Q:

Why in analysis of coupled oscillator, restoring force for uncoupled condition is taken in account?

If the pendulums were free & either one were displaced a small distance $x$, the restoring force would be $m{\omega_0}^2 x$. But in the present situation the coupling spring is stretched a distance $2x$ & exerts a restoring force of $2kx$, $k$ being the spring constant. Thus the equation of motion is: 

$$ m\dfrac{{d^2}x}{d{t^2}} + m{\omega_0}^2 x + 2kx = 0.$$ 
Now, the pendulums are coupled. Then why does the equation contain the term $m{\omega_0}^2x$ which is meant for uncoupled condition i.e. when each pendulum is free?

A:

Because if the coupling is not there it still will behave like $$ m \frac{{\rm d}^2x}{{\rm d}t^2} + m \omega_0^2 x = 0$$ 
The coupling just acts as an additional effect of $k (2 x)$ magnitude.

