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PPT – Sound PowerPoint presentation | free to download - id: 594d2-ZTk2M

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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.
- 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 doesnt necessarily mean that they

exist. - Since we note that vibrations are involved in

sound, we can try the wave idea.

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!

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

pressure and the molecules positions!

Sound a travelling wave

- We have already considered waves on a string. We

were able to work with Newtons 2nd law to get a

wave equation for this. Can we do the same for

sound? - YES! We use the fluid equivalent of Newtons

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).

Speed of sound

- Newtons 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/r1/2
- (this is just like v T/m1/2 for a string).

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.

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.

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 .

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/r1/2 P/r1/2 . But

from the ideal gas law, - PV nRT, P nRT/V by definition,
- r m/V. Thus, v P/r1/2 nRT/m1/2 . We

can replace the m/n (total mass per total moles)

by M, the molar mass) - v gRT/M1/2 , where g CP/CV 1.4 for a

diatomic gas like air has to be introduced due to

thermodynamic considerations.

Sound in air

- v gRT/M1/2
- 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 (75oF24oC 297 K), v

1.4 8.3 J/mole K 297 K / .029 kg/mole1/2

345 m/s 770 mph. - At higher altitudes we have lower temperatures

and hence lower speeds.

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.

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?

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.

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).

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.

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.

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.

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 ears 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)!

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 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?

Intensity decibel (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)

10log10(I/Io) where Io is the softest sound we

can hear (10-12 W/m2) .

Examples

- The weakest sound intensity we can hear is what

we define as Io. In decibels this becomes - I(dB) 10log10(10-12 / 10-12) 0 dB .
- The loudest sound without damaging the ear is 1

W/m2, so in decibels this becomes - I(dB) 10log10(1 / 10-12) 120 dB .

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?

Decibels

- Calculating for each
- I(dB) 10log10(1 x 10-6 / 10-12) 60 dB
- I(dB) 10log10(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) 10log10(2I1)

10log10(2)log10(I1) 10.3log10(I1)

3dBI1(dB)

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) 10log(1 x 10-6 / 10-12) 60 dB , and
- I(dB) 10log(1/4 x 10-6 / 10-12) 54 dB.
- (Notice that 4 is two 2s, so the decrease is two

3dBs for a total of 6 dB.)

The Doppler Effect

- 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!).

Electromagnetic Waves

- For waves on a string and sound waves, we can get

a wave equation from Newtons Second Law. - So far in Physics 251 weve 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?

Electromagnetic Waves

- The basic equations for electric and magnetic

fields are the basic four equations weve dealt

with in this course - Gausss Law for Electric Fields
- Gausss Law for Magnetic Fields
- Amperes Law
- Faradays Law
- Together these four laws are called
- Maxwells Equations .

Electric Field Wave Equation

- Weve written Maxwells 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 moeo ?2Ey/?t2
- Compare this to the wave equation for a string

T ?2y/?x2 m ?2y/?t2 .

Electric Field Waves

- ?2Ey/?x2 moeo ?2Ey/?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).

Electric and Magnetic Waves

- Maxwells 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 EM wave carry energy? If so, how does

that energy relate to the amplitude of the field

oscillation?

Electric Field and Energy

- Recall the energy stored in a capacitor
- Energy (1/2)CV2 where V Ed and for a

parallel plate capacitor where the field exists

between the plates C KA / 4pkd. - Thus, Energy (1/2)KA / 4pkdEd2
- (1/2)eVolE2 where Vol Ad, and
- e K(1/4pk) . Note that E2 is proportional to

the energy per volume!

Magnetic Field and Energy

- Recall that the energy stored in an inductor is
- Energy (1/2)LI2 where L for a solenoid is
- L m N2 A/Length and B for a solenoid is
- B m(N/Length)I .
- Thus, Energy (1/2)mN2A/LengthBLength/mN 2
- (1/2)VolB2/m where Vol ALength . Note

that B2 is proportional to the energy per volume!

Energy and EM Waves

- Energy/Vol (1/2)eE2 and
- Energy/Vol (1/2)B2/m
- Note that Energy density (Energy/Vol) when moving

(m/s) becomes Power/Area, or Intensity. - Also from Maxwells equations, we have for EM

waves that Eo/Bo c ?1/(eomo) . - Putting this together, we have I E B / mo .

Intensity and the Poynting Vector

- Maxwells 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.

Momentum

- From Maxwells 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.

Radiation Pressure

- 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.