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Here's what I'm doing: I'm using MATLAB to model a ball placed within a potential energy well. I'm then driving this ball with an external driving force. The function for the external driving force and the shape of the potential energy curve are both defined by me. Once I drive the ball within the potential with my customized driving force, I look at the ball's motion and take a Fourier Transform to see what frequency components make up it's motion.

If I place the ball in a parabolic potential well and drive it with some frequency, I get two peaks in the Fourier Transform: one for the frequency at which I'm driving the ball, and another peak that corresponds to the resonant freuqency of my potential well (in this case a parabola). The peak at 1Hz corresponds to my driving force

What's interesting is that if I drive it with two different frequencies, then most of the time I get 3 peaks, two corresponding to the two different frequencies that I'm driving the ball with, and the last one being the constant resonant frequency of the potential. Here's an example where I drive with 1Hz and 3Hz

This rule holds for most pairs of driving frequencies, with one big exception. Any two frequencies of the form (f, (resonanting frequency^2)/f) leads to only 2 peaks! The resonant peak goes away! In the case where the mass of the ball is 1, the resonant frequency is sqrt(2), thus for mass = 1, any two driving frequencies of the form (f,2/f) lead to no amplitude at sqrt(2), so take for example 1Hz and 2Hz: Driving with the sum of 1Hz and 2Hz

Where did the resonant peak go? Take another example for driving with 1.15Hz and (2/1.15)Hz: enter image description here

What is the explanation for the disappearance of this peak? Why must the two driving frequencies be of the form listed above???

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Sometimes you only get two peaks, because you happen to "get lucky" with your choice of all the other parameters in the differential equation you are solving, and one term in the general solution happens to be zero.

In the general situation, you are solving $$\ddot x + \omega^2 x = a e^{i\omega_1 t} + b e^{i\omega_1 t}$$ with initial conditions $$x(0) = x_0,\qquad \dot x(0) = v_0.$$ where $a$ and $b$ are complex constants.

The general solution is of the form $$x = A e^{i\omega_1 t} + B e^{i\omega_2 t} + C e^{i\omega t}$$ for some complex constants $A$, $B$, and $C$.

To satisfy the parts of the equation involving $\omega_1$ and $\omega_2$ for all times, you must have $$\begin{align} A(\omega^2 - \omega_1^2) &= a\\ B(\omega^2 - \omega_2^2) &= b\end{align}$$ which give the values of $A$ and $B$.

To satisfy the boundary conditions you must also have $$\begin{align} \Re(A + B + C) &= x_0 \\ \Im(\omega_1 A + \omega_2 B + \omega C) &= -v_0\end{align}$$

which give the real and imaginary parts of $C$.

You didn't tell us the amplitudes and phase angles of your two driving frequencies, or your initial conditions, but if you plug them all into the equations you will find the $C$ happens to be $0$ - i.e. there is nothing in the solution at frequency $\omega$.

If you make some "arbitrary" change - e.g. change your forcing functions from sine to cosine or vice versa - the third frequency will reappear in the output.

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  • $\begingroup$ Hmmm so I came to the conclusion that I was solving: mẍ = -2x + sin(ax)+sin(bx), is there significant benefit to working with complex numbers like you did? Also, I started at 0 with velocity 0. Both driving forces were sin waves of amplitude 1 with no phase. I believe that you'd get zero for C if you plugged everything in, but is there an intuitive explanation for why the frequencies must be related to each other in such a way (f,2/f)? $\endgroup$ – dljs Jul 2 '19 at 16:27

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