Unit 11 Part 2 --Sound Notes



Unit 11, Part 2—Sound

A sound wave is just a type of wave that our ears can detect and our brain can interpret information from (just like our eyes and brain can detect and interpret light waves, which you’ll learn about in the next unit). A sound wave is a longitudinal (mechanical) wave and has all the general properties of waves that we discussed in the last chapter. Since sound is a longitudinal wave, it consists of regions of matter that are compact (compressions) and regions of matter that are loosely packed (rarefactions). A sound wave moves through matter just like you saw a longitudinal wave move through a slinky in lab. If a sound wave is moving through air, the regions of compression consist of air molecules that are very close to each other and will therefore create a region of high air pressure. The regions of not-so-dense air (rarefactions) create a region of low pressure. As the air molecules of the high-pressure region expand and press into the surrounding regions (low pressure), the air molecules will collide and the sound (energy) will be transferred through the matter. Note: Even though sound is a longitudinal wave, all of my drawings (and your book’s) will portray a sound wave as a transverse wave as it’s a lot easier to draw this way…..

Since sound is a mechanical wave, it must have a material medium (matter) to exist. Sound can move through air, water, steel…anything that is “matter”. However, sound can’t travel in space as there aren’t any molecules in space to transmit the energy in the wave. So, all those movies that show people screaming in outer space or show a loud noise when a spaceship blows up are just Hollywood’s way to entertain. In real life, you can’t hear anything in space (unless you’re inside a spaceship or space suit that has an atmosphere).

Like all waves, the velocity of a sound wave only depends on the material it’s traveling through. In general, the denser the material, the faster the sound wave will travel (the molecules of the material are closer together so they can collide faster and transfer the energy faster). That’s why in all those western movies, a cowboy would put his ear to a metal train-track to listen for the train. The sound of the train travels much faster through the dense metal tracks than the not-so-dense air. The speed of sound in air is generally around 343m/s (memorize this!), but the air temperature, atmospheric pressure, and humidity will affect it. The speed of sound in a vacuum (space) is 0m/s…it can’t move if it can’t exist

Sound waves, like all waves, will experience “interference” (either constructive or destructive) if they move through each other. The interference between sound waves will create regions of really loud sound (in constructive interference) and regions of very low sound/no sound at all (in destructive interference). Recording studios get around the interference problem by coating the room with a super-absorbent material (insulation) that absorbs sound waves so they won’t hit a wall, reflect, and possibly cause interference. By doing this, the only sound waves that hit the microphone are the ones that are supposed to and the music comes out perfect (as long as the person behind the mike has talent…).

Pitch vs. Loudness

The pitch of a sound refers to the frequency of the sound wave as humans detect it. A high pitch (like a flute) corresponds to a high frequency sound wave, and a low pitch (like a tuba) corresponds to a low frequency sound wave. For you music people, the note “C” on a piano (one of them) has a frequency of 262 Hz and the next-in-line note “E” has a frequency of 327 Hz. That particular C-note is a lower pitch than the next E simply because of their differing frequencies.

The loudness of a sound depends on the amplitude of the sound wave. Very loud sounds have a lot of energy (think how loud noises can damage your ear), therefore their waves must have large amplitudes. Remember from the last chapter that the amplitude of any mechanical wave tells how much energy it has (the larger the amplitude, the more energy in the wave). The amplitude of a sound wave (which is a longitudinal mechanical wave), however, isn’t shown in the size of the crests or troughs as there aren’t “crests” or “troughs” in a longitudinal wave. Instead, the amplitude of a longitudinal wave is seen in the pressure variations of the wave. Regions of either higher or lower than normal air pressure in a longitudinal wave are regions of large amplitude, which are regions of large energy. The further away the pressure in the wave is from “regular air pressure”, the larger the amplitude and the more energy in the longitudinal wave.

We use the decibel scale to measure the energy in / intensity level of a sound wave. The scale ranges from 0dB (the unit for decibel is dB) to 120dB. Our ears are sensitive from around 0dB (threshold of hearing) to ~130dB (threshold of pain).

