非小角单摆(数学摆)的周期

参考:

体系

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*}

\(\omega_0=\sqrt{g/l}\),上式即为

\begin{equation} \frac{d^2\phi}{dt^2}+\omega_0^2\sin\phi=0 \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) \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) \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)} \label{dphidt} \end{equation}

\(t=0\)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} \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 \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} \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} \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} \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:

\(\phi_0\) \(0^{\circ}\) \(3^{\circ}\) \(5^{\circ}\) \(10^{\circ}\) \(30^{\circ}\) \(45^{\circ}\)
\(T/T_0\) 1 0.0002 1.0005 1.0019 1.0174 1.0369

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。

posted @ 2016-10-18 19:24  瞿立建  阅读(1995)  评论(0编辑  收藏  举报