Sound. What is sound? How do we make sound? Why does sound move that fast? What parameters does the speed of sound depend on? How do we work with the pitch and the volume of sound?. Sound: a form of energy. Sound is a form of energy that moves.
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What is sound?
How do we make sound?
Why does sound move that fast? What parameters does the speed of sound depend on?
How do we work with the pitch and the volume of sound?
Sound is a form of energy that moves.
Is this energy carried by particles (that we call phonons), or is it carried by waves?
[The fact that we can call particles of sound “phonons” doesn’t necessarily mean that they exist.]
Since we note that vibrations are involved in sound, we can try the wave idea.
Sound is transmitted through air.
Is sound transmitted through water (and other liquids)? Can you pipe sound into a swimming pool? YES! In the navy, they use sound in sonar to listen for and find things.
Is sound transmitted through solids (like a knock on a door)? YES! Geologists use this to “look” for oil!
Is sound transmitted through vacuum? No!
In waves on a string, the pieces of the string pulled each other via the tension in the string.
In sound, the molecules of the gas, liquid, or solid will pull on each other via the pressure in the material.
What is waving (or oscillating)? Both the pressureand the molecules’ positions!
We have already considered waves on a string. We were able to work with Newton’s 2nd law to get a wave equation for this. Can we do the same for sound?
YES! We use the fluid equivalent of Newton’s Second Law to get a wave equation. With this we have two adjustments: we need a Bulk Modulus (B in Nt/m2) instead of a Tension (Nt), and we have a volume density (r in kg/m3) instead of a linear density (kg/m).
Newton’s Second law in fluid form gives us the “wave equation” for sound. From this, we get for the molecular displacement, y:
y(x,t) = A sin(kx +/- wt)
where v = w/k = (2pf / [2p/l]) = lf
and from the wave equation, v = [B/r]1/2
(this is just like v = [T/m]1/2 for a string).
Note that the speed of sound depends on B and r, and that it relates l and f. Thus, changing the frequency does NOT change the speed, v; instead it will change the wavelength. Changing the material (changing B and/or r) will change the speed.
For sound, then, we see that the pitch is related to frequency (f, or equivalently, l), while the volume is related to the amplitude, A.
In oscillations, we saw that the energy of a mass (piece of the string) was related to w2A2.
The power (Energy/time) of the wave down the string was related to w2A2v.
For sound, however, we need the idea of power/area which we call Intensity.
This intensity is also related like power:
I a w2A2v(here A is amplitude).
But as sound spreads out, the area for this power increases, and so the Intensity falls off. For a point source, the area for the power is that of a sphere (4pr2). For a point source of sound, this takes the form of an inverse square law for I: I = P / 4pr2 .
For an ideal gas, the bulk modulus, B, is simply equal to the pressure, P. Thus, the speed of sound in air is: v = [B/r]1/2= [P/r]1/2. But from the ideal gas law,
PV = nRT, P = nRT/V; by definition,
r = m/V. Thus, v = [P/r]1/2 = [nRT/m]1/2 . We can replace the m/n (total mass per total moles) by M, the molar mass):
v = [gRT/M]1/2 , whereg =CP/CV =1.4for a diatomic gas like air has to be introduced due to thermodynamic considerations.
v = [gRT/M]1/2
For air, g = 1.4, R = 8.3 Joules/mole-K, T is the temperature in Kelvin, and M (a mixture of N2 and O2) is .029 kg/mole.
Thus at room temperature (75oF=24oC = 297 K), v =[1.4 * 8.3 J/mole K * 297 K / .029 kg/mole]1/2= 345 m/s = 770 mph.
At higher altitudes we have lower temperatures and hence lower speeds.
A standard human ear can hear frequencies from about 20 Hz to about 20,000 Hz. As you get older, however, both ends tend to shrink towards the middle. This will be demonstrated during class, and you can hear for yourself what the different frequencies sound like and what your limits are.
How do we understand what people say? Does it have to do with frequency or intensity?
