SOUND AND HEARING. = BkA and Bk is constant gives pmax1 / A1 = pmax2 / A2 p Pa p. = BkA and solve for A. fba. 10 Pa) (1480 m s) 10 Pa) (1000 Hz)

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1 OUND AND HEARING 6 6 IDENTIFY and ET UP: Eq() gies the waelength in terms o the requency Use Eq(6) to relate the pressure and displacement amplitudes EXECUTE: (a) λ = / = (44 m/s)/000 Hz = 044 m (b) p = BkA and Bk is constant gies p / A = p / A p 8 0 Pa A = A = 0 m 0 m p = 0 0 Pa (c) p = BkA = π BA/ λ p 0 0 Pa pλ = πba= constant so pλ = pλ and λ = λ = (044 m) 69 m = p 0 Pa = / λ = (44 m/s)/69 m = 0 Hz EVAUATE: The pressure amplitude and displacement amplitude are directly proportional For the same displacement amplitude, the pressure amplitude decreases when the requency decreases and the waelength increases 6 IDENTIFY: Apply p = BkA and sole or A π π π BA ET UP: k = and = λ, so k = and p = λ p (0 0 Pa) (480 m s) EXECUTE: A = = = 0 m 9 πb π( 0 Pa) (000 Hz) EVAUATE: Both and B are larger, but B is larger by a much greater actor, so / B is a lot smaller and thereore A is a lot smaller 6 IDENTIFY: Use Eq(6) to relate the pressure and displacement amplitudes ET UP: As stated in Example 6 the adiabatic bulk modulus or air is B = 4 0 Pa Use Eq() to calculate λ rom, and then k = π / λ EXECUTE: (a) = 0 Hz Need to calculate k: λ = / and k = π / λ so k = π / = ( π rad)(0 Hz)/44 m/s = 74 rad/m Then p = BkA= (4 0 Pa)(74 rad/m)( m) = 778 Pa This is below the pain threshold o 0 Pa (b) is larger by a actor o 0 so k = π / is larger by a actor o 0, and p = BkA is larger by a actor o 0 p = 778 Pa, aboe the pain threshold (c) There is again an increase in, k, and p o a actor o 0, so p = 778 Pa, ar aboe the pain threshold EVAUATE: When increases, λ decreases so k increases and the pressure amplitude increases π π π BA 64 IDENTIFY: Apply p = BkA k = =, so p = λ ET UP: = 44 m/s p (44 m/s)(00 Pa) EXECUTE: = = = 86 0 Hz 6 πba π(4 0 Pa)(00 0 m) EVAUATE: Audible requencies range rom about 0 Hz to about 0,000 Hz, so this requency is audible 6 IDENTIFY: = λ Apply Eq(67) or the waes in the liquid and Eq(68) or the waes in the metal bar d 0 m ET UP: In part (b) the wae speed is = = 4 t 90 0 s 6-

2 6- Chapter 6 EXECUTE: (a) Using Eq(67), B ρ (λ ) ρ, = = so [ ] 0 B = = (8 m)(400 Hz) (00 kg m ) 0 Pa 4 0 (b) Using Eq(68), Y = ρ = ( t) ρ = (0 m) (90 0 s) (6400 kg m ) = Pa EVAUATE: In the liquid, = 00 m/s and in the metal, = 80 m/s Both these speeds are much greater than the speed o sound in air 66 IDENTIFY: = d/ t Apply Eq(67) to calculate B ET UP: ρ = 0 kg/m EXECUTE: (a) The time or the wae to trael to Caracas was 9 min 9 s = 79 s and the speed was m/s imilarly, the time or the wae to trael to Keo was 680 s or a speed o 8 0 m/s, and the time to trael to Vienna was 767 s or a speed o m/s The aerage or these three measurements is 4 0 m/s Due to ariations in density, or relections (a subject addressed in later chapters), not all waes trael in straight lines with constant speeds (b) From Eq(67), B = ρ, and using the gien alue o ρ = 0 kg/m and the speeds ound in part (a), the alues or the bulk modulus are, respectiely, 9 0 Pa, 4 0 Pa and 0 Pa EVAUATE: These are larger, by a actor o or, than the largest alues in Table 67 IDENTIFY: d = tor the sound waes in air an in water ET UP: Use water = 48 m/s at 0 C, as gien in Table 6 In air, = 44 m/s EXECUTE: ince along the path to the dier the sound traels m in air, the sound wae traels