Sound and Human Hearing

NOTE: There are short answer analysis questions in the Participation section the informal lab report. Remember to include these answers in your lab notebook as they will be part of your participation grade. PHYS 0212 Sound and Human Hearing

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Sound and Human Hearing This experiment has five parts: 1. Determine the velocity of sound in air with an open-closed tube 2. Determine the velocity of sound in air with an open-open tube 3. Study the relationship between frequency and musical scales ♫ 4. Measure the beat frequency between two sources that are close in pitch 5. Measure the harmonics of your voice The first two parts of this lab will use standing waves to determine the velocity of sound. First, let us consider standing waves on a string. Waves on a string are transverse – the displacement is perpendicular to the direction of travel. y Travel x Displacement PHYS 0212 Sound and Human Hearing

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The equation for a wave is:

y  x, t   A sin  kx  t 

This gives you the displacement y at a position x and at a time t. y

x



 is the wavelength

2  is the angular frequency:   2 f  T

k is the wave number:

k

2



The frequency f is the number of crests that pass a given point per second. The units are Hertz (Hz) where 1 Hz = 1/s. The period T is the time for one complete oscillation.

PHYS 0212 Sound and Human Hearing

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When a wave encounters a barrier, it reflects and travels in the reverse direction. y

x

The incident and reflected waves will overlap and their displacements will add together.

y  x, t   A sin  kx  t   A sin  kx  t  Incident wave

Reflected wave

We can use the trigonometric identity:

sin  A  B   sin A cos B  cos A sin B

To rewrite this as:

y  x, t   2 A sin  kx  cos t 

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y  x, t   A sin  kx  t   A sin  kx  t  y  x, t   A  sin kx cos t  cos kx sin t   A  sin kx cos t  cos kx sin t  y  x, t   A  sin kx cos t  cos kx sin t  sin kx cos t  cos kx sin t  y  x, t   2 A sin  kx  cos t  Note that y = 0, when kx is a multiple of :

kx  n , n  0,1, 2,

If the boundary is placed at x = L, so that L is the length of the string, then

kL  n , n  1, 2,3,

n n  L   n  k 2  2

This says that if L is equal to an integer multiple of a half-wavelength, both ends of the string (x = 0 and x = L) will always have zero displacement.

y  0, t   0 These two points are called nodes and the resulting y ( L, t )  0 wave is called a standing wave. PHYS 0212 Sound and Human Hearing

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L

n=1 Node

 2 Node

Antinode

X=0

X=L

 2  4 The nodes are half a wavelength apart. The antinode is the x position where the maximum amplitude occurs. It is half way between two nodes, so it is one quarter wavelength from either node.

PHYS 0212 Sound and Human Hearing

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L2

n=2

 2

Node

Node Node Antinode

Antinode

 2

 2 v f

v  f

Relationship between wave velocity, frequency and wavelength.

n L  2

n v    2 f 

 v  f  n   2L  PHYS 0212 Sound and Human Hearing

 v  fn  n   2 L   These frequencies are called harmonics. 7

n 1

Each harmonic is an integer multiple of the fundamental.

 v  f1    2 L   Fundamental Frequency n2  v  f2  2    2 f1  2L  2nd Harmonic n3  v  f3  3    3 f1  2L  3rd Harmonic PHYS 0212 Sound and Human Hearing

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Sound Waves

Rarefaction

Compression

Sound waves are longitudinal pressure waves – the displacement is parallel to the direction of travel.

 Compression = Region of high pressure

Rarefaction = Region of low pressure

Mathematically, we can treat sound waves just like we did the transverse waves on a string. Instead of displacement y, we will use pressure P:

PHYS 0212 Sound and Human Hearing

P  x, t   A sin  kx  t  9

Node

Antinode

The Open-Closed Tube

n 1

L

 4

 4

n2  4

 2

2

L3

 2

n3  4

L



 2 PHYS 0212 Sound and Human Hearing





 4

4

  L  2   2 4 L5



4 10

n 1 L



4 Fundamental

n2 L3



L   2n  1 , n  1, 2,3, 4

L

 2

n

 4

 4

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Node

Antinode

Note: The antinode at the end of the tube may extend a little outside the tube and this changes the equation. This is called the end correction.

