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Chapter 12. Sound. Speed of Sound. Varies with the medium v = \/ B/ r Solids and liquids Less compressible Higher Bulk modulus Move faster than in air. MaterialSpeed of Sound (m/s) Air (20 o C) 343 Air (0 o C) 331 Water1440 Saltwater1560 Iron/Steel ~5000.

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Chapter 12

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Chapter 12

Chapter 12

Sound


Speed of sound

Speed of Sound

  • Varies with the medium

  • v = \/ B/r

  • Solids and liquids

    • Less compressible

    • Higher Bulk modulus

    • Move faster than in air


Chapter 12

MaterialSpeed of Sound (m/s)

Air (20oC) 343

Air (0oC) 331

Water1440

Saltwater1560

Iron/Steel ~5000


Speed of sound temperature

Speed of Sound: Temperature

  • Speed increases with temperature (oC)

  • v ≈ (331 + 0.60T) m/s

  • What is the speed of sound at 20oC?

  • What is the speed of sound at 2oC?


Speed of sound example 1

Speed of Sound: Example 1

How many seconds will it take the sound of a lightening strike to travel 1 mile (1.6 km) if the speed of sound is 340 m/s?

v = d/t

t = d/v

t = 1600 m/(340 m/s) ≈ 5 seconds

(count five seconds for each mile)


Pitch

Pitch

  • Pitch – frequency (not loudness)

  • Audible range 20 Hz – 20,000 Hz

    InfrasonicAudibleUltrasonic

    20 Hz 20,000 Hz

    Earthquakes50,000 Hz (dogs)

    Thunder100,000Hz(bats)

    Volcanoes

    Machinery


Intensity

Intensity

  • Intensity = Loudness

  • Louder = More pressure

  • Decibel (dB) – named for Alexander Graham Bell

  • Logarithmic scale

  • Intensity level =b


Chapter 12

b = 10 log I

Io

Io = 1.0 X 10-12 W/m2

= lowest audible intensity


Chapter 12

Example

  • Rustle of leaves = 10 dB

  • Whisper = 20 dB

  • Whisper is 10 times as intense

    Example

  • Police Siren = 100 dB

  • Rock Concert = 120 dB


Decibels example 1

Decibels: Example 1

How many decibels is a sound whose intensity is 1.0 X 10-10 W/m2?

b = 10 log I=10 log (1.0 X 10-10 W/m2)

Io (1.0 X 10-12 W/m2)

b = 10 log (100) = 20 dB


Decibels example 2

Decibels: Example 2

What is the intensity of a conversation at 65 dB

  • = 10 log I

    Io

  • = log I

  • Io

    65 = log I

    10 Io


Chapter 12

6.5 = log I

Io

6.5 = log I – log Io

log I = 6.5 + log Io

log I = 6.5 + log (1.0 X 10-12 W/m2)

log I = 6.5 – 12 = -5.5

I = 10-5.5 = 3.16 X 10-6


Decibels example 3

Decibels: Example 3

What is the intensity of a car radio played at 106 dB?

(Ans: 1.15 X 10-11 W/m2)


Intensity and distance

Intensity and Distance

  • Inverse-squared radius

  • Intensity decreases proportionally as you move away from a sound

    I a1orI1r12 = I2r22

    r2


Distance example 1

Distance: Example 1

The intensity level of a jet engine at 30 m is 140 dB. What is the intensity level at 300 m?

140 dB = 10 log I/Io

14 = log I/Io

14 = log I – log Io

log I = 14 + log Io = 2

I = 100 W/m2


Chapter 12

I = 100 W/m2

I1r12 = I2r22

I2 = I1r12/r22

I2 = (100 W/m2)(30 m)2/(300 m)2

I2 = 120 dB


Distance example 2

Distance: Example 2

If a particular English teacher talks at 80 dB when she is 10 m away, how far would you have to walk to reduce the sound to 40 dB? (Hint: Find the raw intensity of each dB first).

ANS: 1000 m


Musical instruments

Musical Instruments

  • Octave = a doubling of the frequency

    C(middle)262 Hz

    D294 Hz

    E330 Hz

    F349 Hz

    G392 Hz

    A440 Hz

    B494 Hz

    C524 Hz


Stringed instruments

Stringed Instruments

  • Set up a vibrating column of air


Chapter 12

v = lf

L = nln

2

v = FT

m/L


Stringed instr example 1

Stringed Instr.: Example 1

A 0.32 m violin string is tuned to play an A note at 400 Hz. What is the wavelength of the fundamental string vibration?

L = nln

2

L = 1l1

2

l1 = 2L = 0.64 m


Chapter 12

What are the frequency and wavelength of the sound that is produced?

Frequency = 440 Hz

v = lf

  • = v/f = 343 m/s/440 Hz = 0.78 m

    Note that the wavelength differs because the speed of sound in air is different than the speed of the wave on the string.


Stringed instr example 2

Stringed Instr.: Example 2

A 0.75 m guitar string plays a G-note at 392 Hz. What is the wavelength in the string?

