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Physics · Ch 11 — Waves

Formation of Beats

11.7.2

Formation of Beats

When two sound waves of only very slightly different frequencies f1f_1 and f2f_2 (but comparable, equal amplitude) are superposed at a fixed point, the listener perceives not two separate steady tones but a single tone whose loudness itself rises and falls periodically over time -- this periodic throbbing or swelling-and-fading of loudness is called beats, and is a form of temporal (time-domain) interference, in contrast to the purely spatial (position-dependent) interference examined in the previous subsection. The number of loudness maxima heard per second, called the beat frequency, is simply the magnitude of the difference between the two source frequencies, n=∣f1−f2∣n=|f_1-f_2|. This result can be derived mathematically by superposing y1=Asin⁡(ω1t)y_1=A\sin(\omega_1t) and y2=Asin⁡(ω2t)y_2=A\sin(\omega_2t) (both taken with equal amplitude AA, at a fixed point x=0x=0) using the sum-to-product trigonometric identity for sine functions; the resulting expression factors neatly into a slowly-varying amplitude envelope, yP=2Acos⁡ ⁣(2πf1−f22t)y_P=2A\cos\!\big(2\pi\tfrac{f_1-f_2}{2}t\big), multiplying a rapidly-oscillating carrier tone at the average frequency, sin⁡(2πfavgt)\sin(2\pi f_{avg}t) with favg=(f1+f2)/2f_{avg}=(f_1+f_2)/2. Because ∣f1−f2∣|f_1-f_2| is much smaller than f1+f2f_1+f_2 (the two source frequencies being only slightly different), the envelope term yPy_P varies far more slowly in time than the carrier term does, and it is this slow rise and fall of the envelope -- not the underlying fast carrier oscillation -- that the ear registers and perceives as beats. Working through the times at which yPy_P reaches successive maxima (where $\ …

Figure 11.33Two waves superimpose with different frequencies such that there is a time alternation in constructive and destructive interference (they are periodically in and out of phase)

What this figure shows. Two sinusoidal waves of very slightly different frequency are drawn superposed on the same time axis, and beneath them the combined resultant is drawn as a rapidly-oscillating carrier wave whose overall envelope (its outer boundary) itself swells and shrinks slowly and periodically, alternating between wide (loud) and narrow (quiet) sections labelled along the time axis. The figure gives the direct visual meaning of the phrase "beats": because the two component waves drift in and out of phase with each other over time (first reinforcing, then cancelling, then reinforcing again), the amplitude of their sum rises and falls in a slow, regular rhythm distinct from the fast oscillation of either individual wave, and it is this slow swelling-and-fading rhythm, not the fast underlying tone, that a listener perceives as the throbbing beat. …

Misc Example 11.18Beat frequency from two given wavelengths

Worked out. Two sound waves of wavelengths λ1=5 m\lambda_1=5\ \text{m} and λ2=6 m\lambda_2=6\ \text{m} both travel through the same gas at a common speed v=330 m/sv=330\ \text{m/s}, and the task is to find the number of beats produced per second. Each wavelength gives its own frequency via f=v/λf=v/\lambda: the shorter wavelength gives the higher frequency f1=330/5=66 Hzf_1=330/5=66\ \text{Hz}, and the longer wavelength gives the lower frequency f2=330/6=55 Hzf_2=330/6=55\ \text{Hz}. The beat frequency is then simply their difference, ∣f1−f2∣=∣66−55∣=11|f_1-f_2|=|66-55|=11 beats per second, directly applying the beat-frequency formula to two waves …

Misc Example 11.19Beat frequency from two tuning-fork wave equations

Worked out. Two vibrating tuning forks produce waves described by the equations y1=5sin⁡(240πt)y_1=5\sin(240\pi t) and y2=4sin⁡(244πt)y_2=4\sin(244\pi t), and the task is to find the number of beats produced per second. Comparing each equation with the standard form y=Asin⁡(2πft)y=A\sin(2\pi f t) lets the frequency be read directly off the coefficient of t: for the first wave, 2πf1=240π2\pi f_1=240\pi gives f1=120 Hzf_1=120\ \text{Hz}, and for the second, 2πf2=244π2\pi f_2=244\pi gives f2=122 Hzf_2=122\ \text{Hz}. The beat frequency is then ∣f1−f2∣=∣120−122∣=2|f_1-f_2|=|120-122|=2 beats per second, showing how the beat frequency can be extracted directly from two given wave equations without ever needing to kno …