假设傅科摆位于纬度\(\alpha\), 摆长为\(l\), 绳末悬挂着一个质量为\(m\)的大小忽略不计的小球.地球以角速度\(\omega\)自转. 建立\(O\)-\(\!\)-\(xyz\)坐标系, 其中\(O\)为摆的平衡点, \(x\)轴指向东, \(y\)轴指向北, \(z\)轴竖直向上指向天空(如图1所示).假设来自空气的阻尼力正比于摆的速度, 即\(f=kv\), \(k\)为阻尼系数. 而科里奥利力 大小为 \(2mv\omega \sin \alpha \).

图1

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	图1

我们已假定摆绳足够长, 而摆角足够小, 因此不考虑阻尼力和科里奥利力时, 摆在\(O\)-\(\!\)-\(xy\)平面内做简谐振动. 即方程

\[ \left.\begin{array} \ddot {x} = - \dfrac{g}{l}x \\ \ddot {y} = - \dfrac{g}{l}y \end{array} \right\} \ \ (1)\]

考虑阻尼力和科里奥利力

\[\left.\begin{array} \ddot {x} + \dfrac{k}{m}\dot {x} + \dfrac{g}{l}x - 2\omega \dot {y}\sin\alpha = 0 \\ \ddot {y} + \dfrac{k}{m}\dot {y} + \dfrac{g}{l}y + 2\omega \dot {x}\sin\alpha = 0 \end{array} \right\} \ \ (2)\]

将方程组(2) 的第2个式子乘以复数i, 再加到第一个式子里去, 方程(2)可转变为一个复微分方程

\[ \ddot {x} + {\rm i}\ddot {y} + \dfrac{k}{m}\left( {\dot {x} + {\rm i}\dot {y}} \right) +\dfrac{g}{l}\left( {x + {\rm i}y} \right) + \\ \qquad 2\omega \left( {{\rm i}\dot {x} - \dot {y}}\right)\sin \alpha = 0 \ \ (3)\]

作变量代换

\[\xi = x + {\rm i} y\]

得到

\[\ddot {\xi } + \left( {2{\rm i}\omega \sin \alpha + \dfrac{k}{m}} \right)\dot {\xi } + \dfrac{g}{l}\xi = 0\ \ (4)\]

方程(4) 是二阶线性齐次常微分方程, 所以方程(4)的通解为

\[\xi = A {\rm e}^{\lambda _1 t} + B {\rm e}^{\lambda _2 t} \ \ (5)\]

其中\(\lambda \)满足

\[\lambda ^2 + \left( {2{\rm i}\omega \sin \alpha + \dfrac{k}{m}} \right)\lambda + \dfrac{g}{l} = 0 \ \ (6)\]

方程(6)的解是

\[\left.\begin{array}{l} \lambda _1 = - \left( {{\rm i}\omega \sin \alpha + \dfrac{k}{2m}} \right) - \left({P - Q{\rm i}} \right) \\ \lambda _2 = - \left( {{\rm i}\omega \sin \alpha + \dfrac{k}{2m}} \right) + \left({P - Q{\rm i}} \right) \end{array}\!\! \right\} \ \ (7)\]

其中\(P, Q\)满足

\[4\left( {P - Q{\rm i}} \right)^2 = \left( {2{\rm i}\omega \sin \alpha + \dfrac{k}{m}} \right)^2 -4\dfrac{g}{l}\]

即

\[\left.\begin{array}{l} P^2 - Q^2 = - \omega ^2\sin ^2\alpha + \dfrac{k^2}{4m^2} - \dfrac{g}{l} \\ 2PQ = \dfrac{k}{m}\omega \sin \alpha \end{array}\right\}\]

