The equation of motion for a single pendulum subjected to gravity, without taking friction into account is:

$\displaystyle I\ddot\theta=-m g \ell \sin \theta $

For two pendula coupled by a spring a coupling term is added to the equation of motion of each pendulum. The force acting on the spring is equal to $ k\Delta l$, where $ \Delta l=(c\sin\beta-c\sin\alpha)$, is the deviation from the spring equilibrium length. The rotational force is thus $ kc^2(\sin \beta - \sin \alpha)$.

With this added term the equations of motion for the two coupled pendula become:

$\displaystyle I\ddot\alpha$ $\displaystyle =-mg\ell\sin\alpha -kc^2(\sin\alpha -\sin\beta)$    
$\displaystyle I\ddot\beta$ $\displaystyle =-mg\ell\sin\beta +kc^2(\sin\alpha -\sin\beta)$    

These equations cannot be solved analytically.

For small angles $ \alpha$ and $ \beta$ the harmonic approximation can be used: $ \sin \alpha \approx \alpha$ and $ \sin \beta \approx \beta$. The equations of motion become:

$\displaystyle I\ddot{\alpha}$ $\displaystyle =-mg\ell\, \alpha -kc^2(\alpha -\beta)$    
$\displaystyle I\ddot{\beta}$ $\displaystyle =-mg\ell\, \beta +kc^2(\alpha -\beta)$    

These two equations can be decoupled by adding and subtracting the two equations to give:

$\displaystyle I(\ddot\alpha+\ddot\beta)$ $\displaystyle =-mg\ell (\alpha+\beta)$    
$\displaystyle I(\ddot\alpha-\ddot\beta)$ $\displaystyle =-(mg\ell +2kc^2) (\alpha -\beta)$    

With solutions:

$\displaystyle \alpha + \beta$ $\displaystyle = A_+ \cos (\omega_+ t+ \psi_+), \quad \omega_+=\sqrt{mg\ell/I}$    
$\displaystyle \alpha - \beta$ $\displaystyle = A_- \cos (\omega_- t + \psi_-), \quad \omega_-=\sqrt{(mg\ell+2kc^2)/I}$    

and $ A_+$, $ A_-$, $ \psi_+$ and $ \psi_-$ are determined by the initial conditions. Note that $ \omega_+=\omega_0$, where $ \omega_0$ is the frequency for the single harmonic oscillator.

In physical terms $ \omega_+$ and $ \omega_-$ are the frequencies with which the pendula move either in phase, ( $ \alpha(t)=\beta(t)$) or out of phase ( $ \alpha(t)=-\beta(t)$), as illustrated below.

oscillating in fase
In fase motion
 oscillating out of fase
Out of fase motion

The two modes can be combined to give the solution for $ \alpha$ and $ \beta$

$\displaystyle \alpha$ $\displaystyle = A_+\cos(\omega_+t + \psi_+)/2+ A_-\cos(\omega_-t+\psi_-)/2$    
$\displaystyle \beta$ $\displaystyle = A_+\cos(\omega_+t + \psi_+)/2- A_-\cos(\omega_-t+\psi_-)/2$    

In general, the two coupled pendula behave in a complex fashion. A classical example, is found for the initial angles $ \alpha_0=R$ and $ \beta_0=0$ and setting all initial velocities to zero, as illustrated below.
oscillating more complex
One pendulum at rest, the other at an angle R
f
The four equations for the initial conditions at $ t=0$ are:

$\displaystyle A_+ \cos \psi_+ + A_-\cos\psi_- = R$    
$\displaystyle A_+ \cos \psi_+ - A_-\cos\psi_- = 0$    
$\displaystyle -\omega_+A_+ \sin \psi_+ - \omega_-A_-\sin\psi_- = 0$    
$\displaystyle -\omega_+A_+ \sin \psi_+ + \omega_-A_-\sin\psi_- = 0$    

The solutions are $ \psi_+=\psi_-=0$ and $ A_+=A_-=R/2$. The equations of motion become:

$\displaystyle \alpha=R/2(\cos\omega_+t+\cos\omega_-t)$    
$\displaystyle \beta =R/2(\cos\omega_+t-\cos\omega_-t)$    

Using trigonometric identities for the sum and difference of cosine functions, this can be rewritten to give:

$\displaystyle \alpha= R\cos\frac{(\omega_- +\omega_+)t}{2}\cos\frac{(\omega_--\omega_+)t}{2}$ (2)
$\displaystyle \beta = R\sin\frac{(\omega_- +\omega_+)t}{2}\sin\frac{(\omega_--\omega_+)t}{2}$    

The solutions are plotted below:
graph of the oscillator
One pendulum at rest, the other at an angle R=30 degrees

Equations (2) show that the motion of $ \alpha$ and $ \beta$ is constructed out of a multiplication of 2 sines or cosines with different angular frequency ( $ [\omega_-+\omega_+]/2$ and $ [\omega_--\omega_+]/2$).

The oscillation with long period $ 4\pi/(\omega_--\omega_+)$ constitutes an envelope for the oscillation with smaller period $ 4\pi/(\omega_-+\omega_+)$, so called beats.
In the weak coupling limit:

$\displaystyle \lim_{kc^2\rightarrow0}(\omega_- + \omega_+)/2$ $\displaystyle = \omega_+$    
$\displaystyle \lim_{kc^2\rightarrow0}(\omega_- - \omega_+)/2$ $\displaystyle = 0$    

At $ t=(2n+1)2\pi/(\omega_--\omega_+),\,n=0,1,\ldots$ the first pendulum comes to rest whereas the second comes to rest at $ t=n2\pi/(\omega_--\omega_+),\,n=0,1,\ldots$.