SOUND - LONGITUDINAL WAVE c dp, d, dVx

[Vx + dVx = dVx]

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

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

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

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

Continuity

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

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.

& 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.

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 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.

Speed of sound (m/s) steel 5050 seawater 1540 water 1500 air (sea level) 340 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. 11.14 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,

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?

ct  vt Sin  = ct/vt = 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 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 1870-71 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.”

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?

Tip speed ~ 1400 ft/s M ~ 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?

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 ~ 220.6 + 1900m (1.9K/2000m) = 222 K

= {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

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?

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

* 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

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 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.

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