# Why two vibrations of different frequencies and amplitudes need to be commensurable when the resultant vibration is formed from their superposition?

I was reading chapter 2 of AP French's Vibrations and Waves. In the section "Superposed Vibrations of Different Frequency, Beats", this paragraph confused me :-

"Unless there is some simple relation between $$\omega_1$$ and $$\omega_2$$, the resultant displacement will be a complicated function of time, perhaps even to the point of never repeating itself. The condition for any sort of true periodicity in the combined motion is that the periods of the component motions be commensurable—i.e., there exist two integers $$n_1$$ and $$n_2$$ such that $$T = n_1T_1 = n_2T_2$$ The period of the combined motion is then the value of T as obtained above, using the smallest integral values of $$n_1$$ and $$n_2$$ for which the relation can be written.

I am not able to understand as to why this condition has to be met for the resulting vibration to be a periodic.

(Please be aware that I am very new to physics community and currently covering the undergraduate curriculum on my own, hence I do not have a strong background in it).

I'm assuming you are referring to a forced oscillation problem, e.g. a mass on a spring which is being driven by an external force. If I neglect all dissipative forces (e.g. no friction, air resistance etc.), the relevant differential equation is: $$m\frac{d^2x}{dt^2}=-kx+F(t)$$

This is a well-studied equation, in particular for the case where the external force is a simple sinusoidal, e.g. $$F(t) = A\cos(\omega t)$$. In this case, you get a general solution which will depend on the natural frequency of the spring and on the driving frequency $$\omega$$. In particular, if the natural frequency of the spring is close to the driving frequency, then you get a large response from the system (this is called resonance). There is a nice explanation and plot of this in this answer.

These solutions usually have some initial transient (e.g. when you first turn on the force, the system will take some time to stabilize), dependent on the initial conditions, but for sufficiently large times, you get a sinusoidal solution. For the mathematical detail of this, you can look up any solution to the forced harmonic oscillator. The preiod of the final sinusoidal solution (after the transient) will only depend on the driving frequency $$\omega$$, though the amplitude will depend on the difference between the driving frequency $$\omega$$ and the natural frequency of the system.

Your book is talking about the more complex problem of having a driving force with two frequencies, e.g. $$F(t) = A\cos(\omega_1 t)+ B\cos(\omega_2 t)$$. This can be written as $$F(t) = F_1(t)+F_2(t)$$ with $$F_1(t) = A\cos(\omega_1 t)$$ and $$F_2(t) = B\cos(\omega_2 t)$$.

To understand the solution in this case, note that the underlying differential equations is linear. Thus, if you solve the differential equation with $$F(t) = F_1(t)$$ to get a solution $$x_1(t)$$ and then solve the differential equation with $$F(t) = F_2(t)$$, to get a solution $$x_2(t)$$, then the total solution for $$F(t) = F_1(t)+F_2(t)$$ will be $$x_1(t)+x_2(t)$$.

What does this look like? Well by my comment above, $$x_1(t)$$ will (for sufficiently large times) be periodic with period $$\omega_1$$ and $$x_2(t)$$ will (for sufficiently large times) be periodic with period $$\omega_2$$. Thus, you are summing two periodic solutions.

So what does this sum look like? Well this depends on the ratio of $$\omega_1$$ and $$\omega_2$$. In the easiest case that $$\omega_1 = \omega_2$$, then the period is simply $$\omega_1$$. If, on the other hand, you have $$\omega_2 = 2\omega_1$$, then the resulting period is $$2\omega_1$$. In general, the resulting period is given by the least common multiple of the two periods.

If the periods are irrational and different, no such common multiple exists and the sum will not be periodic.

For a more mathematical description of this last point, see e.g. this question.

• Thank you for your answer. I did some more digging, and many answers suggest the same explanation more or less. Thank you for your time. – Chandan Gupta Feb 13 at 22:59