16.5 The Nature of Sound Waves

THE PRESSURE AMPLITUDE OF A SOUND WAVE

Loudness is another attribute of a sound that depends primarily on the pressure amplitude of the wave. Note that the “equilibrium” position (the horizontal axis) is not at zero pressure, but at the ambient (equilibrium) air pressure: ~100 kPa, 1Pa=1N/m2

And remember that the amplitude is the maximum deviation from the equilibrium value.

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In Class Demo

16.6 The Speed of Sound

Sound travels through gases, liquids, and solids at considerably different speeds. Speeds in solid ~1000’s m/s Liquids: ~1000-1500 m/s Gases: ~100’s m/s For sound: the speed of propagation is also independent of frequency or wavelength The mechanisms of sound propagation are rather different between fluids and solids But all are longitudinal compression waves

2

16.6 The Speed of Sound

In a gas, the propagation of condensations (compressions) and rarefactions occurs by the collision of the gas molecules The compression and decompression of the gas in a sound (“acoustic”) wave occurs adiabatically (i.e. without heat being transfers between the small parcels of gas molecules – and the temperature also fluctuates) see section 15.5 of the textbook (subject was covered last semester): In

adiabatic processes: PiViγ= PfVf

γ (15.5)

The letter γ is the lower case “gamma” in Greek.

For an ideal Gas, The acoustic wave speed is given by m

kTγ=v

Number) s(Avogadro' 1002.6constant) gas ideal (the KJ31.8

/KJ1038.1

23

23

×=

==×= −

A

A

NR

NRk

)H ,N ,O like gases diatomic(57or

Ar) He, like gases atomic-(mono 35

222

=γ

3

v= wave speed in m/s T = absolute temperature in kelvins m = mass of gas molecule in kg (use weighted average for mixtures)

4

16.6 The Speed of Sound

Example: Calculate the speed of sound in air

5

16.6 The Speed of Sound

Example: Calculate the speed of sound in air Air is about 20% O2 and 80% N2 m = (0.2)(mO2) + (0.8)(mN2) mO2 = 2(16.0 g)/NA = (0.0320 kg)/ (6.02x1023) = 5.316E-26 kg (computer notation) mN2 = 2(14.0 g)/NA = (0.0280 kg)/ (6.02E23) = 4.651E-26 kg m = (0.20)(5.316E-26 kg) + (0.80)(4.651E-26 kg) = 4.784E-26 kg For diatomic gases (both components are): γ=7/5 = 1.40 What temperature? Let’s say: T=293 K (preferred room temperature of U.S. men)

m/s 344 vm/sJ/kgs/m J/kg ,s/m kgJ :

J/kg 344J/kg 183.1

kg 104.784K) J/K)(293 1038.1)(40.1( v

2222

26-

23

=

=→==

==

××

==−

Units

mkTγ

Notice v is independent of the ambient pressure meaning that your voice sounds roughly the same on top of Mount Everest as in Death Valley; There is only a weak dependence on the temperature within the range found on Earth

16.7 Sound Intensity

Sound waves carry energy that can be used to do work. The amount of energy transported per second is called the power (P) of the wave. The sound intensity (usually denoted as I) is defined as the power that passes perpendicularly through a surface divided by the area (A) of that surface.

API =

6

Units of power: W (watt) 1 watt = 1 joule/second

16.7 Sound Intensity

Example Sound Intensities: 12x10-5W of sound power passed through the surfaces labeled 1 and 2. The areas of these surfaces are 4.0m2 and 12m2. Determine the sound intensity at each surface.

7

16.7 Sound Intensity

Example Sound Intensities: 12x10-5W of sound power passed through the surfaces labeled 1 and 2. The areas of these surfaces are 4.0m2 and 12m2. Determine the sound intensity at each surface.

8

252

5

11 mW100.3

4.0mW1012 −

−

×=×

==API

252

5

22 mW100.1

12mW1012 −

−

×=×

==API

16.7 Sound Intensity

For a 1000 Hz tone, the smallest sound intensity that the human ear can detect is about 1x10-12W/m2. This intensity is called the threshold of hearing. On the other extreme, continuous exposure to intensities greater than ~1.0 W/m2 can be painful. If the source emits sound uniformly in all directions, the intensity depends on the distance from the source in a simple way.