in water or the same time as the wae traels a distance 0 m 0 m = 08 m in air The depth o the dier is water (08 m) (08 m) 896 m air 48 m/s = = This is the depth o the dier; the distance rom the horn is 908 m 44 m/s 0 m EVAUATE: The time it takes the sound to trael rom the horn to the person on shore is t = = s 44 m/s The time it takes the sound to trael rom the horn to the dier is m 896 m t = + = 000 s s = s These times are indeed the same For three igure 44 m/s 48 m/s accuracy the distance o the horn aboe the water can t be neglected 68 IDENTIFY: Apply Eq(60) to each gas ET UP: In each case, express M in units o kg/ mol For H, γ = 4 For He and Ar, γ = 67 EXECUTE: (a) (b) (c) ( 4)( 84 J mol K)( 00 K) = = ( 0 0 kg mol) H ( 67)( 84 J mol K)( 00 K) = = ( kg mol) He ( 67)( 84 J mol K)( 00 K) ( 99 0 kg mol) Ar 0 0 m s = = m s (d) Repeating the calculation o Example 6 at H = 80 air, He = 94air and Ar = 098 air EVAUATE: is larger or gases with smaller M 0 m s T = 00 K gies air = 48m s, and so γ RT 69 IDENTIFY: = λ The relation o to gas temperature is gien by = M ET UP: et T = 0 C = 9 K EXECUTE: At 0 C, λ = m/s 060 m 60 cm = 0 Hz = = γ RT λ = = M 8 cm = = = = 60 cm λ γr =, which is T M λ λ constant, so = T T λ (9 K) 46 K 84 C T T λ EVAUATE: When T increases increases and or ixed, λ increases Note that we did not need to know either γ or M or the gas

3 ound and Hearing 6- γ RT 60 IDENTIFY: = Take the deriatie o with respect to T In part (b) replace d by Δ and dt by M the expression deried in part (a) / d( x ) / ET UP: = x In Eq(60), T must be in kelins 0 C = 9 K Δ T = C = K dx / d γr dt γr / γrt d dt EXECUTE: (a) = = T = = Rearranging gies =, the desired dt M dt M T M T T result Δ ΔT ΔT 44 m/s K (b) = Δ = = = 09 m/s T T 9 K Δ T Δ EVAUATE: ince = 4 0 and is one-hal this, replacing dt by ΔT and d by Δ is accurate Using T the result rom part (a) is much simpler than calculating or 0 C and or C and subtracting, and is not subject to round-o errors 6 IDENTIFY and ET UP: Use t = distance/speed Calculate the time it takes each sound wae to trael the = 800 m length o the pipe Use Eq(68) to calculate the speed o sound in the brass rod EXECUTE: wae in air: t = 800 m/(44 m/s) = 06 s 0 Y 90 0 Pa wae in the metal: = = = m/s ρ 8600 kg/m 800 m t = = 0047 s m/s The time interal between the two sounds is Δ t = 06 s 0047 s = 008 s EVAUATE: The restoring orces that propagate the sound waes are much greater in solid brass than in air, so is much larger in brass 6 IDENTIFY: Repeat the calculation o Example 6 at each temperature ET UP: 70 C = 00 K and 0 C = 60 K (40)(84 J/mol K)(00 K) (40)(84 J/mol K)(60 K) EXECUTE: = 4 m/s (88 0 kg/mol) (88 0 kg/mol) EVAUATE: The speed is greater at the higher temperature The dierence in speeds corresponds to a 7% increase F Y 6 IDENTIFY: For transerse waes, trans = For longitudinal waes, long = μ ρ ET UP: The mass per unit length μ is related to the density (assumed uniorm) and the cross-section area A by μ = Aρ EXECUTE: long trans Y F Υ F Υ = 0 gies = 0 and = 900 Thereore, F A= ρ μ ρ Αρ 900 EVAUATE: Typical alues o Y are on the order o ΔT 8 0 Pa, so the stress must be about 0 Pa I A is on the 6 order o mm = 0 m, this requires a orce o about 00 N 64 IDENTIFY: The intensity I is gien in terms o the displacement amplitude by Eq(6) and in terms o the pressure amplitude by Eq(64) ω = π Intensity is energy per second per unit area ET UP: For part (a), I = 0 W/m For part (b), I = 0 W/m EXECUTE: (a) I = B A ρ ω I ( 0 W/m ) ω ρb π(000 Hz) (0 kg/m )(4 0 Pa) A = = = p I B 0 m p I = ρb 0 = ρ = ( 0 W/m ) (0 kg/m )(4 0 Pa) = 9 0 Pa = 8 0 atm (b) A is proportional to I, so I, so p 0 W/m 7 A = ( 0 m) = 6 0 m 0 W/m 0 W/m (9 0 Pa) 6 Pa 6 0 atm 0 W/m = = = p is also proportional to in