即

\[\left. P = \left[ {\left( { - \omega ^2\sin ^2\alpha + \dfrac{k^2}{4m^2} -\dfrac{g}{l}} \right)^2 \!+\! \left( {\dfrac{k}{m}\omega \sin \alpha } \right)^2} \right]^{\tfrac 14} \!\cdot\\ \qquad \sin \left( {\dfrac{1}{2}\arctan \dfrac{\dfrac{k}{m}\omega \sin \alpha }{ - \omega ^2\sin^2\alpha + \dfrac{k^2}{4m^2} - \dfrac{g}{l}}} \right) \\ Q = \left[ {\left( { - \omega ^2\sin ^2\alpha + \dfrac{k^2}{4m^2} -\dfrac{g}{l}} \right)^2 \!+\! \left( {\dfrac{k}{m}\omega \sin \alpha } \right)^2} \right]^{\tfrac 14} \!\cdot\\ \qquad \cos \left( {\dfrac{1}{2}\arctan \dfrac{\dfrac{k}{m}\omega \sin \alpha }{ - \omega ^2\sin^2\alpha + \dfrac{k^2}{4m^2} - \dfrac{g}{l}}} \right) \!\!\!\right\} \ \ (8)\]

因此, 方程(5) 可被表示为

\[ \xi = A {\rm e}^{\left[ { - \left( { {\rm i}\omega \sin \alpha + \tfrac{k}{2m}} \right) - \left( {P - Q{\rm i}} \right)} \right] t} + \\ \qquad B{\rm e}^{\left[ { - \left( {{\rm i}\omega \sin \alpha + \tfrac{k}{2m}} \right) + \left( {P - Q{\rm i}} \right)} \right]t} = \\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}\left[ {A{\rm e}^{ - Pt}{\rm e}^{{\rm i}Qt}+ B{\rm e}^{Pt}{\rm e}^{ - {\rm i}Qt}} \right] {\rm e}^{ - {\rm i}\omega t\sin \alpha } \ \ (9)\]

定义\(\omega _1 = Q\), \(\omega _2 = \omega \sin \alpha \), 则

\[ \xi = {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A{\rm e}^{ - Pt} + B{\rm e}^{Pt}} \right)\cos\omega _1 t\cos \omega _2 t + }\right.\\ \qquad \left.{\left( {A{\rm e}^{ - Pt} - B{\rm e}^{Pt}}\right)\sin \omega _1 t\sin \omega _2 t} \right] + \\ \qquad {\rm i}{\rm e}^{ - \tfrac{k}{2m}t}\left[{\left( {A{\rm e}^{ - Pt} - B{\rm e}^{Pt}} \right)\sin \omega _1 t\cos \omega _2 t - }\right. \\ \qquad \left.{\left( {A{\rm e}^{ - Pt} + B{\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right]\ \ (10)\]

虑及\(\xi = x + {\rm i}y\), \(A = A_1 + A_2 {\rm i}\), \(B = B_1 + B_2 {\rm i}\), 这里\(A_1 , A_2, B_1, B_2 \)均为实数, 得到方程(2)的解

\[\left. x = {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_1 {\rm e}^{ - Pt} + B_1 {\rm e}^{Pt}}\right)\cos \omega _1 t\cos \omega _2 t + }\right.\\ \qquad \left.{\left( {A_1 {\rm e}^{ - Pt} - B_1 {\rm e}^{Pt}} \right)\sin \omega _1 t\sin \omega _2 t} \right] -\\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}\left[{\left( {A_2 {\rm e}^{ - Pt} - B_2 {\rm e}^{Pt}} \right)\sin \omega _1 t\cos \omega _2 t - }\right. \\ \qquad \left.{\left( {A_2 {\rm e}^{-Pt} + B_2 {\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right]\\ y = {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_2 {\rm e}^{ - Pt} + B_2 {\rm e}^{Pt}} \right)\cos\omega _1 t\cos \omega _2 t +}\right.\\ \qquad \left.{\left( {A_2 {\rm e}^{ - Pt} - B_2 {\rm e}^{Pt}}\right)\sin \omega _1 t\sin \omega _2 t} \right] +\\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_1 {\rm e}^{ - Pt} - B_1 {\rm e}^{Pt}}\right)\sin \omega _1 t\cos \omega _2 t - }\right. \\ \qquad \left.{\left( {A_1 {\rm e}^{ - Pt} + B_1{\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right] \!\!\right\} \ \ (11)\]

其中 \(A_1, A_2, B_1, B_2 \) 由初始条件决定. 至此, 我们 解出了阻尼傅科摆的运动方程.

