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Standing Waves and Sound

Physics 128 Lecture

Concordia College, Moorhead MN

"If sand waves were sound waves
what song would be in the air now."
- S. Vega 

Lecture Outline: 

  1. Standing Waves
  2. Standing Waves on a String
  3. Standing Waves in Air
  4. Intensity and Sound Level
  5. Hearing Range

Standing waves

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

Standing waves on a String

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. 

L = l/2

l = 2L

f = v/(2L)

L = l

l = L

f = v/(L)

L = 3l/2

l = 2L/3

f = 3v/(2L)

 

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

  1. is zero everywhere.
  2. is positive everywhere.
  3. Is negative everywhere.
  4. depends on the location.
Answer.

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

  1. is zero everywhere.
  2. Is positive everywhere.
  3. Is negative everywhere.
  4. Depends on the location.
Answer.

C. A string is stretched between two fixed points. If the tension in the string is increased, the normal mode frequencies

  1. increase.
  2. decrease.
  3. stay the same.

Answer.

Standing Waves in Air Columns

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.

L = l/4

l = 4L

f = v/(4L)

L = 3l/4

l = 4L/3

f = 3v/(4L)

L = 5l/4

l = 4L/5

f = 5v/(4L)

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?

  1. 123 m/s
  2. 259 m/s
  3. 431 m/s
  4. 757 m/s
  5. 965 m/s
Answer.

Intensity and Sound Level

The power of a sound wave is proportional to the amplitude of the wave squared ,and the frequency of the wave squared. We are usually interested in the Intensity of the wave. The intensity I of a wave, or the power per unit area, is the rate at which energy is transported by the wave through a unit area A perpendicular to the direction of travel of the wave. The intensity of a periodic sound wave is proportional to the square of the amplitude and the square of the frequency of the wave.

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?

  1. 90 dB
  2. 100 dB
  3. 160 dB
  4. 800 dB
  5. 8000 dB
Answer.

Hearing Range

Sound waves with frequencies between 20 Hz and 20 kHz are called the audible frequencies. The range of human hearing falls roughly within this range. We hear low frequencies as being low pitched and high frequencies as being high pitched. Audio samples of various frequency sounds are linked below. (Be careful of the volume - it will depend on the settings on your computer.) Sounds above the audible frequencies are called ultrasonic and sounds below the audible frequencies are called infrasonic or subsonic.

Answers to ConcepTests:

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.

Back to question. 

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.

Back to question.

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.

Back to question.

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.

Back to question.

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.

Back to question.


 

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This document is based on work by A. Aarons and E. Mazur. Some material is from Peer Instruction: A Users Guide by E. Mazur published by Prentice Hall.
This page was created by Bryan A. Luther, e-mail: mailto:luther@gloria.cord.edu 
Last Revised: Dec., 1999.