I say fifty, maybe a hundred horses… What you say, Red Eagle? Topic 4.4 Extended B – The speed of sound

I say fifty, maybe a hundred horses… What you say, Red

Eagle?

Topic 4.4 ExtendedB – The speed of sound

FYI: As you observe the animation note the following:(a) There is a pulse velocity v.(b) There is a compression CREST (aka a CONDENSATION).(c) There is a decompression TROUGH (aka a RAREFACTION).(d) The particles of the medium are displaced PARALLEL to v.

Sound waves may be generated by a plucked string, but they do not travel to our ears via strings.Rather, they travel to our ears via air molecules which have been set in oscillation by the vibrating string (or speaker, or explosion, etc.).


Let's have a microscopic look at air molecules set in motion by a moving pulse generator:

Pul

se G

ener

ator

FYI: This type of wave travels through solids, liquids, and gases. It is a LONGITUDINAL WAVE, unlike that of the plucked string (which is a TRANSVERSE WAVE).

FYI: Solids can have not only longitudinal waves, but they can also have transverse waves. Why is it that fluids (liquids and gases) do NOT have transverse waves?

Stepping back, we can imagine at a higher level a pressure wave traveling through the air column like this:

Note that the wave is reflected just like in a string.


Rather than representing sound waves as compressions and rarefactions, we will represent them as sine waves, whose crests are condensations, and whose troughs are rarefactions.

CondensationRarefaction

FYI: Keep in mind, though, that sound waves are LONGITUDINAL.

The following spectrum diagram illustrates the wide frequency ranges for sound.


Infrasonic

Audible

Ultrasonic

Upper Limit

20 kHz

5 GHz

20 Hz

Humans can hear sounds in the audible range.Other animals can hear higher and lower frequencies than humans.For example, cats, dogs and bats can hear into the ultrasonic range.Earth quake waves are in the infrasonic range.We can use vibrating crystals called transducers and powered by electricity to generate sounds in the ultrasonic range.

Ultrasound is used for sonar, cleaning, and imaging.


Since X-rays are harmful to a developing fetus, ultrasound is used.

Sound waves in the 4.2 GHz range can be used to obtain microscopic images:

The blue conductors are 2 m in width.

Recall that the speed of sound in a string is given by


v =F

The Wave Velocity of a Stretched String

where F is the tension (an elastic property) and is the linear mass density (an inertial property).Tension is related to the elastic properties of the string - how it acts when stretched or compressed.Linear mass density is related to the inertial properties of the string - how resistant it is to acceleration.Thus we can generalize the above equation to

v =elastic property of materialinertial property of material

THE SPEED OF SOUND IN SOLIDS AND LIQUIDS


v =elastic property of materialinertial property of material

For sound travel through solids and liquids the inertial property is volume mass density measured in (kg/m3).

For solids, the elastic property is given by Young's modulus .


For liquids, the elastic property is given by the bulk modulus B.Thus

v =

The Speed of Sound in a Solid

v =B

The speed of Sound in a Liquid.

is Young's modulus

B is bulk modulus


The average density of earth's crust 10 km below the continents is 2.7 g/cm3. The speed of longitudinal seismic waves at that depth is 5.4 km/s. What is Young's modulus for the crust at that depth?


v =

, then = v2.Since

v = 5.4 kms

But 1000 mkm

= 5400 m/s

and = 2.7 gcm3

1 kg1000 g

1003 cm3

13 m3= 2700 kg/m3

so that = v2 = (5400)2(2700)

= 7.91010 Pa

FYI: A pascal (Pa) is a unit of pressure and is equal to a n/m2. You can use unit cancellation to verify that = v2 has these units.


For sound travel through gases the speed is still inversely proportional to the square root of the density. But the formula is more complicated because of the level of compressibility of gases.For air, the approximate speed of sound in m/s is given by

THE SPEED OF SOUND IN GASES

v = 331 + 0.6TCThe Speed of Sound in Air TC is air temperature in C°

For each C° increase in temperature, the speed of sound increases by 0.6 m/s.Thus at 20°C (room temperature) the speed of sound in air is about

v = 331 + 0.6TC

v = 331 + 0.6(20)

v = 343 m/s


A better approximation is given by

THE SPEED OF SOUND IN GASES

v = 331 1 + TC

273

The Speed of Sound in Air TC is air temperature in C°

This formula is more accurate at higher air temperatures.In order to verify its compatibility with the previous equation, we recalculate the speed of sound in air at 20°C:

v = 331 1 + TC

273

v = 331 1 +20273

v = 342.9 m/s


Table 14.1 from your book is reproduced here, to show the speed of sound in various solids, liquids, and gases.

