Can there exist harmonic oscillator with asymmetric coupling?

Question

In Classical Mechanics textbooks usually, for a coupled harmonic oscillator with two masses,

coupling is taken to be same in both directions (i.e coupling constant w.r.t to m1 is same as that with m2). Is there any physical significance of a Harmonic Oscillator having coupling constant to be $\kappa$ in one direction and $\zeta$ in other direction?

The equations of motion are: $$m_1 \ddot{x_1}+kx_1+\kappa(x_1-x_2)=0$$ and $$m_2\ddot{x_2}+kx_2-\zeta(x_1-x_2)=0$$

Answer 1

For this particular physical system, the terms $\kappa\,(x_2-x_1)$ and $-\zeta\,(x_2-x_1)$ are (1) the force exerted by particle 2 on particle 1 and (2) that exerted by 1 on 2. The forces must be equal and opposite, by Newton III, therefore we must have $\kappa=\zeta$. The two terms describe the tension in the same spring (i.e. the middle one in your diagram), which is another way to see that the constants must be equal.

The two spring constants for the outer springs can of course be different, even though you have them equal (to $k$) in your equations.

Answer 2

Not if we want Lagrangian mechanics to be true. The system consists of springs, so the potential is a quadratic function of $x_1,x_2$. So we may write:

$L = \frac{1}{2}(m_1\dot{x_1}^2+m_2\dot{x_2}^2)+\frac{1}{2}(ax_1^2+2bx_1x_2+x_2^2)$

This gives equations of motion:

$m_1\ddot{x_1}+ax_1+bx_2=0$

$m_2\ddot{x_2}+cx_2+bx_1=0$

It's slightly nicer to rewrite this as:

$m_1\ddot{x_1}+(a+b)x_1+b(x_2-x_1)=0$

$m_2\ddot{x_2}+(c+b)x_1-b(x_2-x_1)=0$

Which is exactly what your equations are, but where we see that $\kappa = \zeta=b$. So minimal action forces us to set the couplings equal.