1 / 51

# Chapter 12 - PowerPoint PPT Presentation

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. Material Speed of Sound (m/s) Air (20 o C) 343 Air (0 o C) 331 Water 1440 Saltwater 1560 Iron/Steel ~5000.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Chapter 12' - moses-marquez

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Chapter 12

Sound

• Varies with the medium

• v = \/ B/r

• Solids and liquids

• Less compressible

• Higher Bulk modulus

• Move faster than in air

Air (20oC) 343

Air (0oC) 331

Water 1440

Saltwater 1560

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

Infrasonic Audible Ultrasonic

20 Hz 20,000 Hz

Earthquakes 50,000 Hz (dogs)

Thunder 100,000Hz(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

• 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

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)

• Intensity decreases proportionally as you move away from a sound

I a1 or I1r12 = 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

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

D 294 Hz

E 330 Hz

F 349 Hz

G 392 Hz

A 440 Hz

B 494 Hz

C 524 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

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 Hz Fundamental (1st Harmonic)

fn = nf1

f2 = 2f1 = 1320 Hz 1st Overtone (2nd Harmonic)

f3 = 3f1 = 1980 Hz 2nd Overtone (3rd Harmonic)

f4 = 4f1 = 2640 Hz 3rd Overtone (2nd Harmonic)

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

4L (4)(0.26 m)

f1 = 330 Hz Fundamental (1st Harmonic)

fn = nf1

f2 = 3f1 = 990 Hz 3rd Harmonic

f3 = 5f1 = 1650 Hz 5th Harmonic

f4 = 7f1 = 2310 Hz 7th 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

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

Beats 0.15 m) or at (4 + 0.15 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

Beats: Example 1 0.15 m) or at (4 + 0.15 m):

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 0.15 m) or at (4 + 0.15 m):

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 0.15 m) or at (4 + 0.15 m):

• Frequency of sound changes with movement

• Moving towards you = frequency increases (higher pitch)

• Moving away = frequency decreases (lower frequency)

Doppler Effect and the Universe 0.15 m) or at (4 + 0.15 m):

• 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 0.15 m) or at (4 + 0.15 m):

Source moving towards stationary observer

f’ = f

1 - vs

v

Source moving away from stationary observer

f’ = f

1 + vs

v

Moving Observer 0.15 m) or at (4 + 0.15 m):

Observer moving towards stationary source

f’ = 1 + vo f

v

Observer moving away from stationary source

f’ = 1 - vo f

v

Doppler: Example 1 0.15 m) or at (4 + 0.15 m):

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? 0.15 m) or at (4 + 0.15 m):

f’ = f

1 + vs

v

f’ = 1600 Hz = 1600 Hz = 1491 Hz

[1 + (25/343)] 1.07

Doppler: Example 2 0.15 m) or at (4 + 0.15 m):

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 + 0.15 m) or at (4 + 0.15 m):vo f

v

f’ = [1+(3.50/343)]5000 Hz

f’ = (1.01)(5000 Hz) = 5051 Hz