Of course, we can talk loudly or softly, which means we can talk with high or low intensity.
We can also sing our words at different pitches (frequencies).
So what goes into talking?
Along the same lines: both a piano and a guitar can play the same note, but we can tell whether a piano or a guitar did play that note. What is going on?
It turns out that both talking and musical instruments are based on resonance: standing waves are set up in the mouths of people and in the instruments.
We can have the same fundamental frequency set up on a string [ #l/2 = L ] in both a guitar and a piano. But this indicates that there are several wavelengths that obey this. These several wavelengths are called the harmonics, with the longest wavelength (#=1) being the fundamental (longest wavelength, shortest frequency).
Although a guitar and a piano may have the same fundamental frequency, the higher harmonics may resonate differently on the different instruments based on their shape.
In the same way, we can form different words at the same fundamental frequency by changing the shape of our mouth.
It turns out that the ear is a great Fourier Analyzer - that is, it can distinguish many different frequencies in a sound. (The eye is not like this at all!)
It is hard to make computers listen to and understand speech because the computer has to be taught how to Fourier Analyze the sounds and interpret that analysis.
The volume of sound is related to the intensity but it is also related to frequency because the efficiency of the ear is different for different frequencies.
The ear hears frequencies of about 2,000 Hz most efficiently, so intensities at this frequency will sound louder than the same intensity at much lower or higher frequencies.
The ear is a very sensitive energy receiving device. It can hear sounds down as low as 10-12 W/m2. Considering that the ear’s area is on the order of 1 cm2 or 10-4 m2, that means the ear can detect sound energy down to about 10-16 Watts!
The ear starts to get damaged at sound levels that approach 1 W/m2 .
From the lowest to the highest, this is a range of a trillion (1012)!
Even though we can hear sound down to about 10-12 W/m2, we cannot really tell the difference between a sound of 10-11 +/- 10-12 W/m2 .
The tremendous range we can hear combined with the above fact leads us to try to get a more reasonable intensity measure.
But how do we reduce a factor of 1012 down to manageable size?
One way to reduce an exponential is to take its log: log10(1012) = 12
But this gives just 12 units for the range. However, if we multiply this by 10, we get 120 units which is a nice range to have.
However, we need to take a log of a dimensionless number. We solve this problem by introducing this definition of the decibel: I(dB) = 10*log10(I/Io) where Io is the softest sound we can hear (10-12 W/m2) .
The weakestsound intensity we can hear is what we define as Io. In decibels this becomes:
I(dB) = 10*log10(10-12 / 10-12) = 0 dB .
The loudest sound without damaging the ear is 1 W/m2, so in decibels this becomes:
I(dB) = 10*log10(1 / 10-12) = 120 dB .
It turns out that human ears can tell if one sound is louder than another only if the intensity differs by about 1 dB. This does indeed turn out to be a useful intensity measure.
Another example: suppose one sound is 1 x 10-6 W/m2, and another sound is twice as intense at 2 x 10-6 W/m2. What is the difference in decibels?
Calculating for each:
I(dB) = 10*log10(1 x 10-6 / 10-12) = 60 dB
I(dB) = 10*log10(2 x 10-6 / 10-12) = 63 dB .
Notice that a sound twice as intense in W/m2 is always 3 dB louder!
This is the result of a property of logs: If I2 is twice as intense as I1, then in terms of dB:I2(dB) = 10*log10(2*I1) = 10*[log10(2)+log10(I1)] = 10*[.3+log10(I1)] = 3dB+I1(dB)
For a point source, the intensity decreases as the inverse square of the distance. Thus if a source of sound is twice as far away, its intensity should decrease by a factor of 22 or 4. How much will its intensity measured in dB decrease?
I(dB) = 10*log(1 x 10-6 / 10-12) = 60 dB , and
I(dB) = 10*log(1/4 x 10-6 / 10-12) = 54 dB.
(Notice that 4 is two 2’s, so the decrease is two 3dB’s for a total of 6 dB.)