而参考文献[1] 式(7) 与本文式(10)一致, 但文献[1]的作者在从式(7)推到式(8)的过程中, 忽略了\(A, B\)是复数这一重要问题, 因此文献[1]得到的式(8)是不正确的, 与本文的式(11)不一致.其实, 确定摆在平面的运动轨迹所需的初始条件 共有4个：\( x |_{t = 0} \), \( y |_{t = 0} \), \( \dot {x} |_{t = 0}\), \( \dot {y} |_{t = 0} \), 那么易知\(x, y\)的最终解析表达式里也应出现4个积分常数, 而非2个.

注意到

\[\dfrac{\partial \left( {x^2 + y^2} \right)}{\partial \omega _2 } = 0 \ \ (12)\]

式中的\(\omega _2\) 是摆平面的旋转角频率^[2], 而 \(\omega _1 \)则是摆球在摆平面内运动的角频率.在此我们完整地写出\(\omega _1 \)和\(\omega _2 \)的表达式以示强调.

\[\left.\begin{array}{l} \omega _1 = \left[ {\left( {\dfrac{k^2}{4m^2} - \omega ^2\sin ^2\alpha -\dfrac{g}{l}} \right)^2 \!+\! \left( {\dfrac{k\omega \sin \alpha }{m}} \right)^2} \right]^{\tfrac 14} \cdot\\ \quad \cos \left( {\dfrac{1}{2}\arctan\dfrac{\dfrac{k\omega \sin \alpha }{m}}{\dfrac{k^2}{4m^2} - \omega ^2\sin ^2\alpha - \dfrac{g}{l}}} \right)\\ \omega _2 = \omega \sin \alpha \end{array}\!\! \right\} \ \ (13)\]

其中\(k\)为阻尼系数, \(m\)为摆球质量, \(\omega \) 为地球自转角频率, \(\alpha \) 为纬度, \(g\)为重力加速度, \(l\)为摆线长. 据上式易知\(\omega _1 \) 为如下因素所决定：\(k/m\); \(\omega _2 \) 则只由地球自转速度\(\omega\)和纬度\(\alpha \)所决定.

接下来我们将讨论单摆的一些理想情形是如何包含在式(11)里的.

(1) 若\(k=0\), \(\omega \sin \alpha =0\), 简化为简谐振动

\[\left.\begin{array}{l} x = C_1 \cos \left( {\omega _1 t + \varphi _1 } \right) \\ y = C_2 \sin \left( {\omega _1 t + \varphi _2 } \right) \end{array} \right \} \ \ (14)\]

其中\(\omega _1 = \left( {g / l} \right)^{0.5}\), \(C_1 , C_2, \varphi _1, \varphi _2 \) 为由初始条件决定的常数.

(2) 若只有 \(\omega \sin \alpha =0\), 则简化为阻尼摆

\[\left.\begin{array}{l} x = {\rm e}^{ - \tfrac{k}{2m}t}C_1 \cos \left( {\omega _1 t + \phi _1 } \right) \\ y ={\rm e}^{ - \tfrac{k}{2m}t}C_2 \sin \left( {\omega _1 t + \phi _2 } \right) \end{array}\!\! \right \} \ \ (15)\]

其中 \(\omega _1 = \sqrt {\left| {\dfrac{k^2}{4m^2} - \dfrac{g}{l}} \right|} \) .

(3) 若只有\(k=0\), 则简化为无阻尼傅科摆

\[\left.\begin{array}{l} x = \left[ {\left( {A_1 + B_1 } \right)\cos \omega _1 t\cos \omega _2 t +}\right. \\ \qquad \left.{\left( {A_1 - B_1 } \right)\sin \omega _1 t\sin \omega _2 t} \right] - \\ \qquad \left[{\left( {A_2 - B_2 } \right)\sin \omega _1 t\cos \omega _2 t - }\right. \\ \qquad \left.{\left( {A_2 +B_2 } \right)\cos \omega _1 t\sin \omega _2 t} \right] \\ y = \left[ {\left( {A_2 + B_2 } \right)\cos \omega _1 t\cos \omega _2 t + }\right. \\ \qquad \left.{\left( {A_2 - B_2 } \right)\sin \omega _1 t\sin \omega _2 t} \right] + \\ \qquad \left[ {\left( {A_1 - B_1 } \right)\sin \omega _1 t\cos \omega_2 t - }\right. \\ \qquad \left.{\left( {A_1 + B_1 } \right)\cos \omega _1 t\sin \omega _2 t} \right] \end{array} \right\} \ \ (16)\]

其中, \(\omega _1 = \sqrt {\left( {\omega \sin \alpha } \right)^2 + \dfrac{g}{l}} \) .

