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Chapter 12

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Chapter 12

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 (20oC) 343

Air (0oC) 331

Water1440

Saltwater1560

Iron/Steel ~5000

- 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?

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 – 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 = Loudness
- Louder = More pressure
- Decibel (dB) – named for Alexander Graham Bell
- Logarithmic scale
- Intensity level =b

b = 10 log I

Io

Io = 1.0 X 10-12 W/m2

= lowest audible intensity

Example

- Rustle of leaves = 10 dB
- Whisper = 20 dB
- Whisper is 10 times as intense
Example

- Police Siren = 100 dB
- Rock Concert = 120 dB

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

What is the intensity of a conversation at 65 dB

- = 10 log I
Io

- = log I
- Io
65 = log I

10 Io

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

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

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

- Inverse-squared radius
- Intensity decreases proportionally as you move away from a sound
I a1orI1r12 = I2r22

r2

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

I = 100 W/m2

I1r12 = I2r22

I2 = I1r12/r22

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

I2 = 120 dB

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

- Octave = a doubling of the frequency
C(middle)262 Hz

D294 Hz

E330 Hz

F349 Hz

G392 Hz

A440 Hz

B494 Hz

C524 Hz

- Set up a vibrating column of air

v = lf

L = nln

2

v = FT

m/L

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

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.

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

What is the wavelength in the air?

v = lf

- = v/f
- = 343 m/s/392 Hz
- = 0.875 m

- Flute or Organ
- Behaves like a string
- The longer the tube, the lower the frequency (pitch)

v = lf

L = nln

2

fn = nv = nf1

2L

Remember n = harmonic

- Clarinet
- Does not behave like a string
- Only hear odd harmonics

v = lf

L = nln

4

fn = nv = nf1

4L

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

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)

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)

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

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)

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)

- Two waves can interfere constructively or destructively
- Point C = constructive interference
- Point D = destructive interference

Constructive intereference

d = n l

Destructive Interference

d = n l

2

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)

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

- 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

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

Ans: 5 Hz

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

- Frequency of sound changes with movement
- Moving towards you = frequency increases (higher pitch)
- Moving away = frequency decreases (lower frequency)

- Universe is expanding
- Evidence (Hubble’s Law)
- Only a few nearby galaxies are blueshifted
- Most are red-shifted

- Universe will probably expand forever

Source moving towards stationary observer

f’ = f

1 - vs

v

Source moving away from stationary observer

f’ = f

1 + vs

v

Observer moving towards stationary source

f’ = 1 + vo f

v

Observer moving away from stationary source

f’ = 1 - vo f

v

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

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

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

f’ = 1 + vo f

v

f’ = [1+(3.50/343)]5000 Hz

f’ = (1.01)(5000 Hz) = 5051 Hz