Constructive and destructive interference of waves can produce beats, which are alternating loud and soft patterns. Beats occur when two sounds with nearly identical frequencies are played together. The beat frequency is the difference between the two sound frequencies. In the example, Kevin hears 12 beats over 4 seconds between his guitar strings tuned to 280Hz and an unknown frequency. Using the beat frequency formula, the unknown frequency is calculated to be 283Hz since it is described as higher than 280Hz.

1. BEATS
BY: NANCY MANHAS

3. BEATS - AN INTRODUCTION
• Wave interference occurs when two individual
waves meet along the same medium; resulting
in two major types of phenomena, destructive
and constructive interference
• Constructive interference can be described as a
meeting between two of the exact same pulses
at an upward-upward/downward-downward
position
• Destructive interference can be described as a
meeting between two of the exact same
sinusoidal pulses at an upward-downward
position
• Alternating constructive and destructive
interferences, produces an soft-loud-soft module
that is interpreted as beats

4. BEAT FREQUENCY-
APPLICATIONS
Scenario 1:
Kevin is tuning his guitar strings. After finishing his E4 string to a
frequency of 280Hz, he begins to adjust the B3 string. When
Kevin strikes the two strings together, he hears 12 beats every 4
seconds. As well, he notices that the frequency of the B3 string is
higher than the frequency of the E4 string. Determine the
frequency of the B3 string.

5. SOLUTION #1:
• In order to solve this
question, we must first
understand that the beat
frequency is an average of
the frequencies. fbeat = favg =
f1 - f2
• What the observer hears is
the absolute value
difference between these
two frequencies, |f1 - f2|

6. • Since we know that in order for a beat frequency to occur the two
frequencies between the waves must be nearly the same
• Therefore, we can make an assumption that our second frequency will be
similar to that of our first one.
• Using the following equation, beat f = |f1 - f2|, we can find out what the frequency
of the second string is
• First, the equation tells us that the frequency hear is 12 beats every 4 seconds, the
beat frequency can be determined
• fbeat = # of beats/time : 12 beats/4 seconds = 3Hz is heard every second to the
observer.
• We now know fbeat, and can now determine the frequency of the B3 string, using
the following formula: beat f = |f1 - f2|

7. SOLUTION CONTINUED…
• beat f = |f1 - f2| f1 = 280 Hz
+/- 3Hz = 280Hz - f2
Since the frequency can either be +/- 3, due to no indication delivered by beats (shown by the absolute value brackets), we
must determine both the above or below frequencies:
+3 Hz = 280Hz - f2
f2 = 280Hz - 3Hz
f2 = 277 Hz
-3 Hz = 280 Hz - f2
f2 = 280 - (-3Hz)
f2 = 280 + 3Hz
f2 = 283 Hz

8. • Therefore, f2 of the B3 string is either 283Hz or 277Hz
• However, since we are told that Kevin noticed the B3 string to have
a higher frequency when strung, the B3 string must be of higher
value.
• As a result the B3 string must be 283Hz, in order for it to
synonymous to the definition of a beat frequency.

9. CITATIONS
• http://hyperphysics.phy-
astr.gsu.edu/hbase/sound/beat.html
• http://www.physicsclassroom.com/class/sound/Lesso
n-3/Interference-and-Beats
• Nelson Textbook: Physics for Scientists and
Engineers (Revised custom of Volume 1)