由式(8), 式(9)和式(11), 我们注意到

\[\left.\begin{array}{l} x = x\left( {t, \dfrac{k}{m}, \dfrac{g}{l}, \omega \sin \alpha, A, B} \right) \\ y = y\left( {t, \; \; \dfrac{k}{m}, \dfrac{g}{l}, \omega \sin \alpha, A, B} \right) \end{array} \right\} \ \ (17)\]

其中\(A, B\)为积分常数(复数), \(k\)为阻尼系数, \(m\)为摆球质量, \(\omega \) 为地球自转速度, \(\alpha \)为纬度, \(g\)为重力加速度, \(l\)为摆长. 所以在不同的条件下, 会得到不同的运动轨迹. 接下来就将对此进行讨论.

作为积分常数, \(A, B\)由初始条件决定, 即\(\left. x \right|_{t = 0} \), \(\left. y \right|_{t = 0} \), \(\left. \dot{x} \right|_{t = 0} \), \(\left. \dot {y} \right|_{t = 0} \) . 由式(11), 我们有

\[\left.\begin{array}{l}\!\! x = {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_1 {\rm e}^{ - Pt} + B_1 {\rm e}^{Pt}}\right)\cos \omega _1 t\cos \omega _2 t + }\right.\\ \qquad \left.{\left( {A_1 {\rm e}^{ - Pt} - B_1 {\rm e}^{Pt}} \right)\sin \omega _1 t\sin \omega _2 t} \right] - \\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_2 {\rm e}^{ - Pt} - B_2 {\rm e}^{Pt}}\right)\sin \omega _1 t\cos \omega _2 t - }\right. \\ \qquad \left.{ \left( {A_2 {\rm e}^{ - Pt} + B_2{\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right] \\ y = {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_2 {\rm e}^{ - Pt} + B_2 {\rm e}^{Pt}}\right)\cos \omega _1 t\cos \omega _2 t + }\right. \\ \qquad \left.{\left( {A_2 {\rm e}^{ - Pt} - B_2{\rm e}^{Pt}} \right)\sin \omega _1 t\sin \omega _2 t} \right] + \\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_1 {\rm e}^{ - Pt} - B_1 {\rm e}^{Pt}}\right)\sin \omega _1 t\cos \omega _2 t - }\right. \\ \qquad \left.{\left( {A_1 {\rm e}^{ - Pt} + B_1{\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right]\end{array} \right\} \ \ (18)\]