The material below is not require of CP: Intensity (I, in W/m2) and intensity level (β, in dB) will be covered in lecture only---they will NOT be covered in these notes. Here are the equations

[pic] [pic], where Io (the threshold of hearing) = 1 x 10-12 W/m2

The Doppler Effect

Have you ever been standing on the sidewalk listening to the sound of an approaching ambulance or train? As it moves toward you it has a relatively high pitch, but the moment it passes you the pitch suddenly drops down to a lower pitch. Remember that the frequency of any wave (which can be visualized as how close the waves are together) determines the pitch of the sound. As an ambulance puts out a sound wave by blasting its siren, the sound from its siren will radiate away from it in all directions at the same velocity. However, if the ambulance begins to move forward as it is producing its sound, the sound waves that are radiating away from the front of the ambulance begin to get all bunched up due to the sound source (the ambulance) moving in the same direction the waves are trying to move. The sound waves that are radiating out from the back of the ambulance are moving away from the ambulance at some velocity, but the ambulance is moving away from them in the opposite direction that they are moving so they appear to move farther and farther apart. Therefore, if an observer (like you) is in front of the approaching ambulance, the sound waves that reach the observer will appear to have a high frequency (the waves will be scrunched together to mimic a small wavelength which is a high frequency). But as the ambulance passes the observer, the sound waves all of a sudden appear to have a lower frequency (due to the wavelengths appearing to increase which leads to a lower frequency). This observed difference in frequencies of sound waves will hit an observer’s ears and are then detected as a high pitch as the source approaches, and a lower pitch as the ambulance moves passed

(sound waves radiating out

in all directions from the source)

Sound source remaining still. Sound source moving to the right.

In the picture on the left, the source is motionless so all the sound waves coming from it radiate out in all directions at a uniform rate with a uniform velocity-- the waves appear to have a uniform frequency at all areas surrounding the sound source. As the source begins to move (as in the picture to the right), the sound source moves up on the waves that are moving away from the front of the sound source. This makes the waves appear to have a higher frequency (translated to a higher pitch in your ears) in front of the moving source. But the sound waves that are coming from the back of the sound source keep getting further and further apart (sounding like a lower frequency, lower pitch to your ears) since the source is moving in the opposite direction that the waves are moving. IMPORTANT: The sound that the source is emitting is all the same frequency. However, due to the motion of either the object or the observer, the frequency appears to shift creating different pitches.

The material below is not required for CP: In order to calculate the frequency of the sound that you are actually hearing (whether it’s you or the source that’s moving) you would use the following equation:

f ’ = f [pic]

Where:

f’ = the perceived frequency (the one you’re hearing)

f = the frequency actually emitted by the source

V = speed of sound (usually going to be 343 m/s unless otherwise noted)

Vo = speed of the observer (not velocity)

Vs = speed of the source (not velocity)

And… You use the top symbols in the equation ( the + vs –) if the two objects are moving toward each other.

You use the bottom symbols if the two objects are moving away from each other.

Ex: If the observer is stationary (V0 = 0m/s) and the source is moving toward the observer, the observed frequency (f’) will naturally be higher than the emitted frequency (f ). In order to get a larger f’ than f, the fraction in the parentheses must be larger than “one”. For that to work, the top part of the fraction (numerator) must be larger than the bottom (denominator). The only way this can be is if you add the two terms in the numerator together and subtract the two terms in the denominator. So, if the two objects are moving toward each other, use the top symbols (remember: toward—top). If they’re moving away from each other, use the bottom symbols.

If neither the source nor the observer are moving, the Vo and the Vs are both 0m/s. That will leave the fraction as v/v, which is equal to one. Multiplying the emitted frequency by one will leave the observed frequency (f’) as identical to the emitted (f).

Standing Waves

Review:

Incident wave- The wave heading into a boundary.

Reflected wave- The wave that occurs after the incident wave hits a boundary and is “reflected” back.

Standing wave- A standing wave is a combination of incident and reflected waves that together visually produce a single wave with stationary parts that don’t move and non-stationary parts that consistently change from a crest to a trough back and forth. Standing waves are the result of both constructive and destructive interference (review the last chapter if you need to). This is how it happens: When two waves of equal amplitude and wavelength pass through each other in opposite directions, there will be regions where the waves are always out of phase (region of destructive interference where a crest and a trough overlap) and regions where the waves are always in phase (regions of constructive interference where a crest will overlap with a crest or a trough will overlap with a trough). The regions of destructive interference will produce regions in the standing wave that never move, regions that are stationary. These regions are called “nodes”. The regions of constructive interference produce regions in the standing wave that have very large amplitudes and continuously vibrate up and down so at one moment it forms a crest and the next moment it forms a trough. These regions are called antinodes and are regions of maximum displacement.

