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推导小角度下的双摆振动频率

对于质量相等、摆长相等且在小角度近似下的双摆系统,分析两球的运动方程,求解系统的特征频率(角频率)。 假设两个小球质量均为 $m$,摆线长度均为 $l$,且满足小角度近似 $sin \theta \approx \theta$, $cos\theta =1-\frac{\theta^2}{2}$。 写出系统的动能 $T$: $$ T = \frac{1}{2}ml^2\left( 2\dot{\theta_1}^2+\dot{\theta_2}^2 + 2\dot{\theta_1}\dot{\theta_2} \right) $$ 系统势能 $V$: $$ V=\frac{1}{2} mgl\left(2\theta_1^2 +\theta_2^2 \right) -3mgl $$ 写出拉格朗日方程: $$ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{\theta_i}} \right) - \frac{\partial L}{\partial \theta_i} = 0 $$ 角度 $\theta_i$ 有两个 $\theta_1$,$\theta_2$,有两个拉格朗日方程。 $$ 2\ddot{\theta_1} + \ddot{\theta_2}+2\frac{g}{l}\theta_1=0 $$ $$ \ddot{\theta_1} + \ddot{\theta_2}+\frac{g}{l}\theta_2=0 $$ 假设运动满足简谐振动: $\theta_1 = Acos(\omega t)$,$\theta_2 = Bcos(\omega t)$,代入可得: $$ 2g/l-2\omega^2)A -\omega^2 B=0 $$ ...

March 10, 2025 · P. JIA

小角度下双摆的拍现象

介绍 在 Small amplitude oscillations of a double pendulum 一文中,作者实验了小角度下的双摆模型,用 Tracker 软件分析两个摆球的运动,对两球的运动进行傅立叶分析,得到两球的运动可以由两个正弦函数叠加而成。值得一提的是,作者提到,当下摆球的质量远小于上摆球的质量时,可能会出现一种特殊的现象,两种模式的振荡频率大致相同,从而能观察到拍频现象,一个摆的能量会逐渐转移到另一个摆。 An exception can occur if the lower mass is much smaller than the upper mass and then the two modes oscillate at about the same frequency, in which case beating is observed where the energy in one pendulum is gradually tranferred to the other pendulum. 推导 假设摆线长度为 $l_1$ 和 $l_2$,摆球的质量为 $m_1$ 和 $m_2$,摆线与竖直方向的夹角分别为 $\theta_1$ 和 $\theta_2$。 ...

March 6, 2025 · P. JIA