16.5 The Nature of Sound Waves
THE PRESSURE AMPLITUDE OF A SOUND WAVE
Loudness is another In Class Demo 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.
1 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 γ γ adiabatic processes: PiVi = PfVf (15.5) The letter γ is the lower case “gamma” in Greek.
5 For an ideal Gas, γ kT γ = (mono - atomic gases like He, Ar) The acoustic wave = 3 speed is given by v 7 m or (diatomic gases like O2 , N2 , H2 ) 5
−23 v= wave speed in m/s k =1.38×10 J K = R / N A T = absolute temperature in kelvins R = 8.31J K (the ideal gas constant) m = mass of gas molecule in kg = × 23 (use weighted average for mixtures) N A 6.02 10 (Avogadro's Number) 3 16.6 The Speed of Sound
Example: Calculate the speed of sound in air
4 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) 23 mO2 = 2(16.0 g)/NA = (0.0320 kg)/ (6.02x10 ) = 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) γ kT (1.40)(1.38×10−23 J/K)(293 K) v = = m 4.784×10-26 kg = 1.183 J/kg = 344 J/kg Units : J = kg m2 / s2 , J/kg = m2 / s2 → J/kg = m/s v = 344 m/s
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 5 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.
P I = A
Units of power:
W (watt)
1 watt = 1 joule/second
6 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.
P 12×10−5 W = = = × −5 2 I1 2 3.0 10 W m A1 4.0m
P 12×10−5 W = = = × −5 2 I2 2 1.0 10 W m A2 12m 8 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.
P I = 4π r 2
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 9 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” X dBX is the decibel “level” of whatever β = (10 dB)log quantity X that we are measuring relative X REF to a predefined reference XREF 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?
10 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”
X dBX is the decibel “level” of whatever β = (10 dB)log quantity X that we are measuring relative X REF to a predefined reference XREF 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 11 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
β = (10 dB)log(I / Io ) −12 2 Io = 1.00 ×10 W m Where we have chosen the threshold of human The curve represents the intensity levels at which hearing at 1000 Hz to be the reference intensity: I 0 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.
12 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. I β = (10 dB)log Io
I I β = 1 β = 2 1 (10 dB)log 2 (10 dB)log Io Io
I I I I I β − β = 2 − 1 = 2 o = 2 2 1 (10 dB)log (10 dB)log (10 dB)log (10 dB)log Io Io I1 Io I1
I2 3.0 dB = (10 dB)log I1
I2 I2 0.30 0.30 = log =10 = 2.0 I1 I1 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
16 16.9 The Doppler Effect The effect of moving source is MOVING SOURCE 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. MOVING OBSERVER
λ′ = λ − vsT v v v v + vo fsv + fsvo vo fo = = = f = = = f 1+ λ′ λ − − o s vsT v fs vs fs λ fsλ fsλ 1 v = o fo fs fo = fs 1+ 1− vs v v 17