4 6-4 Chapter 6 7 (c) area = (00 mm) = 0 m Part (a): ( 0 W/m )( 0 m ) = 0 J/s 8 Part (b): ( 0 W/m )( 0 m ) = 80 0 J/s EVAUATE: For aint sounds the displacement and pressure ariation amplitudes are ery small Intensities or audible sounds ary oer a ery wide range 6 IDENTIFY: Apply Eq(6) and sole or A λ = /, with = B/ ρ ET UP: ω = π For air, EXECUTE: (a) The amplitude is B = 4 0 Pa 6 Ι (00 0 W m ) A = = = 9 ρβω (000 kg m )(8 0 Pa)( π(400 Hz)) m 9 B ρ (8 0 Pa) (000 kg m ) The waelength is λ = = = = 044 m 400 Hz 9 (b) Repeating the aboe with B = 4 0 Pa and the density o air gies A = 66 0 m and λ = 000 m EVAUATE: (c) The amplitude is larger in air, by a actor o about 60 For a gien requency, the much less dense air molecules must hae a larger amplitude to transer the same amount o energy p 66 IDENTIFY and ET UP: Use Eq(67) to eliminate either or B in I = B EXECUTE: From Eq (9), Using Eq (67) to eliminate B, B ρ = Using Eq(67) to eliminate = ( ) = I = p ( ρ) = p ρ I, Bρ p B p ρb EVAUATE: The equation in this orm shows the dependence o I on the density o the material in which the wae propagates 67 IDENTIFY and ET UP: Apply Eqs(6), (6) and (6) EXECUTE: (a) ω = π = ( π rad)(0 Hz) = 94 rad/s π π ω 94 rad/s k = = = = = 74 rad/m λ 44 m/s B = 4 0 Pa (Example 6) Then p 6 = BkA= (4 0 Pa)(74 rad/m)(00 0 m) = 9 Pa (b) Eq(6): I = ωbka I 6 = (94 rad/s)(4 0 Pa)(74 rad/m)(00 0 m) = 48 0 W/m (c) Eq(6): β = (0 db)log( I/ I0), with I 0 = 0 W/m ( ) β = (0 db) log (48 0 W/m )/( 0 W/m ) = 966 db EVAUATE: Een though the displacement amplitude is ery small, this is a ery intense sound Compare the sound intensity leel to the alues in Table 6 68 IDENTIFY: Apply β = (0 db)log( I/ I0) In part (b), use Eq(64) to calculate I rom the inormation that is gien ET UP: I 0 = 0 W/m From Table 6 the speed o sound in air at 00 C is 44 m/s The density o air at that temperature is EXECUTE: (a) (b) 0kg m 000 μw/m β = (0 db) log 7 db = 0 W/m p (00 N m ) I = = = 7 0 W m Using this in Equation (6), ρ (0 kg m )(44 m s) 7 0 W m β = (0 db) log = 744 db 0 W m EVAUATE: As expected, the sound intensity is larger or the jack hammer 69 IDENTIFY: Use Eq(6) to relate I and β = (0 db)log( I/ I ) Eq(64) says the pressure amplitude and displacement amplitude are related by p 0 π p = BkA = B A ET UP: At 0 C the bulk modulus or air is 4 0 Pa and = 44 m/s I 0 = 0 W/m

5 EXECUTE: (a) p (44 m/s)(60 0 Pa) I = = = 44 0 W/m B (4 0 Pa) 44 0 W/m (b) β = (0 db)log 64 db = 0 W/m p (44 m/s)(60 0 Pa) = = = 8 0 m ound and Hearing 6- (c) A π B π(400 Hz)(4 0 Pa) EVAUATE: This is a ery aint sound and the displacement and pressure amplitudes are ery small Note that the displacement amplitude depends on the requency but the pressure amplitude does not 60 IDENTIFY and ET UP: Apply the relation β β = (0 db) log ( I/ I) that is deried in Example 60 EXECUTE: (a) β 4I ( ) Δ = (0 db)log = 60 db I (b) The total number o crying babies must be multiplied by our, or an increase o kids EVAUATE: For I, = αi where α is some actor, the increase in sound intensity leel is Δ β = (0 db)log α For α = 4, Δ β = 60 db 6 IDENTIFY and ET UP: et reer to the mother and to the ather Use the result deried in