Superposition of two harmonic oscillations

This post is about backward engineering properties of combined oscillations from the resulting superposed motion. In some way it is a simplified form of the harmonic analysis. And my curiosity for it was induced by a problem from Savchenko et al.

This problem has taken my mind for quite some time so I decided to write it up here.

Problem statement Given the graph (Fig. 1) of a motion resulting from a superposition of two harmonic oscillations determine amplitudes and frequencies of the constituent oscillations (i.e. of the oscillations being combined to get the net motion).

Fig. 1: Resulting motion from the superposition of two harmonic oscillations.

Solution

According to the graph the resulting motion can be expressed as:

$$ x(t) = a_1\sin\omega_1 t + a_2\sin\omega_2 t \tag{1} $$

Our task is to determine $a_1, a_2, \omega_1, \omega_2$.

Next we apply transformations to the equation to present it in the following form

$$ x(t) = \hat A (t) \sin(\omega_0 t + \hat\phi) $$

The ideas for the transformations are taken from this lecture.

Let's express $\omega_{1,2}$ as

$$ \omega_1 = \omega - \varepsilon \\ \omega_2 = \omega + \varepsilon $$

Then substituting these into equation (1) and expanding the sines of sum, we get:

$$ x(t) = (a_1 +a_2)\cos\varepsilon t \cdot \sin\omega t + (a_2 -a_1)\sin\varepsilon t \cdot \cos\omega t $$

Now (here, I think, is the most important statement) we can find $\hat A(t)$ and $\hat\phi(t)$ such that:

$$ \left\{\begin{array}{l} (a_1 +a_2)\cos\varepsilon t = \hat A(t)\cos\hat\phi(t) \\ (a_2 -a_1)\sin\varepsilon t = \hat A(t)\sin\hat\phi(t) \end{array}\right.\tag{2} $$

Indeed, using equations (2), we can construct $\hat A(t)$ and $\hat\phi(t)$:

$$ \left\{\begin{array}{l} \hat A(t)^2 = a_1^2 + a_2^2 + 2 a_1a_2 \cos 2\varepsilon t \\ \tan\hat\phi(t) = \frac{a_2 - a_1}{a_2 + a_1}\tan\varepsilon t \end{array}\right.\tag{3} $$

Then using the expressions for $\hat A(t)$ and $\hat\phi(t)$ we can get the expression for the result motion as follows:

$$ x(t) = \hat A(t) \cos(\omega t - \hat\phi(t)) $$

From (3) and the graph of the resulting motion we can get the amplitudes of the constituents:

$$ \left\{\begin{array}{l} \hat A_{\max} = a_1 + a_2 = A\\ \hat A_{\min} = a_2 - a_1 = B \end{array} \right.\Rightarrow \left\{\begin{array}{l} a_1 = \frac{A-B}{2}\\ a_2 = \frac{A+B}{2}\\ \end{array} \right.\tag{4} $$

From (3) we get that the frequency of variation of the amplitude $\hat A(t)$ is $2 \varepsilon$. If we neglect the variation of $\hat\phi(t)$ (this is a point maybe where more rigor is needed) we can approximate the higher frequency variation of the resulting motion as $\omega$. Summarizing the above statements we have for the frequencies of the constituents:

$$ \left\{\begin{array}{l} 2 \varepsilon = \Omega \\ \omega = \omega_0 \\ \end{array} \right.\Rightarrow \left\{\begin{array}{l} \omega_2 - \omega_1 = \Omega \\ \omega_2 + \omega_1 = 2\omega_0\\ \end{array} \right.\Rightarrow \left\{\begin{array}{l} \omega_1 = \omega_0 - \frac{\Omega}{2} \\ \omega_2 = \omega_0 + \frac{\Omega}{2} \end{array} \right.\Rightarrow \left\{\begin{array}{l} \omega_1 = \frac{2\pi}{\tau} - \frac{\pi}{T} \\ \omega_2 = \frac{2\pi}{\tau} + \frac{\pi}{T} \end{array} \right.\tag{5} $$

The last terms in (4) and (5) are the solutions of the problem.

Discussion

In this section I would like to describe my way to the above solution.

At first I wrote a sum of two harmonics and was trying to see if there is something simple relating amplitudes or frequencies of the two terms to the graph of the net motion. But I could not see an easy way to relate those, and decided to work out the formula for the net graph and from there, maybe, I could get some kind of sum of two harmonic functions.

The formula I have come up with for the net motion using the graph is following:

$$ x(t) = \left[ \frac{A-B}{2} \sin\Omega t + \frac{A+B}{2} \right] \sin\omega_0 t $$

But there is no way this could be brought to the sum of 2 harmonics. It is easy to see when the multiplication is performed there is a product of sines and that leads to two more harmonics.. I even calculated a quick Fourier transform for the function and got the confirmation that this function has energy localized at three frequencies: $\omega_0$, $\omega_0 \pm \Omega$.

The time was passing and as I got more curious I started googling and reading about superposition of harmonic oscillations. And I found just what I needed in a physics lecture on acoustics (link), that brought me easily to answer the question about the amplitudes, although to work out the frequencies I had to think a bit more.