TD Ondes - Ex.5

Ex 5.1 - Beats

In acoustics, a beat is an interference between two sounds of slightly different frequencies, perceived as periodic variations in volume whose rate is the difference between the two frequencies.1

In the simplest case, each of the two sounds can be described by a monochromatic wave ( = a wave that can be described by a single frequency). We have then: \[\begin{aligned} u_1 (x, t) &= A \cos(\omega_1 t - k_1 x + \phi_1) \\ u_2 (x, t) &= A \cos(\omega_2 t - k_2 x + \phi_2)\end{aligned}\] with the condition \(\omega_1 \neq \omega_2\).

The amplitude of the wave is the maximum value it attains during an oscillation. In our case, the amplitudes of the waves coincide and they have value \(A\).

The phase of the wave is the argument of its oscillating function. In our case, the two phases are:2

\[\begin{aligned} \psi_1(x, t) &= \omega_1 t - k_1 x + \phi_1.\\ \psi_2(x, t) &= \omega_2 t - k_2 x + \phi_2.\nonumber \label{eq:phase}\end{aligned}\]

Beats can be perceived only when the amplitude of the two starting waves are the same, and their frequencies are very close, meaning that \((\omega_1 - \omega_2)/(\omega_1 + \omega_2) \ll 1\). While the amplitude condition is mathematically necessary, as we will show later, the frequency condition just helps to make the phenomenon more evident to our ears. We will not assume it in the following unless specified otherwise.

  1. Definition taken from, where you can find also nice animations and diagrams.

  2. The parameters \(\phi_1\) and \(\phi_2\) are sometimes referred to as phase offsets.

Point 1

Two waves are in phase when their phases coincide modulus a multiple of \(2\pi\) or, equivalently, when they differ by an even number of \(\pi\).1 \[\psi_1(x, t) = \psi_2(x, t) + 2n\pi, \qquad n \in \mathbb Z \qquad \qquad \textit{condition for two waves in phase} \label{eq:inphase}\]

Thus, for \(t = 0\) and \(x = 0\):2 \[\phi_1 = \phi_2 + 2n\pi \label{eq:phaserelation}\]

  1. This is because the phase is the argument of a periodic function with period \(2\pi\).

  2. Why would it be wrong to write \(t = x = 0\)?

Point 2

Two waves are in anti-phase1 when their phases differ by \(\pi\) modulus a multiple of \(2\pi\) or, equivalently, when they differ by an odd multiple of \(\pi\). \[\psi_1(x, t) = \psi_2(x, t) + (2m+1)\pi, \qquad m \in \mathbb Z \qquad \qquad \textit{condition for two waves in anti-phase} \label{eq:anti-phase}\]

We have to find the time \(t_a\) at which the two waves are in anti-phase for \(x = 0\). Putting the condition we found in Eq. \ref{eq:inphase} \[\begin{aligned} &\psi_1 (x=0, t_a) = \psi_2 (x=0, t_a) + (2m+1)\pi\\ &\omega_1 t_a + \phi_1 = \omega_2 t_a + \phi_2 + (2m+1)\pi\\ &\omega_1 t_a + \phi_2 + 2n\pi = \omega_2 t_a + \phi_2 + (2m+1)\pi\\ &\omega_1 t_a - \omega_2 t_a = (2m+1)\pi - 2n\pi\\ &(\omega_1 - \omega_2) t_a = (2m - 2n + 1)\pi\\ &t_a = \frac{[2(m-n)+1]\pi}{\omega_1 - \omega_2}\\ &t_a = \frac{(2p+1)\pi}{\Delta\omega}, \qquad p \in \mathbb Z \qquad \Delta\omega \equiv \omega_1 - \omega_2 \qquad \qquad \textit{all times at which anti-phase at $x=0$}\end{aligned}\]

The first anti-phase situation is given by \(p=0\). Thus the time we are looking for is \[t_{a, 0} = \frac{\pi}{\Delta\omega}. \label{eq:time_anti}\]

The time \(t_i\) at which the waves are back in phase can be calculated similarly, taking into account the definition given by Eq. \ref{eq:inphase}. Do the calculations as an exercise. The solution is \(t_{i, 1} = \frac{2\pi}{\Delta\omega_1}\).

  1. French: opposition de phase