1. SOUND - LONGITUDINAL WAVEc
dp, d, dVx

2. [Vx + dVx = dVx]

3. (at any position, no properties
are changing with time)
(V and  are only functions of x)
(dVxA << ddVxA)

4. only CV if accelerating
then 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 and
bottom of control volume), dRx =0. So FSx = -Adp

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

6. Total forces = normal surface forces
Change in momentum flux
From continuity eq.

7. Continuity

8. MOMENTUM EQ.
CONTINUITY EQ.
c2 = dp/d

9. 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.

10. & if isentropic, p/k = const Then: c2 = dp/d = p/s
FOR IDEAL GAS (p = RT):
& if isentropic, p/k = const
Then: c2 = dp/d = p/s
= d([const]k)/d
= [const]kk-1 = [const]k(k/)
= kp/ = kRT/
= kRT
c = (kRT)1/2 ~ 340 m/s STP
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.

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

12. 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

13. Newton (1686)* was the first to predict the
velocity of sound waves in air. He used
Boyles Law and assumed constant temperature.
FOR IDEAL GAS (p = RT):
p/ = const if constant temperature
Then: 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
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.

14. Speed of sound (m/s) steel 5050 seawater 1540 water 1500
air (sea level)
If isothermal then c =(RT)1/2. Newton in 1686 assumed isothermal and was about 18% too low since for air k ~ 1.4. Laplace in 1816 pointed out the need to regard the gas as isothermal.
Note that although Eq says that c is only a function of T, because ideal gas also function of p/, in fact here lies the physics c =(p/ )1/2 is like (k/m)1/2 for speed of mass on spring. Boyle referred to the “springyness of air” which is like k and  is like m,

15. V = 0; M = 0
V < c; M < 1
V = c; M = 1
V > c; M > 1 .

16. 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?

19. ct 
vt
Sin  = ct/vt = 1/M
 = sin-1 (1/M)

20. 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
Max\ch figured out – (1) Early artillerymen knew that two bangs could be heard downrange from a gun when a high speed projectile was fired but only one from a low speed projectile. (2) During the Franco-Prussian war of it was found that the novel French Chassepot high-speed bullets caused large crater shaped wounds. The Germans claimed that the French were using explosive projectiles – banned by by the International Treaty of Petersburg. Mach showed that the crater was attributed to the highly pressurized air caused by the bullets bow wave.
“I do not believe in atoms.”

21. POP QUIZ (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?

22. Tip speed ~ 1400 ft/s
M ~ 1400/1100 ~ 1.3

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

24. M = V/c c = (kRT)1/2 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 719
24km T(K) = 220.6
26km T(K) = 222.5
25,900m ~ m (1.9K/2000m) = 222 K

25. = {1.4*287 [(N-m)/(kg-K)] 222 [K]}1/2 = 299 m/s
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

26. M = 1.35 3 km T = 303 oK Wind = 10 m/s What is airspeed of aircraft?
What is time between seeing aircraft overhead and hearing it?

27. M = 1.35 3 km T = 303 oK Note: Mach number and angle depends
on relative velocity
i.e. airspeed
Wind = 10 m/s
M = 1.35
3 km
T = 303 oK
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

28. * note: if T not constant, Mach line would not be straight
D = Vearth t
Wind = 10 m/s
M = 1.35 
3000m
3 km
T = 303 oK
(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/s
D = Veartht = 461 [m/s] t; D = 3000[m]/tan()
t = 5.9 s

29. The END

30. 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.

31. 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 constant
m = mass of air molecule
Csound = (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.

32. Not really linear, although not apparent at the scale of this plot.
For standard atm. conditions
c= 340 m/s at sea level
c = 295 m/s at 11 km