L = nln

2

L = 1l1

2

l1 = 2L = 1.50 m


Chapter 12

What is the wavelength in the air?

v = lf

  • = v/f

  • = 343 m/s/392 Hz

  • = 0.875 m


Open tubes

Open Tubes

  • Flute or Organ

  • Behaves like a string

  • The longer the tube, the lower the frequency (pitch)


Chapter 12

v = lf

L = nln

2

fn = nv = nf1

2L

Remember n = harmonic


Closed tube

Closed Tube

  • Clarinet

  • Does not behave like a string

  • Only hear odd harmonics


Chapter 12

v = lf

L = nln

4

fn = nv = nf1

4L

Remember n = harmonic (1, 3, 5, 7, 9…)


Tubes example 1

Tubes: Example 1

What will be the fundamental frequency and first three overtones for a 26 cm organ pipe if it is open?

fn = nv

2L

f1 = 1v

2L

f1 = (1)(343 m/s) = 660 Hz

(2)(0.26 m)


Chapter 12

f1 = 660 HzFundamental (1st Harmonic)

fn = nf1

f2 = 2f1 = 1320 Hz1st Overtone (2nd Harmonic)

f3 = 3f1 = 1980 Hz2nd Overtone (3rd Harmonic)

f4 = 4f1 = 2640 Hz3rd Overtone (2nd Harmonic)


Chapter 12

Perform the same calculation if the tube is closed.

fn = nv=(1)(343 m/s)

4L(4)(0.26 m)

f1 = 330 HzFundamental (1st Harmonic)

fn = nf1

f2 = 3f1 = 990 Hz3rd Harmonic

f3 = 5f1 = 1650 Hz5th Harmonic

f4 = 7f1 = 2310 Hz7th Harmonic


Tubes example 2

Tubes: Example 2

How long must a flute (open tube) be to play middle C (262 Hz) as its fundamental frequency?

fn = nv

2L

f1 = v

2L

L = v=(343 m/s) = 0.655 m (65.5 cm)

2f1(2)(262 Hz)


Tubes example 3

Tubes: Example 3

If the tube is played outdoors at only 10oC, what will be the frequency of that flute?

v = (331 + 0.60T) m/s

v = 331 + (0.6)(10) = 337 m/s

fn = nv

2L

f1 = v=(337 m/s)=257 Hz

2L(2)(0.655m)


Interference of waves

Interference of Waves

  • Two waves can interfere constructively or destructively

  • Point C = constructive interference

  • Point D = destructive interference


Chapter 12

Constructive intereference

d = n l

Destructive Interference

d = n l

2


Interference example 1

Interference: Example 1

Two speakers at 1.00 m apart, and a person stands 4.00 m from one speaker. How far must he be from the other speaker to produce destructive interference from a 1150 Hz sound?

v = l f

  • = v/f = (343 m/s)/(1150 Hz) = 0.30 m

    d = n l=(1)(0.30 m) = 0.15 m

    2(2)


Chapter 12

Since the difference must be 0.15 m, he must stand at (4 – 0.15 m) or at (4 + 0.15 m):

3.85 m or 4.15 m


Beats

Beats

  • Occur if two sources (tuning forks) are close, but not identical in frequency

  • Superposition (interference) pattern produces the beat.

  • Beat frequency is difference in frequencies


Beats example 1

Beats: Example 1

One tuning fork produces a sound at 440 Hz, a second at 445 Hz. What is the beat frequency?

Ans: 5 Hz


Beats example 2

Beats: Example 2

A tuning fork produces a 400 Hz tone. Twenty beats are counted in five seconds. What is the frequency of the second tuning fork?

f = 20 beats=4 Hz

5 sec

The second fork is 404 Hz or 396 Hz


Doppler effect

Doppler Effect

  • Frequency of sound changes with movement

  • Moving towards you = frequency increases (higher pitch)

  • Moving away = frequency decreases (lower frequency)


Doppler effect and the universe

Doppler Effect and the Universe

  • Universe is expanding

  • Evidence (Hubble’s Law)

    • Only a few nearby galaxies are blueshifted

    • Most are red-shifted

  • Universe will probably expand forever


Moving source

Moving Source

Source moving towards stationary observer

f’ = f

1 - vs

v

Source moving away from stationary observer

f’ = f

1 + vs

v


Moving observer

Moving Observer

Observer moving towards stationary source

f’ = 1 + vo f

v

Observer moving away from stationary source

f’ = 1 - vo f

v


Doppler example 1

Doppler: Example 1

A police siren has a frequency of 1600 Hz. What is the frequency as it moves toward you at 25.0 m/s?

f’ = f

1 - vs

v

f’ = 1600 Hz = 1600 Hz = 1726 Hz

[1 – (25/343)] 0.927


Chapter 12

What will be the frequency as it moves away from you?

f’ = f

1 + vs

v

f’ = 1600 Hz = 1600 Hz = 1491 Hz

[1 + (25/343)] 1.07


Doppler example 2

Doppler: Example 2

A child runs towards a stationary ice cream truck. The child runs at 3.50 m/s and the truck’s music is about 5000 Hz. What frequency will the child hear?

f’ = 1 + vo f

v


Chapter 12

f’ = 1 + vo f

v

f’ = [1+(3.50/343)]5000 Hz

f’ = (1.01)(5000 Hz) = 5051 Hz


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