‘Beats’ is an interesting phenomenon arising from the interference of waves. When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a sound of similar frequency (the average of two close frequencies), but we hear something else also. We hear audibly distinct waxing and waning of the intensity of the sound, with a frequency equal to the difference in the two close frequencies.

To see this mathematically, let us consider two harmonic sound waves of nearly equal angular frequency \(\omega_1\) and \(\omega_2\) and fix the location to be \(x=0\) for convenience. Eq. (15.2) with a suitable choice of phase \((\phi=\pi / 2\) for each) and, assuming equal amplitudes, gives
\(s_1=a \cos \omega_1 t\) and \(s_2=a \cos \omega_2 t \quad \dots(15.45)\).
Here we have replaced the symbol \(y\) by \(s\) since we are referring to longitudinal not transverse displacement. Let \(\omega_1\) be the (slightly) greater of the two frequencies. The resultant displacement is, by the principle of superposition,
\(
s=s_1+s_2=a\left(\cos \omega_1 t+\cos \omega_2 t\right)
\)
Using the familiar trigonometric identity for \(\cos A+\cos B\), we get
\(
=2 a \cos \frac{\left(\omega_1-\omega_2\right) t}{2} \cos \frac{\left(\omega_1+\omega_2\right) t}{2} \dots(15.46)
\)
which may be written as:
\(s=\left[2 a \cos \omega_b t\right] \cos \omega_a t \dots(15.47)\)
If \(\left|\omega_1-\omega_2\right|<<\omega_1, \omega_2, \omega_a>>\omega_b\), where
\(\omega_b=\frac{\left(\omega_1-\omega_2\right)}{2}\) and \(\omega_a=\frac{\left(\omega_1+\omega_2\right)}{2}\)
Now if we assume \(\left|\omega_1-\omega_2\right|<<\omega_1\), which means \(\omega_a>>\omega_b\),
we can interpret Eq. (15.47) as follows. The resultant wave is oscillating with the average angular frequency \(\omega_a\); however, its amplitude is not constant in time, unlike a pure harmonic wave. The amplitude is the largest when the term \(\cos \omega_b t\) takes its limit +1 or -1. In other words, the intensity of the resultant wave waxes and wanes with a frequency which is \(2 \omega_{\mathrm{b}}=\omega_1-\omega_2\).
Since \(\omega=2 \pi f\), the beat frequency \(f_{\text {beat }}\), is given by
\(
f_{\text {beat }}=f_1-f_2 \dots(15.48)
\)
Fig. 15.16 illustrates the phenomenon of beats for two harmonic waves of frequencies \(11 \mathrm{~Hz}\) and \(9 \mathrm{~Hz}\). The amplitude of the resultant wave shows beats at a frequency of \(2 \mathrm{~Hz}\).

Example 1: Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency \(5 \mathrm{~Hz}\). The tension of the string B is slightly increased and the beat frequency is found to decrease to \(3 \mathrm{~Hz}\). What is the original frequency of B if the frequency of A is \(427 \mathrm{~Hz}\)?

Solution:

Increase in the tension of a string increases its frequency. If the original frequency of \(\mathrm{B}\left(f_B\right)\) were greater than that of \(\mathrm{A}\left(f_A\right)\), further increase in \(f_B\) should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that \(f_B<f_A\). Since \(f_A-f_B=5 \mathrm{~Hz}\), and \(f_A=427 \mathrm{~Hz}\), we get \(f_B=422 \mathrm{~Hz}\).

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