Table 14.1 The Speed of Sound in Various Media (typical values)Medium Speed (m/s)

Solids...

Aluminum 5100

Copper 3500

Iron 4500

Glass 5200

Polystyrene 1850

Granite* 6000Steel* 5941

Medium Speed (m/s)

Liquids...

Mercury 1400

Ethyl Alcohol 1125

Water (0°C) 1402

Seawater* 1522 Water (20°C) 1482

Medium Speed (m/s)

Gases...

Air (100°C) 387

Air (0°C) 331

Air (20°C)* 343

Oxygen (0°C) 316

Hydrogen (0°C) 1284Helium (0°C) 965

*Media from other sources.

In this section we will show the gruesome details in the derivation of the speed of sound in liquids:

In a taut string, potential energy is associated with the periodic stretching of tensioned string elements.


v =B

The speed of Sound in a Liquid. B is bulk modulus

In a fluid, potential energy is associated with the periodic expansions and contractions of small volume elements V of the fluid.The bulk modulus is defined as ratio of the change in pressure p to the corresponding fractional change in volume V/V:

B = -p

V/VThe Bulk Modulus of a Liquid

The ratio V/V has no units, so the units if B are those of pressure: n/m2.

FYI: The harder a liquid is to compress, the smaller the fractional change in volume. How does this affect the bulk modulus?

FYI: The harder a liquid is to compress, the larger the bulk modulus. How does this affect the speed of sound through that liquid?

FYI: The minus sign in the formula for bulk modulus ensures that B is always positive. Can you see how?

Consider a simplified pulse of fluid moving down the tube as shown:


As the compression zone moves through the fluid at the wave velocity v, it runs into non-compressed fluid.

compression zone

We now analyze (using Newton's 2nd law) a small element of that non-compressed fluid, shown immediately to the right of the compression zone:

v x

A

x

A

mass element

mass element

The mass of the non-compressed fluid element is given by

mass = density·volumem = Vm = ·x·A m = Ax


All of the fluid within the mass element is accelerated to the wave speed v in the time it takes the compression zone to move through the distance x. But

compression zone

v x

A

mass element

m = Ax

v = xt

so that x = vt which we may substitute:

m = Ax m = Avt


We know that the average acceleration of the mass element is

compression zone

v x

A

mass element

m = Avt

a = vt

so that Newton's 2nd law,

F = ma , becomes

F = ma

F = (Avt) vt

F = (Av)(v)

F = (Av2)(v/v) Why?


Recall that pressure = force/area, so that

compression zone

v x

mass element

F = pA

F = (Av2)(v/v)

The fluid element therefore has two forces acting on it in the direction of its acceleration, shown above.

(p + p)A pA

Thus the sum of the forces acting on the fluid element is given by F = (p + p)A - pA

F = Ap and we can substitute:

F = (Av2)(v/v) Ap = (Av2)(v/v)

F = pA + pA - pA

p = (v2)(v/v)


Just as we could express the mass in terms of the wave velocity, we can express the volume V in terms of the wave velocity v:

compression zone

v x

mass element

p = (v2)(v/v)

V = AxV = Avt

V = Avt

If the volume of the mass element is V = Avt, then the change in volume of the mass element is

Since the change in volume of the mass element is negative, but its change in velocity is positive, we put a negative sign in the formula to get

V = -Avt


compression zone

v x

mass element

p = (v2)(v/v)

Now we can look at the fractional change in volume of the fluid element, and write it in terms of the wave velocity:

V = -Avt

V = Avt

VV

= -AvtAvt

= - vv

so thatv/v = - V/V


compression zone

v x

mass element

p = (v2)(v/v)

Substitution yields

v/v = -V/V

p = (v2)(v/v)

p = (v2)(-V/V)

v2 = - pV/V

v2 = B

v =

B

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I say fifty, maybe a hundred horses… What you say, Red Eagle? Topic 4.4 Extended B – The speed of sound