Standing Waves and SoundStanding Waves and Sound"If sand waves were sound waves
what song would be in the air now."
- S. Vega
Lecture Outline:
Two waves with the same frequency, wavelength, and amplitude traveling in opposite directions will interfere and produce standing waves. Let the harmonic waves be represented by the equations below
Adding the waves and using a trig identity we find
This is a standing wave -- a stationary vibration pattern. It has nodes - points where the medium doesn't move, and antinodes - points where the motion is a maximum. A standing wave on a string might look like
Consider a string of length L that is fixed at both ends. The string has a set of natural patterns of vibration called normal modes. This can be determined very simply. First remember that the ends are fixed so they must be nodes. This means a certain number of wavelengths or half wavelengths can fit on the string determined by the length of the string. The first three possible standing waves are shown below.
The wavelengths can be related to the length. The frequency, then, is found from the wavelength.
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In general we can write
The lowest frequency is called the fundamental frequency or the 1st harmonic. The higher frequencies are called overtones. Integer multiples of the 1st harmonic are labeled as the 2nd, 3rd, etc., harmonics. The diagram below shows the 3rd harmonic.
A. A string is fixed at both ends and plucked so it vibrates in a standing wave mode as shown below. Let upward motion correspond to positive velocities. When the string is in position b, the instantaneous velocity of points along the string
B. A string is fixed at both ends and plucked so it vibrates in a standing wave mode as shown below. Let upward motion correspond to positive velocities. When the string is in position c, the instantaneous velocity of points along the string
C. A string is stretched between two fixed points. If the tension in the string is increased, the normal mode frequencies
Just as we have standing waves on strings we can have standing sound waves in columns of air. The organ pipe is the basic example. At the open end of a pipe we expect a displacement antinode. If the end is closed we would expect a displacement node. Using the same sort of arguments as we did for the string we can find the normal modes of the air column in the pipe.
For two open ends the first three are
By using the same techniques as for the string it can be quickly shown that
just as it is for the two fixed end string.
For a closed end and an open end:
The frequencies can be found by the same techniques used on the string.
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Notice that the "closed at one end" case only exhibits the odd harmonics. The frequencies of the normal modes for the two cases can be written as:
D. The frequency of a person's voice increases by about a factor of 3 when he breaths in helium. Which of the following is the speed of sound in helium gas?
Intensity is usually express in terms of a quantity called the sound level.
The units of the sound level are decibels (dB). The threshold of hearing is about 0 db, a normal conversation is about 50 dB, a police siren is about 120 dB. The threshold of pain is about 120 dB. Sounds louder than this can cause damage to the ear.
The intensity of a sound wave falls off as the distance from the source increases. If the source radiates sound in all directions the sound wave can be thought of a an expanding sphere of pressure. The sound intensity as a function of distance can be easily calculated in this special case. The result is
E. A siren has a sound level of 80 dB at a range of 20 meters. If the intensity of the sound is increased by a factor of 100 what is the new sound level?
A. The answer is 1. At b the string is at one of the two extreme positions. It is momentarily at rest as the various parts of the string change the direction of their velocities.
B. The answer is 4. The velocity depends on location. Some parts of the string will be moving up while others will be moving down.
C. The answer is 1. Increasing the tension causes the speed of the waves to increase. This causes the frequency to increase as the frequency is equal to the speed divided by the wavelength and the wavelength is determined by the length of the string.
D. The answer is 5. The wavelength is fixed by the length and proportions of the person's voice box, throat and mouth. An increase in frequency must then correspond to a proportional increase in speed of the wave. 343*3 = 1029 is close to 965.
E. The answer is 2. If the intensity goes up by 100 and the change in sound level is 10*log(100) = 10(2) = 20. The new sound level is 80 + 20 = 100 dB.
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