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

3. FYI: As you observe the animation note the following: (a) There is a pulse velocity v.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.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.). Topic 4.4 Extended B – The speed of sound  Let's have a microscopic look at air molecules set in motion by a moving pulse generator: Pulse Generator 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?

4.  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. Topic 4.4 Extended B – The speed of sound  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.

5.  The following spectrum diagram illustrates the wide frequency ranges for sound. Topic 4.4 Extended B – The speed of 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.

6.  Ultrasound is used for sonar, cleaning, and imaging. Topic 4.4 Extended B – The speed of sound  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.

7.  Recall that the speed of sound in a string is given by Topic 4.4 Extended B – The speed of sound v = F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 material inertial property of material T HE S PEED OF S OUND I N S OLIDS AND L IQUIDS

8. Topic 4.4 Extended B – The speed of sound v = elastic property of material inertial property of material  For sound travel through solids and liquids the inertial property is volume mass density  measured in (kg/m 3 ).  For solids, the elastic property is given by Young's modulus . T HE S PEED OF S OUND I N S OLIDS AND L IQUIDS  For liquids, the elastic property is given by the bulk modulus B.  Thus v =  The Speed of Sound in a Solid v = BB The speed of Sound in a Liquid.  is Young's modulus B is bulk modulus

9. Topic 4.4 Extended B – The speed of sound The average density of earth's crust 10 km below the continents is 2.7 g/cm 3. 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? T HE S PEED OF S OUND I N S OLIDS AND L IQUIDS v = , then  = v 2 . Since v = 5.4 km s But 1000 m km = 5400 m/s and  = 2.7 g cm 3 1 kg 1000 g 100 3 cm 3 1 3 m 3 = 2700 kg/m 3 so that  = v 2   = (5400) 2 (2700)  = 7.9  10 10 Pa FYI: A pascal (Pa) is a unit of pressure and is equal to a n/m 2. You can use unit cancellation to verify that  = v 2  has these units.

10. Topic 4.4 Extended B – The speed of sound  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 T HE S PEED OF S OUND I N G ASES v = 331 + 0.6T C The Speed of Sound in Air T C 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 + 0.6T C v = 331 + 0.6(20) v = 343 m/s

11. Topic 4.4 Extended B – The speed of sound  A better approximation is given by T HE S PEED OF S OUND I N G ASES v = 331 1 + T C 273 The Speed of Sound in Air T C 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 + T C 273 v = 331 1 + 20 273 v = 342.9 m/s

12. Topic 4.4 Extended B – The speed of sound  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* 6000 Steel* 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) 1284 Helium (0°C) 965 *Media from other sources.

13.  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. Topic 4.4 Extended B – The speed of sound v = B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/VpV/V The Bulk Modulus of a Liquid  The ratio  V/V has no units, so the units if B are those of pressure: n/m 2. 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?

14.  Consider a simplified pulse of fluid moving down the tube as shown: Topic 4.4 Extended B – The speed of sound  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x A xx A mass element  The mass of the non-compressed fluid element is given by mass = density·volume m =  V m =  ·  x·A m =  A  x

15. Topic 4.4 Extended B – The speed of sound  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x A mass element m =  A  x v = xtxt so that  x = v  t which we may substitute: m =  A  x m =  Av  t

16. Topic 4.4 Extended B – The speed of sound  We know that the average acceleration of the mass element is compression zone v xx A mass element m =  Av  t a = vtvt so that Newton's 2nd law,  F = m a, becomes  F = m a  F = (  Av  t) vtvt  F = (  Av)(  v)  F = (  Av 2 )(  v/v) Why?

17. Topic 4.4 Extended B – The speed of sound  Recall that pressure = force/area, so that compression zone v xx mass element F = pA  F = (  Av 2 )(  v/v)  The fluid element therefore has two forces acting on it in the direction of its acceleration, shown above. (p +  p)A pApA  Thus the sum of the forces acting on the fluid element is given by  F = (p +  p)A - pA  F = A  p and we can substitute:  F = (  Av 2 )(  v/v) A  p = (  Av 2 )(  v/v)  F = pA +  pA - pA  p = (  v 2 )(  v/v)

18. Topic 4.4 Extended B – The speed of sound  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x mass element  p = (  v 2 )(  v/v) V = A  x V = Av  t  V = A  v  t  If the volume of the mass element is V = Av  t, 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 = -A  v  t

19. Topic 4.4 Extended B – The speed of sound compression zone v xx mass element  p = (  v 2 )(  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 = -A  v  t V = Av  t VVVV = -A  v  t Av  t = - vvvv so that  v/v = -  V/V

20. Topic 4.4 Extended B – The speed of sound compression zone v xx mass element  p = (  v 2 )(  v/v)  Substitution yields  v/v = -  V/V  p = (  v 2 )(  v/v)  p = (  v 2 )(-  V/V)  v 2 = - pV/VpV/V  v 2 = B v = BB