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Topic 4.4 Extended B – The speed of sound

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

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Topic 4.4 Extended B – The speed of sound

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

  2. 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). Topic 4.4 ExtendedB – The speed of sound 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? 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: 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.

  3. Condensation Rarefaction Topic 4.4 ExtendedB – The speed of sound 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. FYI: Keep in mind, though, that sound waves are LONGITUDINAL.

  4. Topic 4.4 ExtendedB – The speed of sound The following spectrum diagram illustrates the wide frequency ranges for sound. 5 GHz Upper Limit Humans can hear sounds in the audible range. Other animals can hear higher and lower frequencies than humans. Ultrasonic For example, cats, dogs and bats can hear into the ultrasonic range. 20 kHz 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. Audible 20 Hz Infrasonic

  5. Topic 4.4 ExtendedB – The speed of sound Ultrasound is used for sonar, cleaning, and imaging. Sound waves in the 4.2 GHz range can be used to obtain microscopic images: Since X-rays are harmful to a developing fetus, ultrasound is used. The blue conductors are 2 m in width.

  6. The Wave Velocity of a Stretched String Topic 4.4 ExtendedB – The speed of sound THE SPEED OF SOUND IN SOLIDS AND LIQUIDS Recall that the speed of sound in a string is given by F  v = 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 elastic property of material inertial property of material v =

  7. The Speed of Sound in a Solid The speed of Sound in a Liquid. Topic 4.4 ExtendedB – The speed of sound THE SPEED OF SOUND IN SOLIDS AND LIQUIDS elastic property of material inertial property of material v = 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 =  is Young's modulus B  v = B is bulk modulus

  8. Topic 4.4 ExtendedB – The speed of sound THE SPEED OF SOUND IN SOLIDS AND LIQUIDS 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?   , then  = v2. v = Since 5.4 km s 1000 m km But v = = 5400 m/s 1003 cm3 13 m3 2.7 g cm3 1 kg 1000 g = 2700 kg/m3  = and so that  = v2  =(5400)2(2700)  =7.91010 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.

  9. The Speed of Sound in Air Topic 4.4 ExtendedB – The speed of sound THE SPEED OF SOUND IN GASES 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 v = 331 + 0.6TC 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

  10. TC 273 20 273 v = 331 1 + v = 331 1 + The Speed of Sound in Air Topic 4.4 ExtendedB – The speed of sound THE SPEED OF SOUND IN GASES A better approximation is given by TC 273 v = 331 1 + 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 = 342.9 m/s

  11. Topic 4.4 ExtendedB – 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) Medium Speed (m/s) Medium Speed (m/s) Solids... Gases... Liquids... Granite* 6000 Hydrogen (0°C) 1284 Seawater* 1522 Steel* 5941 Helium (0°C) 965 Water (20°C) 1482 Air (100°C) 387 Glass 5200 Water (0°C) 1402 Aluminum 5100 Mercury 1400 Air (20°C)* 343 Iron 4500 Ethyl Alcohol 1125 Air (0°C) 331 Oxygen (0°C) 316 Copper 3500 Polystyrene 1850 *Media from other sources.

  12. The Bulk Modulus of a Liquid The speed of Sound in a Liquid. 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? Topic 4.4 ExtendedB – The speed of sound In this section we will show the gruesome details in the derivation of the speed of sound in liquids: B  v = B is bulk modulus In a taut string, potential energy is associated with the periodic stretching of tensioned string elements. 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: p V/V B = - 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?

  13. v Topic 4.4 ExtendedB – The speed of sound Consider a simplified pulse of fluid moving down the tube as shown: mass element compression zone A x As the compression zone moves through the fluid at the wave velocity v, it runs into non-compressed fluid. 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: A x The mass of the non-compressed fluid element is given by mass element mass = density·volume m = V m = ·x·A m = Ax

  14. v Topic 4.4 ExtendedB – The speed of sound mass element compression zone A x m = Ax 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 x t v = x = vt so that which we may substitute: m = Ax m = Avt

  15. v Topic 4.4 ExtendedB – The speed of sound mass element compression zone A x m = Avt We know that the average acceleration of the mass element is v t a = F = ma , becomes so that Newton's 2nd law, F = ma v t F = (Avt) F = (Av)(v) Why? F = (Av2)(v/v)

  16. (p + p)A pA v Topic 4.4 ExtendedB – The speed of sound mass element compression zone x F = (Av2)(v/v) Recall that pressure = force/area, so that F = pA The fluid element therefore has two forces acting on it in the direction of its acceleration, shown above. Thus the sum of the forces acting on the fluid element is given by F = (p + p)A- pA F = pA + pA- pA F = Ap and we can substitute: F = (Av2)(v/v) Ap= (Av2)(v/v) p= (v2)(v/v)

  17. v Topic 4.4 ExtendedB – The speed of sound mass element compression zone x 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: V = Ax V = Avt If the volume of the mass element is V = Avt, then the change in volume of the mass element is V = Avt 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

  18. v Topic 4.4 ExtendedB – The speed of sound mass element compression zone x p= (v2)(v/v) V = Avt V = -Avt Now we can look at the fractional change in volume of the fluid element, and write it in terms of the wave velocity: v v -Avt Avt V V = - = so that v/v = - V/V

  19. v Topic 4.4 ExtendedB – The speed of sound mass element compression zone x p= (v2)(v/v) v/v = -V/V Substitution yields p= (v2)(v/v) p= (v2)(-V/V) p V/V v2 = - v2 = B B  v =

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