For a double pendulum system with equal masses, equal string lengths, and under the small-angle approximation, analyze the equations of motion for the two bobs and solve for the characteristic frequencies (angular frequencies) of the system.
Assume both bobs have same mass $m$ and the massless string length $l$. All pivots are assumed to be frictionless.
Under the small angle approximation, $sin \theta \approx \theta$ and $cos\theta =1-\frac{\theta^2}{2}$.
The kinetic energy $T$ of the system can be rewriten as follows.
$$ T = \frac{1}{2}ml^2\left( 2\dot{\theta_1}^2+\dot{\theta_2}^2 + 2\dot{\theta_1}\dot{\theta_2} \right) $$
The potential energy $V$ of the system can be rewriten as follows.
$$ V=\frac{1}{2} mgl\left(2\theta_1^2 +\theta_2^2 \right) -3mgl $$
Apply Lagrangian eqution.
$$ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{\theta_i}} \right) - \frac{\partial L}{\partial \theta_i} = 0 $$
Two Lagrangian equations for $\theta_1$ and $\theta_2$ are as follows.
$$ 2\ddot{\theta_1} + \ddot{\theta_2}+2\frac{g}{l}\theta_1=0 $$
$$ \ddot{\theta_1} + \ddot{\theta_2}+\frac{g}{l}\theta_2=0 $$
Assume the solutions satisfy the Simple-Harmonic Motion with the form of $\theta_1 = Acos(\omega t)$ and $\theta_2 = Bcos(\omega t)$, the following equations are as follows. $$ 2g/l-2\omega^2)A -\omega^2 B=0 $$
$$ -\omega^2 A + (g/l-\omega^2)B =0 $$
There exists a solution such that the coefficient determinant is zero, giving the solution: $\omega = \sqrt{\frac{g}{l}(2\pm\sqrt{2})}$。
Thus, two frequencies of modes of oscillation are $\omega_1 =\sqrt{\frac{g}{l}(2+\sqrt{2})} $ and $\omega_2=\sqrt{\frac{g}{l}(2-\sqrt{2})} $. The motion of the pendulum can be expressed as linear superposition of two oscillations.
When two frequncies are close to each other, a beating phenomenon occurs.
But hwo can the two frequencies be close to each other?
Yes! Changing the mass of two bobs.
As we consider the mass of bobs, the frequencies can be obtained as
$$ \omega^2 = \frac{g}{l} \left[ 1+\mu \pm \sqrt{(1+\mu)\mu} \right] $$
where $\mu = \frac{m_2}{m_1}$[1].
Provided that $g/l = 1$, dependencies of the frequencies $\omega_1$, $\omega_2$ are shown as follows.
As $\mu$ is close to $0$, namely mass of top bob $m_1$ is much greater than that of bottom bob $m_2$, the two frequencies are close, which a beating phenomenon occurs.
Reference: [1] https://math24.net/double-pendulum.htm