Sound

1. 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?

2. Sound: a form of energy 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.

3. Sound: what does it go through? 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!

4. Sound: what is waving? 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!

5. Sound: a travelling wave 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 need to make 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).

6. Speed of sound Newton’s Second law in fluid form gives us the “wave equation” for sound. From this, we get for the molecular displacement, y: y = 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).

7. Volume and pitch 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.

8. Energy, Power and Intensity 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.

9. 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 .

10. Sound in air 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.

11. Sound in air 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.

12. Human Hearing: Pitch 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.

13. Talking 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?

14. 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.

15. Resonance 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).

16. Harmonics 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.

17. Fourier Analysis 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.

18. Human Hearing: Volume 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.

19. Intensity: W/m2 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)!

20. Intensity: need for a new unit Even though we can hear sound down to about 10-12 W/m2, we cannot really tell the difference between a sound of 1.00 x 10-10 W/m2 and 1.01 x 10-10 W/m2 even though the difference is 1.0 x 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?

21. Intensity: dB 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) .

22. Examples The weakestsound intensity we can hear is what we define as Io. In decibels this becomes: Io(dB) = 10*log10(10-12 W/m2 / 10-12 W/m2) = 0 dB . The loudest sound without damaging the ear is 1 W/m2, so in decibels this becomes: I(dB) = 10*log10(1 W/m2 / 10-12 W/m2) = 120 dB .

23. Decibels 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?

24. Decibels Calculating for each: I(dB) = 10*log10(1 x 10-6W/m2 / 10-12 W/m2) = 60 dB I(dB) = 10*log10(2 x 10-6 W/m2 / 10-12 W/m2) = 63 dB . Notice that a sound twice as intense in W/m2 is always 3 dB louder!

25. Distance and loudness 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 10-6W/m2 / 10-12 W/m2) = 60 dB , and I(dB) = 10*log(1/4 x 10-6 W/m2 / 10-12 W/m2) = 54 dB. (Notice that 4 is two 2’s, so the decrease is two 3dB’s for a total of 6 dB.)

26. The Doppler Effect The Doppler Effect is explained nicely in the Computer Homework program (Vol 2, #0) entitled Waves and the Doppler Effect. The basic equation derived in the CH program is: freceiver = fsource *(va +/- vr)/(va +/- vs) where va is the velocity of sound in the air, vr is the velocity of the receiver, and vs is the velocity of the source. Choose the +/- signs to make the answer the way you know it has to be: higher if moving towards, lower if moving away.

27. Dopper Effect – with wind If we have a wind blowing, that means the air is moving. The va will have to be replaced with (va +/- vw) where vw is the velocity of the wind. You must see which way the sound is going – it is going from the source towards the receiver. If the wind is in the same direction as the sound, then you use the plus sign; if opposite you use the negative sign.

28. Dopper Effect off of a moving reflector If you have a moving reflector of sound, then all you do is treat the moving reflector first as a moving receiver, then as a moving source; in the equation below, the freceiver is considered stationary with the original source; the vr becomes the vreflector and the vs becomes the vreflector also: freceiver = fsource *(va +/- vr)/(va -/+ vs)