\[\left.\begin{array}{l} \dot {x} = \left\{ { - \dfrac{k}{2m}{\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_1{\rm e}^{ - Pt} + B_1 {\rm e}^{Pt}} \right)\cos \omega _1 t\cos \omega _2 t + \left( {A_1 {\rm e}^{ - Pt} -B_1 {\rm e}^{Pt}} \right)\sin \omega _1 t\sin \omega _2 t} \right]} \right. +\\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}A_1 \left[ { - P{\rm e}^{ - Pt}\cos\left( {\omega _1 -\omega _2 } \right)t - \left( {\omega _1 -\omega _2 } \right) {\rm e}^{ - Pt}\sin \left({\omega _1 -\omega _2 } \right)t} \right]+\\ \qquad \left. { {\rm e}^{ - \tfrac{k}{2m}t}B_1 \left[ {P{\rm e}^{Pt}\cos\left( {\omega _1 +\omega _2 } \right)t - \left( {\omega _1 + \omega _2 } \right) {\rm e}^{Pt}\sin \left({\omega _1 +\omega _2 } \right)t} \right]\; } \right\} - \\ \qquad \left\{ { - \dfrac{k}{2m}{\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_2{\rm e}^{ - Pt} - B_2 {\rm e}^{Pt}} \right)\sin \omega _1 t\cos \omega _2 t - \left( {A_2 {\rm e}^{ - Pt} +B_2 {\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right]} \right. + \\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}A_2 \left[ { - P{\rm e}^{ - Pt}\sin \left({\omega _1 - \omega _2 } \right)t + \left( {\omega _1 - \omega _2 } \right) {\rm e}^{ - Pt}\cos \left({\omega _1 - \omega _2 } \right)t} \right] -\\ \qquad \left. { {\rm e}^{ - \tfrac{k}{2m}t}B_2 \left[ {P{\rm e}^{Pt}\sin \left({\omega _1 + \omega _2 } \right)t + \left( {\omega _1 + \omega _2 } \right) {\rm e}^{Pt}\cos \left( {\omega_1 + \omega _2 } \right)t} \right]} \right\} \\ \dot {y} = \left\{ { - \dfrac{k}{2m}{\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_1{\rm e}^{ - Pt} - B_1 {\rm e}^{Pt}} \right)\sin \omega _1 t\cos \omega _2 t - \left( {A_1 {\rm e}^{ - Pt} +B_1 {\rm e}^{Pt}} \right)\cos \omega _1 t\sin \omega _2 t} \right]} \right. + \\ {\rm e}^{ - \tfrac{k}{2m}t}A_1 \left[ { - P{\rm e}^{ - Pt}\sin \left({\omega _1 - \omega _2 } \right)t + \left( {\omega _1 - \omega _2 } \right) {\rm e}^{ - Pt}\cos \left({\omega _1 - \omega _2 } \right)t} \right] -\\ \qquad \left. { {\rm e}^{ - \tfrac{k}{2m}t}B_1 \left[ {P{\rm e}^{Pt}\sin \left({\omega _1 + \omega _2 } \right)t + \left( {\omega _1 + \omega _2 } \right) {\rm e}^{Pt}\cos \left( {\omega_1 + \omega _2 } \right)t} \right]} \right\} + \\ \qquad \left\{ { - \dfrac{k}{2m}{\rm e}^{ - \tfrac{k}{2m}t}\left[ {\left( {A_2{\rm e}^{ - Pt} + B_2 {\rm e}^{Pt}} \right)\cos \omega _1 t\cos \omega _2 t + \left( {A_2 {\rm e}^{ - Pt} -B_2 {\rm e}^{Pt}} \right)\sin \omega _1 t\sin \omega _2 t} \right]} \right. +\\ \qquad {\rm e}^{ - \tfrac{k}{2m}t}A_2 \left[ { - P{\rm e}^{ - Pt}\cos\left( {\omega _1 -\omega _2 } \right)t - \left( {\omega _1 - \omega _2 } \right) {\rm e}^{ - Pt}\sin \left({\omega _1 -\omega _2 } \right)t} \right] + \\ \qquad \left. { {\rm e}^{ - \tfrac{k}{2m}t}B_2 \left[ {P{\rm e}^{Pt}\cos\left( {\omega _1 +\omega _2 } \right)t - \left( {\omega _1 + \omega _2 } \right) {\rm e}^{Pt}\sin \left({\omega _1 +\omega _2 } \right)t} \right]\; } \right\} \end{array} \!\! \right \} \ \ (19)\]

对于\(t=0\),

$$\left.\begin{array}{l} \left. x \right|_{t = 0} = A_1 + B_1 , \ \ \ \left. y \right|_{t = 0} = A_2 + B_2 \\ \left. \dot {x} \right|_{t = 0} = \left( { - \dfrac{k}{2m} - P} \right)A_1 +\left( { - \dfrac{k}{2m} + P} \right)B_1 - \\ \qquad\left( {\omega _1 - \omega _2 } \right)A_2 + \left( {\omega _1 + \omega _2 } \right)B_2\\ \left. \dot {y}\right|_{t = 0} = \left( { - \dfrac{k}{2m} - P} \right)A_2 + \left( { - \dfrac{k}{2m} + P} \right)B_2 +\\ \qquad\left( {\omega _1 - \omega _2 } \right)A_1 - \left( {\omega _1 + \omega _2 } \right)B_1 \end{array}\!\!\right\}\ \ (20)$$

出于对称性的考虑, 我们不妨令\(\left. y \right|_{t = 0} =0\)(因为无论初始位置如何, 我们都可以将图1的坐标系绕着\(z\)轴旋转, 使得 \(\left. y \right|_{t = 0} = 0\).并且显然在旋转后的坐标系下, 运动方程(2)不变), 再求解方程(20)中的\(A_1, A_2, B_1, B_2 \).