The drawing below is a standing wave:

Antinode Node

One wavelength

Notice in the above standing wave, ½ of a wavelength is the amount of wave between two nodes. The amount of a standing wave between every-other node represents a whole wavelength.

Resonance- Once a standing wave has been created, and more and more incident and reflected waves continue to move through each other, their energy will be consistently added to the standing wave they’re flowing through. The increase in energy that the incident (and reflected) waves bring to the standing wave creates larger and larger amplitudes of the antinodes in the standing wave. This is called resonance. In order for resonance to occur, a standing wave must first be created and then more energy (delivered by the incident waves) must be added to it that will make the amplitude of the standing wave HUGE. When this happens with sound waves, the resonance will create a sound loud enough for people to hear some distance away from the source. This is how instruments with air columns (flutes, saxophones, acoustic guitars, etc) produce “music” that people can hear even if they’re across the room. The air column in a flute, for example, is just a chamber that enables a standing wave to form in it. When the musician playing the instrument continues to add energy to the air column (by blowing into the flute), the waves already inside the instrument will reflect off the interior surface and mix with the incident waves and form a standing wave whose amplitude soon becomes very large. Remember that large amplitudes in sound waves translate to loud sound. The larger the amplitude of the sound wave, the louder the sound. Without the “resonance chamber” in instruments (piano, acoustic guitar, all the brass and reed instruments, etc), no one would be able to hear the music. When acoustical guitar strings vibrate, they produce sound waves in the air that eventually make it to the resonance chamber in the guitar and produce a standing wave (resonance). The same thing happens in pianos. I think this is really neat and I don’t even play an instrument…..

Resonance in “Closed-end” Tubes

For all of the instruments that have a column of air with which to create a standing wave for resonance, the length of the tube determines the wavelength/frequency of the wave(s) that can resonate (form a standing wave) in them. When a standing wave is produced in a resonance tube that has one open-end and one closed-end (like a guitar or a milk jug), there MUST be a node at the closed end and an anti-node at the open-end. The anti-node is the region where the air molecules will vibrate back and forth causing the air molecules around the open end to vibrate back and forth and create a sound wave through the outside air. A sound wave won’t be/can’t be produced outside the instrument if a node (no air movement) is at the open end of the instrument. Therefore, since only certain wavelengths (with certain frequencies) of waves can exactly fit in a tube of a certain length so there is a node at the closed end and an anti-node at the open end, only certain frequencies of sound can be produced by any given instrument. Cool, huh?

In a closed-end tube (one end is closed, the other is open), the following fractions of wavelengths can produce resonance (a standing wave):

( = ¼ wavelength ( = ¾ wavelength (= 5/4 wavelength etc…

(The Fundamental Freq (3rd Harmonic) (5th Harmonic)

Or 1st Harmonic) (1st Overtone) (2nd Overtone)

What the above drawings show you is that in any closed-end tube of a certain length, only certain waves (with certain sized wavelengths) can resonate (produce a standing wave) in them. The waves must exactly fit in the tubes in such a way that there is a node at the bottom (closed-end) and an anti-node at the top (open-end). This makes it so only certain wavelengths of waves (with their corresponding frequencies) can resonate in any given tube of a particular length. For the three closed-end tubes shown above and all other closed-end tubes, resonance will occur every 1/2(, but any particular wave needs to fit at least ¼ of a wavelength in it. Any sound wave that can resonate (produce a standing wave) in a closed-end tube is able to have its sound amplified and heard.

Resonance in Open-end Tubes

Some instruments (like a sax or a flute) have resonating chambers that consist of two open ends, called an “open-end” tube. In the case of an open-end resonating tube, there must be an anti-node at both ends of the chamber (which will make at least one node in the middle). Remember that anti-nodes must be at the open ends if any sound is going to escape the tube.

1/2( 1( 1 ½ ( (3/2() etc…..

