Coupled oscillators

Figure 9: Coupled oscillators in one dimension

Let us first consider a one-dimensional chain of $N$ particles of mass $m$ with equal equilibrium separation $a$. The particles are coupled to massless springs with force constant $k_{c}$, except for the first and last springs at the two ends of the chain which have spring constant $k$. the individual displacement of the particle $i$ from its equilibrium position along the $x$ axis is called $u_{i}$. The ends of the fist and last spring are assumed fixed:

$$u_{0}=u_{N+1}=0.$$

Since the force of an individual mass is determined only by the compression or expansion of the adjacent springs, the equation of motion for particle $i$ is given by:

$$m\frac{d^{2}u_{i}}{dt^{2}} = -k_{c}(u_{i}-u_{i+1})-k_{c}(u_{i}-u_{i-1})$$ (46)

$$= -k_{c}(2u_{i}-u_{i+1}-u_{i-1}).$$ (47)

The equations for particles $i=1$ and $i=N$ next to the walls are given by

$$m\frac{d^2u_1}{dt^2}=-k_c(u_1-u_2)-ku_1,$$ (48)

$$m\frac{d^2u_N}{dt^2}=-k_c(u_N-u_{N-1})-ku_N.$$ (49)

Note that for $k_c=0$ all the equations will decouple and the motion of the particles become independent of their neighbors. The above equations describe longitudinal oscillations, i. e. motion along the direction of the chain. The equations for transverse motion are equivalent.