# 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?

• Maybe you should include a picture or explain how everything is coupled, because it's hard to answer this right now. – Javier Feb 9 '15 at 19:26
• @Javier Badia: Two vertical simple pendulums whose bobs are attached with a spring are stretched in opposite directions initially. – user36790 Feb 10 '15 at 2:52

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.
• Yes, it is an external force applied to the pendulum. Without $2kx$, which can be switched on and off, the term $m\omega_0^2x$ is an external force acting permanently. – Vladimir Kalitvianski Feb 9 '15 at 21:47
• @ja72: Sir, the two pendulums are stretched in the opposite directions. So, on one pendulum, the restoring force would be $kx$; and the restoring force on the other side is $-kx$. Then how can it be $2kx$?? – user36790 Feb 10 '15 at 4:22
• The only possible coupled motion is when one displaces $+x$ the other one $-x$. The spring length is thus $(+x)-(-x)=2x$ – ja72 Feb 10 '15 at 13:27