For a cylindrical tube, the end correction scales with the diameter (d) of the tube:

L



 d  n   2 4 2

Measure and record the diameter of the tube! PHYS 0212 Sound and Human Hearing

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Plot L versus n (mode number):



 d  L  n   2 4 2

Slope =

 2

 d  Intercept =     4 2

v f PHYS 0212 Sound and Human Hearing

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Basic Procedure Measure the frequency from the sine wave generator. Slowly increase the length of the tube until the amplitude is at a maximum. Record the Length (L) and the mode number (n).

Repeat for all the harmonics in the tube. Repeat with a different frequency.

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Antinode

Antinode

The Open-Open Tube

n 1

L

 2

 2

n2  2

2

 2

n3  2

L2



 2

L3

 2

 2 PHYS 0212 Sound and Human Hearing

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

L  n , n  1, 2,3, 2

2L  n

v  v  f    n   2L n  2L  v

Antinode

Antinode

Note: The antinodes at the ends of the tube may extend a little outside the tube and this changes the equation. This is called the end correction.

For a cylindrical tube, the end correction scales with the diameter (d) of the tube:

v   f  n  2 L  1.4 d  

Measure and record the diameter of the tube! PHYS 0212 Sound and Human Hearing

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Plot f versus n (mode number):

v   f  n  2 L  1.4 d  

v Slope = 2 L  1.4d Intercept =

PHYS 0212 Sound and Human Hearing

0

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Fast Fourier Transform Natural sounds are a sum of many different frequencies:

P  t   A1 sin  2 f1t   A2 sin  2 f 2t  

 AN sin  2 f N t 

A Fourier Transform F is a mathematical operation that determines how much each frequency contributes to the total sound.

F f 

DON’T PANIC





f  t ei 2 ft dt

DON’T PANIC

A Fast Fourier Transform (FFT) is a discrete and optimized version of a Fourier transform that produces a plot of the relative amplitude versus frequency.

n 1

f1

n2

f2

n3

n4

f3

f4

PHYS 0212 Sound and Human Hearing

You can determine the harmonics of the tube by finding the peak frequencies on the FFT graph.

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Jean Baptiste Joseph Fourier (1768-1830)

Basic Procedure Remove the cap from the tube. Place the Vernier microphone inside the tube. Collect data and find the harmonic frequencies from the FFT. Make a plot of f versus n and determine the velocity of sound from the slope.

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Musical Scales An octave is an interval over which the frequency doubles.

D

F

E

G

B

A

D

C

E

F

G A

C A B 220 Hz 440 Hz 880 Hz The notes of an octave are divided up into 8 whole steps: A, B, C, D, E, F, G. There are also half steps (sharps and flats) between these notes: F ♯ , G♭

G

G ♯ , A♭

F

C♯,

D♭

D

C

A♯, B♭ A 1

2

Thus there are 12 intervals in an octave.

B 3

4

D♯, E♭

E

5

6

7

8

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10

11

12 20

An equal temperament scale may be calculated using the following equation:

f  f 0  2m 12  m= 0 m= 1 m= m= m= m=

2 3 4 5

m= 6 m= 7 m= 8 m= 9 m = 10 m = 11 m = 12

f   440 Hz  20 12

A = 440 Hz

f   440 Hz  21 12

A♯ = 466 Hz

f   440 Hz  22 12

B = 494 Hz

f   440 Hz  23 12

C = 523 Hz

f   440 Hz  2 f   440 Hz  25 12

D = 587 Hz

f   440 Hz  26 12

D♯ = 622 Hz

4 12

C♯

= 554 Hz

f   440 Hz  27 12

E = 659 Hz

f   440 Hz  28 12

F = 698 Hz

= 740 Hz f   440 Hz  2 f   440 Hz  210 12 G = 784 Hz 9 12

F♯

Starting frequency: f 0 Step: m (integer)