Example 6 or the dierence in sound intensity leel or the two sounds Relate intensity to distance rom the source using Eq(6) EXECUTE: From Example 6, β β = (0 db)log( I/ I) Eq(6): I / I = r / r or I / I = r / r Δ β = β β = (0 db)log( I / I ) = (0 db)log( r / r ) = (0 db)log( r / r ) Δ β = (0 db) log(0 m/00 m) = 40 db EVAUATE: The ather is times closer so the intensity at his location is times greater I I I 6 IDENTIFY: β = (0 db) log β β = (0 db)log ole or I I I 0 ET UP: I log y = x then y = 0 x et β = 70 db and β = 9 db I I I EXECUTE: 700 db 90 db = 0 db = (0 db) log log = and = 0 = 0 I I I EVAUATE: I < I when β < β 6 (a) IDENTIFY and ET UP: From Example 6 Δ β = (0 db)log( I/ I) et Δ β = 0 db and sole or I/ I EXECUTE: 0 db = 0 dblog( I/ I) so = log( I/ I) and I/ I = 00 (b) EVAUATE: According to the equation in part (a) the dierence in two sound intensity leels is determined by the ratio o the sound intensities o you don t need to know I, just the ratio I/ I 64 IDENTIFY: For an open pipe, = For a stopped pipe, = = λ 4 ET UP: = 44 m/s For a pipe, there must be a displacement node at a closed end and an antinode at the open end 44 m/s EXECUTE: (a) = = = 090 m (94 Hz) (b) There is a node at one end, an antinode at the other end and no other nodes or antinodes in between, so λ = and λ = 4 = 4(090 m) = 6 m 4 (c) = = = (94 Hz) = 97 Hz 4 EVAUATE: We could also calculate or the stopped pipe as 44 m/s = = = 97 Hz, which agrees with λ 6 m our result in part (a) 6 IDENTIFY and ET UP: An open end is a displacement antinode and a closed end is a displacement node ketch the standing wae pattern and use the sketch to relate the node-to-antinode distance to the length o the pipe A displacement node is a pressure antinode and a displacement antinode is a pressure node

6 6-6 Chapter 6 EXECUTE: (a) The placement o the displacement nodes and antinodes along the pipe is as sketched in Figure 6a The open ends are displacement antinodes Figure 6a ocation o the displacement nodes (N) measured rom the let end: undamental 060 m st oertone 00 m, 090 m nd oertone 00 m, 060 m, 00 m ocation o the pressure nodes (displacement antinodes (A)) measured rom the let end: undamental 0, 0 m st oertone 0, 060 m, 0 m nd oertone 0, 040 m, 080 m, 0 m (b) The open end is a displacement antinode and the closed end is a displacement node The placement o the displacement nodes and antinodes along the pipe is sketched in Figure 6b Figure 6b ocation o the displacement nodes (N) measured rom the closed end: undamental 0 st oertone 0, 080 m nd oertone 0, 048 m, 096 m ocation o the pressure nodes (displacement antinodes (A)) measured rom the closed end: undamental 0 m st oertone 040 m, 0 m nd oertone 04 m, 07 m, 0 m EVAUATE: The node-to-node or antinode-to-antinode distance is λ / For the higher oertones the requency is higher and the waelength is smaller 66 IDENTIFY: A pipe closed at one end is a stopped pipe Apply Eqs(68) and (6) to ind the requencies and Eqs(69) and (6) to ind the highest audible harmonic in each case ET UP: For the open pipe n =,, and or the irst three harmonics and or the stopped pipe n =,, and EXECUTE: (a) = and n = n 44 m/s = = 8 Hz = 764 Hz, = 46 Hz, 4 = 8 Hz (040 m) (b) = and 4, n = n n =,,, 44 m/s = = 9 Hz = 7 Hz, = 9 Hz, 7 = 7 Hz 4(040 m) 0,000 Hz 0,000 Hz (c) open pipe: n = = = closed pipe: = = 04 But only odd n are present, so 8 Hz 9 Hz n = 0 EVAUATE: For an open pipe all harmonics are present For a stopped pipe only odd harmonics are present For pipes o a gien length, or a stopped pipe is hal what it is or an open pipe

