# Spring-mass system with complex spring constant

Suppose a system containing a mass $m$ on frictionless surface, attached by a spring to a wall. The spring's constant is complex, given by $K = K_1 + K_2i$, with $K_1 \gg K_2$. Write the equation of motion, and show it has the dynamic of a damped oscillator.

So, Newton's second law: $$m\ddot{x} = -(K_1+K_2i)x \Leftrightarrow \ddot{x} = -\frac{K_1+K_2i}{m}x$$ Solve for $x = e^{\lambda t}$, we get $$\lambda^2 = -\frac{K_1 + K_2i}{m} \Leftrightarrow \lambda = \pm\frac{1}{\sqrt{m}}\sqrt{-(K_1+K_2i)}$$ $$\lambda = \pm\sqrt{\frac{K_1}{m}}i\sqrt{1+\frac{K_2}{K_1}i}\approx \pm\sqrt{\frac{K_1}{m}}i\left(1+\frac{K_2}{2K_1}i\right) = \pm\sqrt{\frac{K_1}{m}}i\mp\sqrt{\frac{K_1}{m}}\frac{K_2}{2K_1}$$ This is where I'm stuck. Since the roots are not complex conjugates, we don't get the solution $$x(t) = e^{-\sqrt{\frac{K_1}{m}}\frac{K_2}{2K_1}t}\left(A\cos\left(\sqrt{\frac{K_1}{m}}t\right) + B\sin\left(\sqrt{\frac{K_1}{m}}t\right)\right)$$

Which is what I was guided to find. The solutions will be complex, which is not physical.

Any help?

## 1 Answer

I'm not a physics guru (yet!), but here's my 50 cents:

Since $\lambda$ represents complex vibration frequency, you're not interested in a negative real part of the root (as negative frequency isn't physical, or at least in the naive interpretation).

So instead, you're taking the positive root, and then if $\lambda$ is a solution to the ODE, so is its conjugate.