9

24 rPI

π=

I Intensity of sound at observer P Power “emitted” by an “isotropic” sound source r Distance from source to observer 4πr2 surface area of a sphere of radius r

16.8 Intensity Level: Decibels Biological response to sound intensity (loudness) turns out also to be logarithmic

The decibel (dB) is a measurement unit used when comparing two sound Intensities (or for that matter, ratio of any two values in general).

For example, we often encounter signal strength of electrical signals quoted in “dbmV”

( )

=

REF

logdB 10X

Xβ

10

Example: Signal strength in cable TV systems are measured relative to (noise level) of 1mV(millivolt) across 75 ohms. This measure is called dBmV. Expressed in dBmV, the minimum room-temperature noise in a perfect cable system is -59.17 dBmV. What is the voltage of this minimum noise?

dBX is the decibel “level” of whatever quantity X that we are measuring relative to a predefined reference XREF

16.8 Intensity Level: Decibels

( )

=

REF

logdB 10X

Xβ

11

Example: Signal strength in cable TV systems are measured in dB relative to 1 mV(millivolt) across 75 ohms. This measure is called dBmV. Expressed in dBmV, the minimum room-temperature noise in a perfect cable system is -59.17 dBmV. What is the voltage of this minimum noise? β = -59.17 dB = (10 dB) log[V / (1 mV)] log[V / (1 mV)] = -59.17 / 10 = -5.917 V / (1 mV) = 10-5.917 = 1.21 x 10-6 V = (1 mV)(1.21 x 10-6) V = 1.21 x 10-9 V

dBX is the decibel “level” of whatever quantity X that we are measuring relative to a predefined reference XREF

Biological response to sound intensity (loudness) turns out also to be logarithmic

The decibel (dB) is a measurement unit used when comparing two sound Intensities (or for that matter, ratio of any two values in general).

For example, we often encounter signal strength of electrical signals quoted in “dbmV”

( ) ( )oII /logdB 10=β212 mW1000.1 −×=oI

12

16.8 Intensity Level: Decibels

Because of the way in which the human hearing mechanism responds to intensity, it is appropriate to use a logarithmic scale called the intensity level: β

Where we have chosen the threshold of human hearing at 1000 Hz to be the reference intensity: I0

The curve represents the intensity levels at which sounds of various frequencies have the same loudness. The curves are labeled by their intensity levels at 1000 Hz and are known as the Fletcher–Munson curves.

16.8 Decibels Example Comparing Sound Intensities

Audio system 1 produces a sound intensity level of 90.0 dB, and system 2 produces an intensity level of 93.0 dB. Determine the ratio of intensities I2/I1.

16.8 Decibels Example Comparing Sound Intensities

Audio system 1 produces a sound intensity level of 90.0 dB, and system 2 produces an intensity level of 93.0 dB. Determine the ratio of intensities I2/I1.

16.8 Decibels Example Comparing Sound Intensities

Audio system 1 produces a sound intensity level of 90.0 dB, and system 2 produces an intensity level of 93.0 dB. Determine the ratio of intensities I2/I1.

( )

=

oIIlogdB 10β

( )

=

oII1

1 logdB 10β ( )

=

oII2

2 logdB 10β

( ) ( ) ( ) ( )

=

=

−

=−

1

2

1

21212 logdB 10logdB 10logdB 10logdB 10

II

IIII

II

II

o

o

oo

ββ

( )

=

1

2logdB 10dB 0.3II

0.210 30.0

1

2 ==II

=

1

2log0.30II

16.9 The Doppler Effect

The Doppler effect is the change in frequency or pitch of the sound detected by an observer because the motion of the source and/or motion of the observer relative to the medium of sound propagation. In the following, the subscript “O” refers to “observer” and subscript “S” refers to the “source” Notice however, that even within the SAME problem, the same object can be both the source and the observer, but in different contexts

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16.9 The Doppler Effect

MOVING SOURCE

Tvs−=′ λλ

sssso fvfv

vTv

vvf−

=−

=′

=λλ

−

=vv

ffs

so 11

17

MOVING OBSERVER

+=

+=

+=

λλλ s

os

s

ossoo f

vff

vfvfvvf 1

+=

vvff o

so 1

The effect of moving source is SIMILAR, but not identical to the case of the moving observer. They give difference relationships between the the observed frequency fO and the source frequency fS.