(Fundamental Freq (2nd Harmonic) (3rd Harmonic)

or 1st Harmonic) (1st Overtone) (2nd Overtone)

For an open-end resonating tube/column, a minimum of ½ of any wave’s wavelength must be able to fit in the tube to produce a standing wave. After that, resonance will occur at every 1/2( intervals. When you blow into a flute (an open-end tube), for example, many different frequencies of air are coming out of your mouth and moving into the tube. Only those frequencies of waves in the air that you blew into the tube which have wavelengths that can exactly fit in the tube in such a way so there’s an anti-node at both ends can resonate and be heard. So, it depends on the length of the wavelength and the length of the tube for resonance to occur. This will make more sense (hopefully) when we discuss it in class. It’ a complicated concept, so work with it a bit and do the corresponding problems and hopefully it will make sense. If not, come see me…..

Fundamental Frequency, Overtones, and Harmonics (this material is not required for CP)

The lowest frequency (i.e. largest wavelength) that can resonate in anything (a closed-end tube, an open-end tube, a guitar or piano string, etc) is called the fundamental frequency. Whole number multiples (1x, 2x, 3x, etc) of the fundamental frequency are called harmonics.

Remember, in a closed-end tube (look at the diagrams above that represent them) the fundamental frequency is a wave that is so large it can only fit ¼ of a wavelength in it (with a node at the closed-end and an antinode at the open-end). The fundamental frequency is also known as the first harmonic. Twice the fundamental frequency (2 x 1/4( or 2/4( or 1/2() can’t resonate in a closed-end tube as there would be no way for this ratio of a wave to have a node at the closed-end and an antinode at the open-end. Therefore, there is no 2nd harmonic in closed-end tubes. The 3rd harmonic is 3 x 1/4( (3/4(). This will resonate as 3/4( will fit into the tube with a node at the closed-end and an antinode at the open-end. There is no 4th harmonic in closed-end tubes (4 x 1/4( or 4/4() as this condition won’t allow a node at the closed-end and an antinode at the open-end, etc.

For open-end tubes, the fundamental frequency (i.e. first harmonic) is 1/2(. The 2nd harmonic is 2 x 1/2( (i.e. 1() and this will produce resonance (look at the diagrams above). Three times the fundamental frequency (3/2( or 1 ½ ( ) will also produce resonance (look at the diagram above). Therefore, open-end tubes don’t skip harmonics (unlike closed-end tubes).

Overtones are just the next higher frequency that will resonate in any particular tube. For example, the 1st overtone is the next higher frequency after the fundamental, and the 2nd overtone is the next highest frequency after the first overtone (or two frequencies higher than the fundamental).

Here are some easy practice problems….

1. Sound with a frequency of 261.6 Hz travels through water at a speed of 1435 m/s. What is its wavelength in water?

f = 261.6 Hz (waves per second)

v = 1435 m/s

( = ? v = f( ( = v/f 5.49 m

2. Find the average frequency of a sound wave moving in air at room temp (343m/s) with a wavelength of 0.667m.

f = ?

( = 0.667m

v = 343m/s (this is the normal speed of sound in air at room temp…you’ll need to KNOW this)

v = f( f = v/( 514.24 Hz

3. The human ear can detect sounds with frequencies between 20 and 20,000 Hz. Find the largest and smallest wavelengths the ear can detect.

v = 343m/s v = 343m/s v = f( ( = v/f

f = 20Hz f = 20 x 103 Hz

( = ? (= ?

17.15 m 0.017 m

Sound waves come in many different frequencies with many different wavelengths.

4. A tuning fork with a frequency of 392 Hz is found to cause resonances in a closed-end air column at 0.21m and 0.63m. The air temp is 27 degrees Celcius. Find the velocity of sound at that temperature.

The spacings between the resonances (for both the open- and closed-end tubes) equal ½ a wavelength (resonance will occur every ½ wavelength in both open and closed-end pipes). Therefore ½ wavelength is 0.42m, so 1 whole wavelength is 0.84m. Using v = f(, you’ll come out with v = 329 m/s.

*Notice the speed of sound in air at 27 degrees C is different than the speed of sound in room temp air. You only need to know the speed of sound in regular room temperature air from memory. If you ever need anything else, it will either be given to you or you’ll be asked to calculate it.