7 67 IDENTIFY: For a stopped pipe, the standing wae requencies are gien by Eq(6) ET UP: The irst three standing wae requencies correspond to n =, and (44 m/s) EXECUTE: = = 06 Hz, = = 7 Hz, = = 9 Hz 4(07 m) ound and Hearing 6-7 EVAUATE: All three o these requencies are in the audible range, which is about 0 Hz to 0,000 Hz 68 IDENTIFY: Model the auditory canal as a stopped pipe o length = 40 cm For a stopped pipe, λ = 4, = and 4,, n = n n=,, ET UP: Take the highest audible requency to be 0,000 Hz = 44 m/s 44 m/s EXECUTE: (a) = = = 8 0 Hz λ = 4 = 4(0040 m) = m This requency is 4 4(0040 m) audible 0,000 Hz (b) For = 0,000 Hz, = = 6; the highest harmonic which is audible is or n = (ith harmonic) 80 Hz 4 = = 79 0 Hz EVAUATE: For a stopped pipe there are no een harmonics 69 IDENTIFY: For either type o pipe, stopped or open, the undamental requency is proportional to the wae speed The wae speed is gien in turn by Eq(60) ET UP: For He, γ = / and or air, γ = 7/ γ EXECUTE: (a) The undamental requency is proportional to the square root o the ratio, so M M ( ) 88 (6 Hz) 767 Hz He air He = air = = γ air M He (7 ) 400 γ (b) No In either case the requency is proportional to the speed o sound in the gas EVAUATE: The requency is much higher or helium, since the rms speed is greater or helium 60 IDENTIFY: There must be a node at each end o the pipe For the undamental there are no additional nodes and each successie oertone has one additional node = λ ET UP: = 44 m/s The node to node distance is λ / λ EXECUTE: (a) = so λ = Each successie oertone adds an additional λ / along the pipe, so λn n n = and λ n =, where n =,,, n = = n λn 44 m/s (b) = = = 688 Hz = = 8 Hz = = 06 Hz All three o these requencies are (0 m) audible EVAUATE: A pipe o length closed at both ends has the same standing wae waelengths, requencies and nodal patterns as or a string o length that is ixed at both ends 6 IDENTIFY and ET UP: Use the standing wae pattern to relate the waelength o the standing wae to the length o the air column and then use Eq() to calculate There is a displacement antinode at the top (open) end o the air column and a node at the bottom (closed) end, as shown in Figure 6 EXECUTE: (a) Figure 6 λ /4= λ = 4 = 4(040 m) = 060 m 44 m/s = = = 64 Hz λ 060 m (b) Now the length o the air column becomes (040 m) = 0070 m and λ = 4 = 080 m 44 m/s = = = 0 Hz λ 080 m EVAUATE: maller means smaller λ which in turn corresponds to larger

8 6-8 Chapter 6 6 IDENTIFY: The wire will ibrate in its second oertone with requency 6 pipe, pipe wire = F/ μ pipe wire when wire pipe = The irst oertone standing wae requency or a wire ixed at both ends is 4 = For a stopped wire = wire wire m 7 0 kg ET UP: The wire has μ = = = 8 0 kg/m The speed o sound in air is = 44 m/s 080 m wire 40 N EXECUTE: wire = = 694 m/s 8 0 kg/m (080 m)(44 m/s) wire pipe = = = = wire (694 m/s) wire = gies = 4 wire pipe 0070 m 70 cm EVAUATE: The undamental or the pipe has the same requency as the third harmonic o the wire But the wae speeds or the two objects are dierent and the two standing waes hae dierent waelengths wire pipe Figure 6 (a) IDENTIFY and ET UP: Path dierence rom points A and B to point Q is 00 m 00 m = 00 m, as shown in Figure 6 Constructie intererence implies path dierence = nλ, n =,,, EXECUTE: 00 m = nλ so λ = 00 m / n n n(44 m/s) = = = = n(7 