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BEATS BY: NANCY MANHAS

Nancy Manhas - Beat Frequencies

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Page 1: Nancy Manhas - Beat Frequencies

BEATSBY: NANCY MANHAS

Page 2: Nancy Manhas - Beat Frequencies
Page 3: Nancy Manhas - Beat Frequencies

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

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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.

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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|

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• 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|

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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

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• 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.

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CITATIONS

• http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html

• http://www.physicsclassroom.com/class/sound/Lesson-3/Interference-and-Beats

• Nelson Textbook: Physics for Scientists and Engineers (Revised custom of Volume 1)