\[\left.\begin{array}{l} \left. x \right|_{t = 0} = A_1 + B_1 \\ \left. y \right|_{t = 0} = 0 \\ \left. \dot {x} \right|_{t = 0} + \left( {\dfrac{k}{2m} + P} \right)x =2PB_1 + 2\omega _1 B_2 \\ \left. \dot {y} \right|_{t = 0} - \left( {\omega _1 - \omega _2 } \right)x= - 2\omega _1 B_1 + 2PB_2 \end{array}\!\! \right \} \ \ (21)\]

\[\left.\begin{array}{l} B_1 = \Bigg\{P\left[ {\left. \dot {x} \right|_{t = 0} + \left( {\dfrac{k}{2m}+ P} \right) x |_{t = 0} } \right] - \\ \qquad \omega _1 \left[ { \dot {y} |_{t = 0} - \left( {\omega_1 - \omega _2 } \right) x |_{t = 0} } \right]\Bigg\} \Bigg/ \Big (2P^2 + 2\omega _1 ^2 \Big)\\ B_2 = \Bigg \{\omega _1 \left[ { \dot {x} |_{t = 0} + \left({\dfrac{k}{2m} + P} \right) x |_{t = 0} } \right] +\\ \qquad P\left[ { \dot {y} |_{t = 0} - \left({\omega _1 - \omega _2 } \right)\left. x \right |_{t = 0} } \right]\Bigg\} \Bigg/ \Big (2\omega _1 ^2 + 2P^2\Big) \\ A_1 = x |_{t = 0} - B_1 \\ A_2 = - B_2 \end{array} \right\} \ \ (22)\]

至此, 方程(11) 可被表示为如下形式

\[\left.\begin{array}{l} x = x\left( {\left. {t, \; x} \right|_{t = 0} , \; \left. \dot {x} \right|_{t =0} , \; \left. \dot {y} \right|_{t = 0} , \dfrac{k}{m}, \; \dfrac{g}{l}, \; \omega \sin \alpha } \right) \\ y = y\left( {\left. {t, \; x} \right|_{t = 0} , \; \left. \dot {x} \right|_{t =0} , \left. \dot {y} \right|_{t = 0} , \dfrac{k}{m}, \dfrac{g}{l}, \; \omega \sin \alpha } \right) \end{array} \!\! \right \} \ \ (23)\]

在此定义

\[ {\pmb r} \equiv \left( {x, y} \right) \equiv \\ \quad {\pmb r}\left( {\left. {t, x} \right|_{t = 0}, \left. \dot {x} \right|_{t = 0} , \left. \dot {y} \right|_{t = 0} , \dfrac{k}{m}, \; \dfrac{g}{l}, \omega\sin \alpha } \right) \ \ (24)\]

借助计算机, 可以给定不同的条件, 简要地分析摆的运动轨道. 下文中将假定\(\dfrac{g}{l} = 0.1\) s\(^{ - 2}\) , 这表明重力加速度为10 N/kg, 而摆长约100 m; 以及地球自转角频率为现实生活中的 \(\omega = \dfrac{{2}\pi}{86 164 s} = 7.29\times 10^{ - 5}\) s\(^{ - 1}\). 模拟从\(t=0\)到\(t=2 000\) s 的运动轨道.

(1) 固定 \(\dfrac{k}{m} = 0.005\) s\(^{ - 1}\), \(\alpha = \dfrac{\pi }{3}\), 对不同的

\[\left[ {\left. x \right|_{t = 0} , \; \left. \dot {x} \right|_{t = 0} , \; \left. \dot {y}\right|_{t = 0} } \right]\]

显然

\[ {\pmb r} \left( {\left. {t, {\rm a} x} \right|_{t = 0} , \left. { {\rm a}\dot {x}} \right|_{t = 0}, \left. { {\rm a}\dot {y}} \right|_{t = 0} } \right) =\\ \qquad {\rm a}{\pmb r }\left( {\left. {t, x}\right|_{t = 0} , \; \left. \dot {x} \right|_{t = 0} , \left. \dot {y} \right|_{t = 0} } \right) \ \ (25)\]

a为一个常数. 所以轨道的形状是由\(\left. x \right|_{t = 0}\), \(\left. \dot {x} \right|_{t = 0}\), \(\left. \dot{y} \right|_{t = 0} \) 的相对比例决定的. 对于 \(\left. x \right|_{t = 0} \ne 0\), 我们不妨假定\(\left. x\right|_{t = 0} = 1\). 另外, \(\left. \dot {x} \right|_{t = 0} , \; \left. \dot {y} \right|_{t = 0} \)不应设置得过大, 否则将不满足微小摆角的近似.

