非小角单摆(数学摆)的周期

求解了非小角单摆(数学摆)的周期,破除中学物理5度神话。是不是可以看成单摆,不是看是不是摆角小于5度,而是看计时仪器的精度。

参考:

  • Moshe Gitterman,THE CHAOTIC PENDULUM,Chaper 1
  • 安徽师范大学陆同兴教授:非线性振动初步

如图所示,体系由一个轻质细杆和末端带有的质点组成,细杆长$l$,质点质量为$m$。


图为数学摆

设摆角为$\phi$,忽略所有阻力和摩擦力,有

\begin{equation*} ml^2\frac{d^2\phi}{dt^2}=-mgl\sin\phi \end{equation*}

整理得

\begin{equation*} \frac{d^2\phi}{dt^2}+\frac{g}{l}\sin\phi=0 \end{equation*}

令$\omega_0=\sqrt{g/l}$,上式即为

\begin{equation} \frac{d^2\phi}{dt^2}+\omega_0^2\sin\phi=0 \tag{1}\label{newtondyn} \end{equation}

上式两边乘以$\frac{d\phi}{dt}$,

\begin{equation*} \begin{split} &\frac{d\phi}{dt}\cdot\frac{d^2\phi}{dt^2}+\omega_0^2\sin\phi\frac{d\phi}{dt}=\frac{d\phi}{dt}\cdot\frac{d}{dt}\left (\frac{d\phi}{dt}\right )-\omega_0^2\frac{d\cos\phi}{dt}\\ &\frac{1}{2}\frac{d}{dt}\left (\frac{d\phi}{dt}\right )^2-\omega_0^2\frac{d\cos\phi}{dt}=0 \end{split} \end{equation*}

上式对$t$积分,得

\begin{equation} E=\frac{1}{2}\left (\frac{d\phi}{dt}\right )^2+\omega_0^2(1-\cos\phi) \tag{2}\label{energy} \end{equation}

积分常数$E$的物理意义是摆的能量。$E$选得摆在最低点时势能为0。

在最大角位移$\phi_0$处,角速度$\frac{d\phi}{dt}=0$,代入\eqref{energy}式,得

\begin{equation} E= \omega_0^2(1-\cos\phi_0) \tag{3}\label{energyexpl} \end{equation}

将上式代入\eqref{energy}式,得

\begin{equation} \frac{d\phi}{dt} =\omega_0\sqrt{2(\cos\phi-\cos\phi_0)} \tag{4}\label{dphidt} \end{equation}

设$t=0$时,$\phi=0$,经过1/4周期,$t=T/4$时,摆到达最大位移$\phi=\phi_0$。对\eqref{dphidt}式积分,有

\begin{equation*} \int_0^{T/4}\omega_0 dt=\int_0^{\phi_0}\frac{d\phi}{\sqrt{2(\cos\phi-\cos\phi_0)}} \end{equation*}

\begin{equation} \begin{split} \frac{\omega_0 T}{4}=&\int_0^{\phi_0}\frac{d\phi}{\sqrt{2(\cos\phi-\cos\phi_0)}}\\ =&\int_0^{\phi_0}\frac{d\phi}{2\sqrt{\sin^2(\phi_0/2)-\sin^2(\phi/2)}} \end{split} \tag{5}\label{intphi} \end{equation}

上式用到了$\cos\phi=1-2\sin^2(\phi/2)$。

\begin{equation} \sin(\phi/2)=\sin(\phi_0/2)\sin\psi \tag{6}\label{newvar} \end{equation}

$\phi$的变化范围是从0到$\phi_0$,那么$\psi$的变化范围是从0到$\pi/2$。再考虑到$\cos(\phi/2)/2d\phi=\sin(\phi_0/2)\cos\psi d\psi$,代入\eqref{intphi}有

\begin{equation} \begin{split} \frac{\omega_0 T}{4}=&\int_0^{\phi_0}\frac{d\phi}{2\sqrt{\sin^2(\phi_0/2)-\sin^2(\phi/2)}}\\ =&\int_0^{\phi}\frac{d\phi}{2\sin(\phi_0/2)\cos\psi }\\ =&\int_0^{\pi/2}\frac{d\psi}{\cos(\phi/2)}\\ =&\int_0^{\pi/2}\frac{d\psi}{\sqrt{1-\sin^2(\phi_0/2)\sin^2\psi}} \end{split} \tag{7}\label{intpsi} \end{equation}

将上式中的被积函数展开成级数

\begin{equation*} \begin{split} \left [1-\sin^2(\phi_0/2)\sin^2\psi \right ]^{-0.5}=&1+\frac{1}{2}\sin^2(\phi_0/2)\sin^2\psi + \\ &\frac{3}{8}\sin^4(\phi_0/2)\sin^4\psi +\cdots \end{split} \end{equation*}

代入\eqref{intpsi},右边

\begin{equation} \begin{split} & \int_0^{\pi/2}\frac{d\psi}{\sqrt{1-\sin^2(\phi_0/2)\sin^2\psi}}=\\ & \int_0^{\pi/2}\left [1+\frac{1}{2}\sin^2(\phi_0/2)\sin^2\psi +\frac{3}{8}\sin^4(\phi_0/2)\sin^4\psi \right ] d\psi = \\ & \frac{\pi}{2}+\frac{\pi}{8}\sin^2(\phi_0/2)+\frac{9}{128}\sin^4(\phi_0/2)+\cdots \end{split} \tag{8}\label{right} \end{equation}

代入\eqref{intpsi},考虑到小角摆动周期为$T_0=2\pi/\omega_0$,得数学摆的周期为

\begin{equation} \begin{split} T=&\frac{2\pi}{\omega_0}\left [1+\frac{1}{4}\sin^2(\phi_0/2)+\frac{9}{64}\sin^4(\phi_0/2)+\cdots \right ]\\ \approx & T_0\left [1+\frac{1}{4}\sin^2(\phi_0/2) \right ] \end{split} \tag{9}\label{period} \end{equation}

可见,周期与摆幅$\phi_0$有关。只有当摆幅$\phi_0$非常小的时候,周期才与摆幅无关。下表是周期与摆幅的对应的关系:

$\phi_0$$0^{\circ}$$3^{\circ}$$5^{\circ}$$10^{\circ}$$30^{\circ}$$45^{\circ}$
$T/T_0$10.00021.00051.00191.01741.0369

即便摆幅很小,$\phi_0 \lt 5^{\circ}$,周期也是随摆幅而变的。即便摆幅比较大,如$\phi_0 = 45^{\circ}$,周期也只是比0摆幅极限下的周期$T_0$多了3.7%而已。

中学物理讲单摆有个$5^{\circ}$神话,即要求单摆摆幅要小于$5^{\circ}$。应该是因为中学物理实验室的计时仪器最高能精确到毫秒的原因吧。如果计时仪器只能精确到秒,摆幅$45^{\circ}$也是可以的。

标签: 单摆, 数学摆, 周期

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