Hz,) n =,,, λ 00 m 00 m The lowest requency or which constructie intererence occurs is 7 Hz (b) IDENTIFY and ET UP: Destructie intererence implies path dierence = ( n/ ) λ, n =,,, EXECUTE: 00 m = ( n / ) λ so λ = 400 m / n n n(44 m/s) = = = = n(86 Hz), n =,,, λ 400 m (400 m) The lowest requency or which destructie intererence occurs is 86 Hz EVAUATE: As the requency is slowly increased, the intensity at Q will luctuate, as the intererence changes between destructie and constructie 64 IDENTIFY: Constructie intererence occurs when the dierence o the distances o each source rom point P is an integer number o waelengths The intererence is destructie when this dierence o path lengths is a hal integer number o waelengths λ = = 44 m s (06 Hz) = 67 m ince P is between the speakers, x must be ET UP: The waelength is ( ) in the range 0 to, where = 00 m is the distance between the speakers EXECUTE: The dierence in path length is Δ l = ( x) x= x, or x= ( Δ l)/ For destructie intererence, Δ l = ( n+ ( ))λ, and or constructie intererence, Δ l = nλ (a) Destructie intererence: n = 0 gies Δ l = 08 m and x = 08 m n = gies Δ l = 08 m and x = 4 m No other alues o n place P between the speakers (b) Constructie intererence: n = 0 gies Δ l = 0 and x = 00 m n = gies Δ l = 67 m and x = 07 m n = gies Δ l = 67 m and x = 8 m No other alues o n place P between the speakers (c) Treating the speakers as point sources is a poor approximation or these dimensions, and sound reaches these points ater relecting rom the walls, ceiling, and loor EVAUATE: Points o constructie intererence are a distance λ / apart, and the same is true or the points o destructie intererence 6 IDENTIFY: For constructie intererence the path dierence is an integer number o waelengths and or destructie intererence the path dierence is a hal-integer number o waelengths ET UP: λ = / = (44 m/s)/(688 Hz) = 000 m EXECUTE: To moe rom constructie intererence to destructie intererence, the path dierence must change by λ/ I you moe a distance x toward speaker B, the distance to B gets shorter by x and the distance to A gets longer by x so the path dierence changes by x x = λ/ and x = λ / 4 = 0 m EVAUATE: I you walk an additional distance o 0 m arther, the intererence again becomes constructie

9 ound and Hearing IDENTIFY: Destructie intererence occurs when the path dierence is a hal integer number o waelengths ET UP: = 44 m s, so λ = = ( 44 m/s) ( 7 Hz) = 00 m I r A = 800 m and r B are the distances o the person rom each speaker, the condition or destructie intererence is r ( ) λ, B ra n EXECUTE: Requiring ( ) λ 0 rb = ra + n+ > gies n r A λ ( 800 m) ( 00 m) 4, n = and the closest distance to B is ( )( ) alue o r B occurs when 4, = + where n is any integer + > = = so the smallest EVAUATE: For r B = 00 m, the path dierence is ra rb = 700 m This is λ 67 IDENTIFY: Compare the path dierence to the waelength λ = = 44 m s 860 Hz = 0400 m ET UP: ( ) ( ) r = 800 m m = 00 m EXECUTE: The path dierence is 4 m 0 m = 4 m path dierence = The path dierence is a halinteger number o waelengths, so the intererence is destructie λ EVAUATE: The intererence is destructie at any point where the path dierence is a hal-integer number o waelengths 68 IDENTIFY: = = λ beat ET UP: = 44 m/s, et λ = 60 cm and λ = 6 cm λ > λ so > ( λ λ) (44 m/s)(00 0 m) EXECUTE: = = = = 6 Hz There are 6 beats per λ λ λλ (60 0 m)(6 0 m) second EVAUATE: We could hae calculated and and subtracted, but