定参数为 [1, 0, 0.5] 及[1, 0, 5]. 如图2所示, 当摆以一个\(y\)方向的小速度开始摆动时, 轨道为一个十分漂亮的螺线.这说明科里奥利力影响较小, 摆近似为只有阻尼的情形.而图3中的螺线明显倾斜, 正是因为有更大的速度, 更大的科里奥利力.

而若加上\(x\)方向的速度, 则轨道倾斜, 如图4.

图2

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	图2 参数为(1, 0, 0.5)时摆的轨道

图3

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	图3 参数为(1, 0, 5)时摆的轨道

图4

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	图4 加上\(x\)方向速度后摆的轨道

对于\(\left[ {0, \left. \dot {x} \right|_{t = 0} , \left. \dot {y} \right|_{t = 0} }\right]\)的情形, 初始位置在坐标原点, \(x\)轴、\(y\)轴又是对称的. 不 妨设为\(\left[ {0, \left. \dot {x} \right|_{t= 0} , \; 0} \right]\), 轨道近似于\(\left[ {1, \left. \dot {x} \right|_{t = 0} , \; 0} \right]\), 如图5所示.

在\(y\)轴方向上的偏移十分小, 不过摆平面倒确实旋转了一些, 如图6.

(2) 固定 \(\left[ {\left. x \right|_{t = 0} , \left. \dot {x} \right|_{t = 0} , \left. \dot {y} \right|_{t= 0} } \right] = \left[ {1, 0, 0} \right]\), \(\dfrac{k}{m} =0.005\) s\(^{ - 1}\), 改变\(\alpha \)和改变\(\omega \)效果相同, 因为

\[ {\pmb r} \equiv \left( {x, y} \right) \equiv \ \quad {\pmb r} \left( {\left. {t, \; x} \right|_{t =0} , \left. \dot {x} \right|_{t = 0} , \left. \dot {y} \right|_{t = 0}, \dfrac{k}{m}, \dfrac{g}{l}, \omega \sin \alpha } \right) \ \ (26)\]更低的纬度与更慢的地球转速等效, 它们都给出了更小的科里奥利力以及更小的 \(y\)轴偏移, 如图7和图8所示.

图5

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	图5 参数为(1, 0, 0)时摆的轨道

图6

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	图6 参数为(0, 1, 0)时摆的轨道

图7

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	图7 纬度为10\(^\circ\)N 处摆的轨道

图8

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	图8 纬度为80\(^\circ\)N处摆的轨道

另外, 对于远大于平常的科里奥利力, 我们会得到一些完全不同的有趣轨道 (如图9). 类似于混沌现象, 轨道的形状是千奇百怪的. 不过在此我们不作过多讨论, 毕竟这也与地球实际状况不符.

图9

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	图9 \(w\sin \alpha =0.07\)时摆的轨道

(3) 固定 \(\left[ {\left. x \right|_{t = 0} , \left. \dot {x} \right|_{t = 0} , \left. \dot {y} \right|_{t= 0} } \right] = \left[ {1, 0, 1} \right]\), \(\alpha = \dfrac{\pi }{3}\), \( {k}/{m}\)不同

显然研究\(k/m\)是有意义的. 在不同的介质中, 比如水, 或者对于不同质量的小球, 会得到不同的\(k/m\).

一般说来, 更大的 \(k/m\) 使得摆更快地停止. 如图10所示, \(k/m=0.05\)得到的螺线与\(k/m=0.005\)的图2不同.

图10

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	图10 \(k/m=0.05\) s\(^{-1}\) 时摆的轨道

本文重在完整地解出了阻尼傅科摆的运动方程, 也改正了文献[1]中的错误. 而关于其轨道性质的研究还有待深入进行.

致谢：感谢南京大学吴盛俊教授对本文的帮助与指导.