n}^{2}}\right) $$

$$ = \frac{1}{2}k\left\lbrack {{x}_{n + 1}^{2} - \left( {1 + r + \cdots + {r}^{n - 1}}\right) {a}^{2}}\right\rbrack . $$

由于

$$ {W}_{n + 1} = r{W}_{n} = {r}^{2}{W}_{n - 1} = \cdots = {r}^{n}{W}_{1} = \frac{1}{2}k{r}^{n}{a}^{2}, $$

故有

$$ {x}_{n + 1}^{2} - \left( {1 + r + \cdots + {r}^{n - 1}}\right) {a}^{2} = {r}^{n}{a}^{2} \Rightarrow $$

$$ {x}_{n + 1} = \sqrt{1 + r + \cdots + {r}^{n}}a = \sqrt{\frac{1 - {r}^{n + 1}}{1 - r}}a, $$

于是

$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n + 1} = \sqrt{\frac{1}{1 - r}}a. $$

即若击打次数不限,汽锤至多能将桩打进地下 $\sqrt{\frac{1}{1 - r}}a\mathrm{\;m}$ .