Beats
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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 original 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 the known frequency.
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Beats occur when two waves with nearly identical but not equal frequencies interact and interfere with each other. The interference causes the amplitude of the resulting wave to vary periodically over time. The rate at which the amplitude varies, known as the beat frequency, is proportional to the difference between the individual frequencies. Beats can be observed in the amplitude over time graph as the amplitude alternates between constructive and destructive interference. To determine the amplitude of the resulting wave function from two interfering waves, their individual wave functions are combined using the equation for a resultant amplitude that includes the mean angular frequency and angular frequency difference.
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1) Standing waves are produced by the superposition of two waves of equal amplitude and wavelength travelling in opposite directions. The displacements of the two waves add vectorially, resulting in areas of constructive and destructive interference.
2) On a string fixed at both ends, the fundamental mode has a node at each end and an antinode in the middle. Higher harmonics have additional nodes and antinodes appearing at regular intervals, with frequencies that are odd integer multiples of the fundamental frequency.
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Two violinists are tuning their instruments but realize their violins are out of tune with each other due to beating frequencies between the sound waves. The 1st violin produces a wave with frequency of 250Hz while the 2nd violin produces a wave with frequency of 274Hz, resulting in a beat frequency of 24Hz. When the sound waves are added through superposition, the displacement at a point 150s after the waves are produced is calculated to be 6.60m.
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Two waves of slightly different frequencies produce beats, which are a periodic variation in sound intensity or loudness. When the waves constructively interfere, intensity is maximum and amplitude is at its peak. When they destructively interfere, intensity is minimum and amplitude decreases. Beats allow musicians to tune instruments by matching frequencies until the beats disappear. The question examples calculate mean angular frequency, beat frequency, and intensity ratios using relevant equations from the document.
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3) Decreasing the tension by 2.77N, or 3.96%, will lower the fundamental frequency from 300Hz to 294Hz, the frequency needed to play the song.
nLo2
Beats occur when two sounds waves with similar frequencies are combined. The frequency of the beats is equal to the difference between the frequencies of the two sound waves. If a violin player hears 2 beats per second while tuning a string, the untuned string's frequency must be 2 Hz lower than the tuned string's frequency of 440 Hz. For two animals calling from land and water with wavelengths of 5m and 20m, the beat frequency is calculated to be 5.5 Hz based on the speed of sound in each medium and the wavelength-frequency relationship.
Lo8
Two waves that are close in frequency, when combined, produce a beat pattern where the amplitude increases and decreases over time as heard by the human ear. Beats occur due to the interference created by two waves interacting, with the beat frequency equal to the difference between the two original frequencies. For example, two cellos slightly out of tune at 249Hz and 245Hz would produce a beat frequency of 4Hz.
Lo8 upload
This is a Learning object for physics 101 at UBC.This powerpoint presentation explains the concept of beats and beat frequency and includes word problems to further solidify one's understanding.
LD2
Beats occur when two sounds waves with similar frequencies are combined. The frequency of the beats is equal to the difference between the frequencies of the two sound waves. If a violin player hears 2 beats per second while tuning a string, the untuned string's frequency must be 2 Hz lower than the reference frequency of 440 Hz. For two animals calling from land and water, the beat frequency is calculated using their wavelengths to determine frequencies, then taking the difference, which is 5.5 Hz.
LD2
Beats occur when two sounds waves with similar frequencies are combined. The frequency of the beats is equal to the difference between the frequencies of the two sound waves. If a violin player hears 2 beats per second while tuning a string, the untuned string's frequency must be 2 Hz lower than the reference frequency of 440 Hz. For two animals calling from land and water, the beat frequency is calculated using their wavelengths to determine frequencies, then taking the difference, which is 5.5 Hz.
LD2
Beats occur when two sounds waves with similar frequencies are combined. The frequency of the beats is equal to the difference between the frequencies of the two sound waves. If a violin player hears 2 beats per second while tuning a string, the untuned string's frequency must be 2 Hz lower than the reference frequency of 440 Hz. For two animals calling from land and water, the beat frequency is calculated to be 5.5 Hz based on the difference between their calculated frequencies.
Lo2
Beats occur when two sounds waves with similar frequencies are combined. The frequency of the beats is equal to the difference between the frequencies of the two sound waves. If a violin player hears 2 beats per second while tuning a string, the untuned string's frequency must be 2 Hz lower than the reference frequency of 440 Hz. For two animals calling from land and water, the beat frequency is calculated using their wavelengths to determine frequencies, then taking the difference, which is 5.5 Hz.
Learning object 2
The document discusses beats, which occur when two sounds with similar frequencies interact. Beats are heard as a modulation in volume as the waves alternately interfere constructively and destructively. When trying to tune a flute to a tuner emitting 440Hz, but the flute is playing at 434Hz, beats will be heard at a frequency of 6Hz as the two tones modulate together. The overall tone heard will be the average frequency of 437Hz.
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Beats
- 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)