The Doppler Effect is explained nicely in the Computer Homework program (Vol 4, #5) entitled Waves and the Doppler Effect.
fR = fS*[(v +/- vR) / (v +/- vS)]
where speeds are relative to the air, not the ground, and the +/- signs are determined by directions (use common sense!).
For waves on a string and sound waves, we can get a wave equation from Newton’s Second Law.
So far in Physics 251 we’ve talked about electric and magnetic fields. Can the fields “wave” ?
If so, where do we start to try to get a wave equation for the fields?
The basic equations for electric and magnetic fields are the basic four equations we’ve dealt with in this course:
Gauss’s Law for Electric Fields
Gauss’s Law for Magnetic Fields
Together these four laws are called
Maxwell’s Equations .
We’ve written Maxwell’s Equations in integral form, but they can also be written in differential form using the curl and the divergence (Calculus III topics). By combining these equations we get the following wave equation:
2Ey/x2 = moeo2Ey/t2
Compare this to the wave equation for a string: T 2y/x2 = m 2y/t2 .
2Ey/x2 = moeo2Ey/t2
This has the solution:Ey = Eo sin(kx wt + fo)
and the phase velocity of this wave depends on the parameters of the space that the wave is going through: v = [1/(em)] .
Recall that eo = 1/(4pk) where k = 9 x 109 Nt-m2/Coul2 , and mo = 4p x 10-7 T-m/A .
Thus electric waves should propagate through vacuum with a speed of (you do the calculation).
Maxwell’s Equations also predict that whenever we have a changing Electric Field, we have a changing Magnetic Field. Thus, we really have an Electromagnetic Wave rather than just an isolated Electric Field wave.
Does this E&M wave carry energy? If so, how does that energy relate to the amplitude of the field oscillation?
Recall the energy stored in a capacitor:
Energy = (1/2)CV2where V = E*d and for a parallel plate capacitor where the field exists between the plates: C = KA / 4pkd.
Thus, Energy = (1/2)[KA / 4pkd][Ed]2
= (1/2)eVolE2 where Vol = A*d, and
e = K(1/4pk) . Note that E2 is proportional to the energy per volume!
Recall that the energy stored in an inductor is:
Energy = (1/2)LI2where L for a solenoid is
L = m N2 A/Lengthand B for a solenoid is
B = m(N/Length)I .
Thus, Energy =(1/2)[mN2A/Length][BLength/mN ]2
= (1/2)VolB2/m where Vol = A*Length . Note that B2 is proportional to the energy per volume!
Energy/Vol = (1/2)eE2 and
Note that Energy density (Energy/Vol) when moving (m/s) becomes Power/Area, or Intensity.
Also from Maxwell’s equations, we have for E&M waves that: Eo/Bo = c = [1/(eomo)] .
Putting this together, we have: I = E B / mo.
Maxwell’s Equations also predict that electromagnetic waves will travel in a direction perpendicular to the directions of both the waving Electric and the waving Magnetic Fields, and that the direction of the waving Electric Field must be perpendicular to the direction of the waving Magnetic Field. This is stated in the following:
S = (1/o)E x B
where S is called the Poynting vector and gives the Intensity of the electromagnetic wave.
From Maxwell’s Equations we also predict that electromagnetic waves should carry momentum, where the amount of momentum depends on the energy per speed:p = Energy / c .
(In relativity, we will see that electromagnetic energy can be considered to be carried by photons, where photons have mass. From relativity, E = mc2, and p = mc, so p = E/c.)
If the light is absorbed the material will receive this amount of momentum. If the light is reflected, the material will receive twice this amount of momentum.
Pressure is Force/Area, and Force is change in momentum with respect to time. Hence, electromagnetic radiation should exert a pressure on objects when it hits them.
Radiation Pressure = F/A =
(dp/dt) / A = [d(Energy/speed)/dt] / A =
[Power/speed] / A = [Power/Area] / speed = S/c.
If the radiation reflects, then the momentum is twice and so the radiation pressure is also twice.