5. A 440 Hz tuning fork is held above a closed-end pipe. Find the spacings between the resonances in room temperature air.

f = 440Hz

v = 343 m/s

1/2( = ? This question is a “closed-pipe” resonating question. It is asking you to find the “spacings between the resonances” if the frequency of the resonating sound wave inside of the tube is 440Hz. Remember that resonance will occur every 1/2( in a closed-end pipe. Therefore, this question is really asking you to find what the length of 1/2( is. In order to do that, you need to first find the length of a whole wavelength and then take ½ of that value.

v = f( ( = v/f 1 ( = 0.78m so ½ ( = 0.39m

6. The 480 Hz tuning fork is used with an open-end resonating column to determine the velocity of sound in Helium gas. If the spacing between the resonances are 1.1 m, what is the velocity of sound in He?

f = 480Hz

½( = 1.1m therefore 1( = 2.2m

v = ?

Remember that sound waves will travel at different speeds in different materials. They travel at 343 m/s in air at room temp, but in Helium gas they’ll have a different velocity. Also remember that in an open-end tube, resonance will occur every 1/2(. Therefore, if the “resonance spacings” are 1.1m apart, 1/2( is also 1.1m.

v = f( (480 Hz)(2.2m) 1056.00 m/s

7. The frequency of a tuning fork is unknown. A student uses an air column at room temperature and finds resonances spaced by 0.392m. What is the frequency of the tuning fork?

f = ?

1/2( = 0.392m so 1( = 0.784m

v = 343m/s

v = f( f = v/( 437.50 Hz

8. The auditory canal leading to the eardrum is a closed pipe 0.03m long. Find the lowest frequency (the lowest pitch) sound wave that can resonate in it in room temperature air.

length of closed-end tube = 0.03m

v = 343m/s

f = ?

**This question wants to know the lowest frequency of sound wave that will resonate in your ear canal. Remember that low frequency waves have really long wavelengths. So, you need to actually find the largest wave (biggest wavelength) that can resonate in an ear canal that is 0.03m long. In a closed-end pipe, the smallest fraction of a wave that is needed to fit into the pipe in order to resonate is 1/4(. Therefore, the largest wave that can resonate in the ear canal is one that is so big it can only fit ¼ of itself into the pipe. So, the length of the pipe represents ¼ ( (0.03m) and 1( = 0.12m

¼ ( = 0.03m so 1( = 0.12m

v = 343 m/s

f = ?

v = f( f = v/( 343m/s / 0.12m 2,858.33 Hz

*This is the frequency that our ears are most sensitive to.

9. A bugle can be though of as an open-end pipe. If a bugle were straightened out, it would be 2.65 long. If the speed of sound is 343m/s, find the lowest frequency (the lowest pitch) that can resonate in it.

Length of open tube resonating chamber = 2.65m

v = 343m/s

lowest “resonating” frequency = ?

This problem is very similar to problem #8. The only difference is this is an “open-tube” problem. If we want the lowest frequency, we’re actually looking for the largest wavelength (remember that low frequencies have large wavelengths) that can resonate in a tube 2.65m long. The largest wavelength (and therefore the lowest frequency) of any wave that can fit in an open-end tube and resonate is a wave that is so big it can only fit ½ of a wavelength in it. Therefore, the length of the tube must be equal to 1/2( (2.65m), so 1( = 5.28m

1( = 5.28m

v = 343 m/s

f = ?

v = f( f = v/( 343m/s / 5.28m 64.96 Hz

10. A soprano sax is an open pipe. If all its keys were closed, it is approximately 0.65m long. Using 343 m/s as the speed of sound, find the lowest frequency (the lowest pitch) that can be played on this instrument.

length of open-tube = 0.65m

v = 343m/s

f = ?

This problem is another open-tube problem just like #9. In order to find the lowest frequency of sound that will resonate in it, you really need to find the longest wavelength that can fit in it and produce resonance. The largest wave (and therefore the lowest frequency) that can resonate in an open-end pipe is a wave that is so large that it can only fit ½ of its wavelength can fit in it. Therefore, the length of the tube is equal to ½ ( (0.65m), so a whole wavelength is equal to 1.30m.

1 (= 1.30m

v = 343 m/s

f = ?

v = f( f = v/( 343m/s / 1.30m 263.85 Hz

11. A 5000 Hz sound wave is emitted by an ambulance moving at 3.5 m/s toward a stationary person. What is the frequency actually heard by the person?

f = 5000 Hz

V = 343m/s

Vs = 3.5 m/s

Vo = 0 m/s

f ’ = ? f ’ = f [pic]

*use the Doppler shift equation with the top (toward = top) symbols in both parts.

The observed frequency should be larger than 5000 Hz….and it is. 5051.54 Hz

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