The starting frequency for a scale is arbitrary and there are two common standards in music. The standard used here is the A440, which has middle A as the starting frequency set to 440 Hz.

f0  440 Hz

f   440 Hz  211 12 G♯ = 831 Hz

f   440 Hz  212 12 A = 880 Hz PHYS 0212 Sound and Human Hearing

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In this part of the lab, you will verify that the spacing between notes in a musical scale is indeed exponential.

f  f 0  2m 12  You will then determine the standard (A440 or C256) for a xylophone and a set of tuning forks.

Basic Procedure Measure the frequency with the FFT for each note (C to C) of the xylophone.

Measure the frequency with the FFT for each note (C to C) of the tuning forks. Compare the xylophone and turning fork notes to those predicted by the A440 and C256 standards.

PHYS 0212 Sound and Human Hearing

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Beat Frequency When you add two oscillators together, the result looks like this: 1.5000 1.5000

The result is just the sum of the displacements.

0.5000 0.5000 0.0000 0.0000 -0.5000 -0.5000

2.5000

-1.0000 -1.0000

2.0000

-1.5000 -1.5000

1.0000

1.5000 Seconds Seconds

20 Hz

Amplitude

Amplitude Amplitude

1.0000 1.0000

0.5000 0.0000 -0.5000 -1.0000

1.5000

-1.5000 -2.0000

1.0000

Amplitude Amplitude

-2.5000 0.5000

Seconds

0.0000 -0.5000 -1.0000 -1.5000 Seconds

320 Hz PHYS 0212 Sound and Human Hearing

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Beat Frequency If the two frequencies are close then the result is a beat pattern. 1.5000 1.5000

The beat frequency is just the frequency difference of the two oscillators.

Amplitude Amplitude

1.0000 1.0000 0.5000 0.5000

f B  f 2  f1

0.0000 0.0000 -0.5000 -0.5000

2.5000 2.0000

-1.0000 -1.0000

1.5000

-1.5000 -1.5000 Seconds Seconds

1024 Hz

Amplitude

1.0000 0.5000 0.0000 -0.5000 -1.0000

1.5000

-1.5000 -2.0000

1.0000

Amplitude Amplitude

-2.5000 0.5000

Seconds

0.0000

f B  1047 Hz  1024 Hz  23 Hz

-0.5000 -1.0000 -1.5000 Seconds

1047 Hz PHYS 0212 Sound and Human Hearing

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The beat frequency sounds like a warble between the two sounds. Musicians often tune one instrument to another by adjusting the pitch until the beat frequency disappears.

You can determine the beat frequency by measuring the period between the beats. 2.5000 2.0000 1.5000

Amplitude

1.0000 0.5000 0.0000 -0.5000 -1.0000 -1.5000 -2.0000 -2.5000

43.5 ms

Seconds

f B  1 T  1 0.0435 s  23 Hz PHYS 0212 Sound and Human Hearing

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The Human Voice The human voice is produced by air moving across the vocal cords within the larynx (B). The vocal cords are located just behind the laryngeal prominence (Adam's Apple) (A) and they are basically membranes that partially cover the top of the trachea (C) whenever we use our voice. The esophagus (D) carries food and drink to the stomach.

This is a picture inside the larynx when the vocal cords are abducted (open) for breathing. The cords themselves are the white membranes.

Photograph courtesy of James P. Thomas, M.D.

PHYS 0212 Sound and Human Hearing

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When we use our voice, the vocal cords close over the trachea leaving only a small gap. Air from the lungs passes through this gap causing the vocal cord membranes (the vocal folds) to vibrate.

Photograph courtesy of James P. Thomas, M.D. Photograph courtesy of James P. Thomas, M.D.

