This document discusses beats, which occur when two sounds waves with nearly identical frequencies are heard simultaneously. Constructive and destructive interference causes the amplitude of the combined wave to regularly increase and decrease, creating a modulation effect perceived as a beating sound. The frequency of this amplitude modulation is equal to the difference between the individual frequencies. An example calculation demonstrates how to determine the frequency of one violin if the blended sound of two out-of-tune violins has a known frequency and beat period.

1. Beats
Physics 101 Learning Objects

2. Beats
 Occur from variations in a sound wave’s amplitude.
 This happens we listen to two waves that have very similar (but
slightly different) frequencies at the same time.

3. Beats
 At one point, constructive interference
occurs between the two waves.
 As time passes, the crests of the two
waves become displaced (they occur at
different times), causing for the
amplitude of the resultant wave to
decrease.

4. Beats
 Eventually, the two waves become out of phase with each other, and destructive
interference occurs.
 As the two waves go in and out of phase, the amplitude of the resultant amplitude
slowly increases and decreases.
 This is termed “modulation” of amplitude, and is perceived by human ears as a
beating sound.

5. Calculating the Resultant Wave:
 Consider two waves with the same amplitudes but different wavelengths
(thus, different frequencies). We also imagine that both phases are zero and
that the position is fixed at xo.
 The equations for the two waves are:
s1(xo,t) = smcost(k1xo – ω1t)
s2(xo,t) = smcost(k2xo – ω2t)
 To get the equation for the resultant wave, we combine these two equations.
We receive:
Stotal(xo,t) = 2smcos( 𝜔𝑡)co𝑠(∆𝜔𝑡)
𝜔 =
𝜔1+ 𝜔2
2
“Mean angular frequency”
∆𝜔 =
𝜔1−𝜔2
2
“Angular frequency difference”

6. Calculating the Resultant Wave:
 When humans listen to the sound of two waves that differ only slightly in
frequency, they hear a tone that has the mean frequency of the two waves
which and varies in intensity.
 NOTE: This only happens when the frequency of the two waves are fairly
similar. When there is a large difference in the frequency, humans will
perceive this as two distinct tones.

7. Question: Tuning the Orchestra
 As the concertmaster of an orchestra, it is your job to ensure that every
instrument is in tune at the beginning of a performance. As you play a note on
your violin with a frequency of 440 Hz, you can hear another violin that is
slightly out of tune. You note that the blended sound of the two violins has a
frequency of 444 Hz.
 A) What is the frequency being produced by the other violin?
 B) What is the period of the resultant wave being produced?

8. Question: Tuning the Orchestra
 A) What is the frequency being produced by the other violin?
Recall that the perceived frequency is the mean value of the two sound
waves.
Thus, we may calculate the frequency of the other violin with:
444 Hz =
440 𝐻𝑧+𝑓
2
888 Hz = 440 Hz + f
f = 448 Hz

9. Question: Tuning the Orchestra
 B) What is the period of the resultant wave being produced?
Due to the amplitude of the resultant wave being large for both the positive
and negative crests, the frequency of the amplitude’s modulations is the
same as the difference of the two original wave frequencies.
Beat frequency:
448 Hz – 440 Hz = 8 Hz
Period = 1/Frequency
Period = 1/8
= 0.125 s