# Can there exist harmonic oscillator with asymmetric coupling?

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 $$m_2\ddot{x_2}+kx_2-\zeta(x_1-x_2)=0$$

• Could you specify the equations of motion in order to avoid misunderstandings on how the coupling constants enter? May 31, 2017 at 10:03
• M1*x1''+k1*x1+Kappa(x1-x2)=0 and M2*x2"+k2*x2-zeta(x1-x2)=0. May 31, 2017 at 11:13
• In principle, of course. May 31, 2017 at 12:01
• @Qmechanic So it means that in principle it doesn't violate anything ex: Hooke's Law etc. May 31, 2017 at 12:58
• @ChetanWaghela: As mentioned in the answer below, it violates Newton's Third Law (and therefore momentum conservation as well.) May 31, 2017 at 13:41

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.

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.