# Physical Meaning of two solutions when asking Angular Frequency of a System as a Function of Amplitude

Introduction Take the Forced Harmonic Oscillator of a spring: $$\ddot x + 2\Gamma \dot x +\omega_0^2 x = f_0 \cos(\omega t)$$ where $$\omega_0$$ is the natural frequency $$\sqrt{\frac{k}{m}}$$, $$\omega$$ the angular frequency of the external force, $$f_0 = F_0/m$$ for simplicity, and the dots represent time derivatives $$\dot x = \frac{\mathrm{d}x}{\mathrm{d}t}$$.

There's a homogeneous solution and a particular solution, but the homogeneous solution gets neglected because of damping, so the system has the frequency of the external force $$x(t) = A \cos{(\omega t)}$$

The resulting amplitude, derived via complex numbers, is $$A = \frac{f_0}{\sqrt{(\omega_0^2 - \omega^2 )^2 + 4 \Gamma^2 \omega}}$$

In my problem, for simplicity, damping is ignored; i.e. $$\Gamma =0$$. $$A = \frac{f_0}{\sqrt{(\omega_0^2 - \omega^2 )^2}}$$

The Question

If one is given the data of $$A$$, $$f_0$$, and $$\omega_0$$ it's possible to know the frequency of the system ($$\omega$$), but look at the derivation of the formula for $$\omega$$.

We begin by multiplying both sides by $$\frac{\sqrt{(\omega_0^2 - \omega^2 )^2}}{A}$$:

$$\sqrt{(\omega_0^2 - \omega^2 )^2} = \frac{f_0}{A}$$

Now, by applying the square root, we could had have inside. I have numbered the paths for the function of $$\omega$$

$$\sqrt{(\omega_0^2 - \omega^2 )^2} = \omega_0^2 - \omega^2 \; \; (1)$$ $$\sqrt{(\omega_0^2 - \omega^2 )^2} = \omega^2 - \omega_0^2 \; \; (2)$$

because of $$\sqrt{4} = \pm 2$$. This ends up in two functions for $$\omega$$:

$$\omega = \sqrt{ \omega_0^2 - \frac{f_0}{A} } \;\;(1)$$ $$\omega = \sqrt{ \omega_0^2 + \frac{f_0}{A} } \;\;(2)$$

I'm a physics student, so I'm familiar writing $$(\omega_0^2 - \omega^2)^2$$ in the Amplitude equation, but reading engineering textbooks, the second way $$(\omega^2 - \omega_0^2)^2$$ appears often. People always tend to forget about the two solutions of the square root, so depending on how they wrote the Amplitude equation (physics or engineering), you end up with way different solutions.

What an answer to this would look like

a) What's the physical meaning of having two solutions here? Is it mere mathematical? If so, why could one be preferred over the other?

b) The Physical Way (1) leads to an equation which allows for imaginary frequencies. Should we ignore an imaginary solution? (even though complex numbers pop up in so many places in physics)

c) Also, I tried to search for this in google but totally failed, is there a paper about someone experimenting with this? A graph would be so helpful.

Why I see this as important

The hardest thing for me to understand here is that there are two mathematical solutions, but nature is deterministic at this macroscopic scale, so one must be preferred. Could the fact of ignoring friction have to do with the problem? If I included friction I have to solve for a $$\omega^4$$ equation

I used $$\gamma \equiv 2\Gamma$$.
It's due to the fact that $$A(\omega)$$ isn't injective. You will also need the phase to determine which of the $$\omega$$'s is the one.
$$$$A = \frac{f_{0}}{ \sqrt{ (\omega_{0}^{2} - \omega^{2})^{2} + (\omega\gamma)^{2} } }$$$$ and the phase by $$$$\tan \varphi = \frac{\gamma\omega}{\omega_{0}^{2} - \omega^{2}}$$$$ Therefore, knowing the phase will tell you about the sign of $$\omega_{0}^{2} - \omega^{2}$$.
For the case $$\gamma = 0$$, as @Mechanic pointed out, $$$$\varphi = \begin{cases} 0 & \implies \omega_{0}^{2} - \omega^{2} > 0\\ \pi & \implies \omega_{0}^{2} - \omega^{2} < 0 \end{cases}$$$$
• It's possible to figure out what $\phi$ is, in the undamped case. You need the sign of $\omega_0^{2} -\omega^{2}$. If it's positive, then $\phi$ is 0, else it's $\pi$. Sep 27, 2021 at 4:44