Beat and Beat frequencies in physics

Published on: **Mar 3, 2016**

Published in:
Science

- 1. BEATS BY: NANCY MANHAS
- 2. 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
- 3. 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.
- 4. 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|
- 5. • 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|
- 6. 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
- 7. • 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.
- 8. 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)