c

SOUND - LONGITUDINAL WAVE

dp, d, dVx

[Vx + dVx = dVx]0

(at any position, no properties are changing with time)

(V and are only functions of x)

(dVxA << ddVxA)

dRx represents tangential forces on control volume; because there is no relative motion along wave (wave is on both sides of top andbottom of control volume), dRx =0. So FSx = -Adp

only CV if acceleratingthen this term appears

dRx represents tangential forces on control volume; because there is no relative motion along wave (wave is on both sides of top andbottom of control volume), dRx =0. So FSx = -Adp

du/dy = 0

du/dy = 0

Change in momentum flux

From continuity eq.

Total forces = normal surface forces

Continuity

MOMENTUM EQ.

CONTINUITY EQ.

c2 = dp/d

FOR IDEAL GAS (p = RT): & if isentropic, p/k = const

Then: c2 = dp/d = p/s

= d([const]k)/d = [const]kk-1 = [const]k(k/)

= kp/ = kRT/ = kRT

c = (kRT)1/2 ~ 340 m/s STP

Should the adiabatic approximation be better at low or high frequencies?

Note: the adiabatic approximation is better at lower frequencies than higher frequencies because the heat production due to conduction is weaker when the wavelengths are longer (frequencies are lower).

The adiabatic condition works because for most sound waves

of interest the distance between compression and rarefaction

are so widely separated that negligible transfer of heat takes place.

“The often stated explanation, that oscillations in a sound wave are too rapid to allow appreciable conduction of heat, is wrong.”

~ pg 36, Acoustics by Allan Pierce

Newton (1686)* was the first to predict thevelocity of sound waves in air. He used

Boyles Law and assumed constant temperature.

FOR IDEAL GAS (p = RT): p/ = const if constant temperatureThen: dp/d = d(RT)/d = RT

c = (RT)1/2 ~ isothermal c = (kRT)1/2 ~ isentropic

(k)1/2 too small or (1/1.18) (340 m/s) = 288 m/s

Speed of sound (m/s)steel 5050seawater 1540water 1500air (sea level) 340

V = 0; M = 0 V < c; M < 1

V = c; M = 1 V > c; M > 1

.

As measured by the observer is the speed/frequency of sound coming from the approaching siren greater, equal or less than the speed/frequency of sound from the receding siren?

vt

ct

Sin = ct/vt = 1/M = sin-1

(1/M)

Mach (1838-1916)

First to make shock waves visible.

First to take photographs of projectiles in flight.

Turned philosopher –“psychophysics”: all knowledge is based on sensations“I do not believe in atoms.”

(1) What do you put in a toaster?(2) Say silk 5 times, spell silk,

what do cows drink?(3) What was the first man-made object to break the sound barrier?

POP QUIZ

Tip speed ~ 1400 ft/sM ~ 1400/1100 ~ 1.3

Lockheed SR-71 aircraft cruises at around M = 3.3 at an altitude

of 85,000 feet (25.9 km).

What is flight speed?

Lockheed SR-71 aircraft cruises at around M = 3.3 at an altitude of 85,000 feet (25.9 km).

What is flight speed?

M = V/c

c = (kRT)1/2

Table A.3, pg 71924km T(K) = 220.626km T(K) = 222.525,900m ~ 220.6 + 1900m (1.9K/2000m) = 222 K

Lockheed SR-71 aircraft cruises at around M = 3.3 at an altitude of 85,000 feet (25.9 km; 222 K). What is flight speed?

c = {kRT}1/2 = {1.4*287 [(N-m)/(kg-K)] 222 [K]}1/2 = 299 m/s

V = [M][c] = [3.3][299 m/s] = 987 m/s

The velocity of a 30-ob rifle bullet is about 700 m/s

Vplane / Vbullet = 987/700 ~ 1.41

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

(a) What is airspeed of aircraft?(b) What is time between seeing aircraft overhead and hearing it?

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

(a) What is airspeed of aircraft?V (airspeed) = Mc = 1.35 * (kRT)1/2 1 = 1.35 * (1.4*287 [N-m/kg-K] *303 [K])1/2

= 471 m/s

Note: Mach numberand angle dependson relative velocity

i.e. airspeed

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

(b) What is time between seeing aircraft overhead and hearing it?

= sin-1 (1/M) = sin-1 (1/1.35) = 47.8o

Vearth = 471m/s – 10m/s = 461m/sD = Veartht = 461 [m/s] t; D = 3000[m]/tan() t = 5.9 s

D = Vearth t

3000m

* note: if T not constant, Mach line would not be straight

The END

Let us examine why heat does not have time to flow from a compression to a rarefaction and thus to equalize the temperature.

Mean free path ~ 10-5 cm (at STP).

Wavelength of sound in air at 20kHz is ~ 1.6 cm.

Let us examine why heat does not have time to flow from a compression to a rarefaction and thus to equalize the temperature.

For heat flow to be fast enough, speed must be > (½ )/(1/2 T) = Csound

Thermal velocity = (kBT/m)1/2

kB = Bolzmann’s constantm = mass of air moleculeCsound = (kRT)1/2 = (kkBNavag.T/m)1/2

Although heat flow speed is less, mean free path ~ 10-5 cm (at STP).Wavelength of sound in air at 20kHz is ~ 1.6 cm.

Not really linear, althoughnot apparent at the scale

of this plot.

For standard atm. conditionsc= 340 m/s at sea levelc = 295 m/s at 11 km