Though the human voice is much more subtle and dynamic, the mechanism is similar to blowing air across a stretched rubber band. The more you stretch the rubber band, the higher the pitch (frequency) that is produced. Likewise, the more the vocal cords are stretched, the higher the pitch of the voice. High Pitch

Low Pitch PHYS 0212 Sound and Human Hearing

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Your voice, like all naturally occurring sounds, is composed of many different frequencies. The most prominent frequencies are the harmonics above and below the frequency you are trying to produce. For instance, say you attempt to sing a constant note of C = 128 Hz. A FFT of your voice would reveal peaks at 128 Hz, 256 Hz, 512 Hz and so on. The combination of all the harmonics in a sound is called the timbre. The number of harmonics and their amplitude varies from one person to the next, so the timbre of your voice is unique.

Basic Procedure Sing a note with your voice matched to a tuning fork Produce a FFT plot and find the harmonics Make a plot of fn versus n and show that the slope is f1, as expected from: f n  nf1

Compare the number and amplitudes of your harmonics to those of you lab partner

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NOTE: There are short answer analysis questions in the Participation section the informal lab report. Remember to include these answers in your lab notebook as they will be part of your participation grade.

PHYS 0212 Sound and Human Hearing

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The simplest approach to building a musical scale is to make the spacing between notes linear. That is, to take the frequency range of one octave and just divide by twelve.

880 Hz  440 Hz step   36.67 Hz 12

A = 440 Hz A♯ = 477 Hz B = 513 Hz

Though simple to calculate, this method does not produce intervals of pitch that are pleasing to the human ear.

C = 550 Hz C♯ = 587 Hz D = 623 Hz D♯ = 660 Hz E = 697 Hz F = 733 Hz F♯ = 770 Hz G = 807 Hz

The equal temperament scale, with the pitches spaced exponentially, is the most pleasing to the ear. This is just due to the fact that the ear is constructed to respond logarithmically to changes in pitch in much the same way that it responds logarithmically to changes in intensity.

G♯ = 843 Hz A = 880 Hz PHYS 0212 Sound and Human Hearing

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Another common standard is the C256 with the starting frequency being middle C at 256 Hz. 0 12 C = 256 Hz m = 0 f   256 Hz  2 1 12 C♯ = 271 Hz m = 1 f   256 Hz  2 m= m= m= m=

2 3 4 5

m= 6 m= 7 m= 8 m= 9 m = 10 m = 11 m = 12

f   256 Hz  22 12

f   256 Hz  23 12

D = 287 Hz D♯ = 304 Hz

f   256 Hz  24 12 f   256 Hz  25 12

F = 342 Hz

f   256 Hz  26 12

F♯ = 362 Hz

E = 323 Hz

f   256 Hz  27 12

G = 384 Hz

f   256 Hz  28 12

G♯ = 406 Hz

Notice that the notes in the C256 standard are very close to the those in the A440 standard.

A440

A = 431 Hz f   256 Hz  2 f   256 Hz  210 12 A♯ = 456 Hz

A = 440 Hz f  440 Hz  431 Hz  9 Hz

f   256 Hz  211 12 B = 483 Hz

B = 494 Hz f  494 Hz  483 Hz  11 Hz

9 12

f   256 Hz  212 12 C = 512 Hz

A♯ = 466 Hz f  466 Hz  456 Hz  10 Hz

C = 523 Hz f  523 Hz  512 Hz  11 Hz

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Helpful Hint To see if a particular note comes from the A440 or C256 standard, solve this equation for m:

f  f0  2

m 12



12  f  m ln   ln 2  f 0 

Say you measure a frequency of 1024 Hz from a tuning fork. Is this part of the A440 or C256 standard?

Try 440 Hz:

12  1024 Hz  m ln    14.6 ln 2  440 Hz 

Try 256 Hz:

m

m is not an integer!

12  1024 Hz  m is an integer, so it must be ln   2 4. 0  part of the C256 standard. ln 2  256 Hz 

PHYS 0212 Sound and Human Hearing

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