doing it this way we would hae to be careul to retain enough igures in intermediate calculations to aoid round-o errors 69 IDENTIFY: beat = a b For a stopped pipe, = 4 ET UP: = 44 m/s et a = 4 m and b = 6 m b > a so a > b ( b a) (44 m/s)(00 0 m) EXECUTE: a b = = = = Hz There are beats per 4 a b 4ab 4(4 m)(6 m) second EVAUATE: Increasing the length o the pipe increases the waelength o the undamental and decreases the requency 640 IDENTIFY: = beat 0 = Changing the tension changes the wae speed and this alters the requency F F F0 ET UP: = so =, where F = F0 +Δ F et 0 = We can assume that Δ F/ F0 is ery m m m small Increasing the tension increases the requency, so = beat 0 m EXECUTE: (a) beat 0 ( F0 F F0 ) ΔF/ F0 is small This gies that beat 0 Δ F beat ( Hz) = = = 0 = = +Δ = + ΔF = F0 B / F ΔF m F 0 (b) 068% F Hz EVAUATE: The ractional change in requency is one-hal the ractional change in tension + 64 IDENTIFY: Apply the Doppler shit equation = + ET UP: The positie direction is rom listener to source = 00 Hz = 40 Hz ( m/s)(40 Hz) EXECUTE: = 0 = 0 m/s = gies = = = + 00 Hz 40 Hz > since the source is approaching the listener EVAUATE: / ΔF ΔF + = + when F0 F0 780 m/s

10 6-0 Chapter 6 64 IDENTIFY: Follow the steps o Example 69 ET UP: In the irst step, =+ 00 m/s instead o 00 m/s In the second step, = 00 m/s instead o + 00 m/s EXECUTE: W 40 m/s = = (00 Hz) = 8 Hz Then + 40 m/s + 00 m/s + 40 m/s 00 m/s (8 Hz) 66 Hz 40 m/s = W = = EVAUATE: When the car is moing toward the relecting surace, the receied requency back at the source is higher than the emitted requency When the car is moing away rom the relecting surace, as is the case here, the receied requency back at the source is lower than the emitted requency + 64 IDENTIFY: Apply the Doppler shit equation = + ET UP: The positie direction is rom listener to source = 9 Hz + 44 m/s 0 m/s (a) = 0 = 0 m/s = = (9 Hz) = 7 Hz + 44 m/s + 44 m/s + 0 m/s (b) =+ 0 m/s =+ 0 m/s = = (9 Hz) = 7 Hz + 44 m/s + 0 m/s (c) beat = = 4 Hz EVAUATE: The distance between whistle A and the listener is increasing, and or whistle A < The distance between whistle B and the listener is also increasing, and or whistle B < 644 IDENTIFY and ET UP: Apply Eqs(67) and (68) or the waelengths in ront o and behind the source 44 m/s Then = / λ When the source is at rest λ = = = 0860 m 400 Hz 44 m/s 0 m/s EXECUTE: (a) Eq(67): λ = = = 0798 m 400 Hz + 44 m/s + 0 m/s (b) Eq(68): λ = = = 09 m 400 Hz (c) = / λ (since = 0), so = (44 m/s)/0798 m = 4 Hz (d) = / λ = (44 m/s)/09 m = 7 Hz EVAUATE: In ront o the source (source moing toward listener) the waelength is decreased and the requency is increased Behind the source (source moing away rom listener) the waelength is increased and the requency is decreased 64 IDENTIFY: The distance between crests is λ In ront o the source = / T λ = and behind the source ET UP: T = 6 s = 0 m/s The crest to crest distance is the waelength, so λ = 0 m + λ = EXECUTE: (a) = / T = 06 Hz λ = gies = λ = 0 m/s (0 m)(06 Hz) = 0 m/s + 0 m/s + 0 m/s (b) λ = = = 09 m 06 Hz EVAUATE: I the duck was held at rest but still paddled its eet, it would produce waes o waelength 0 m/s λ = = 0 m In ront o the duck the waelength is decreased and behind the duck the waelength is 06 Hz increased The speed o the duck is 78% o the wae speed, so the Doppler eects are large

11 + 646 IDENTIFY: Apply = + ET UP: = 000 Hz The positie direction is rom the listener to the source = 44 m/s ound and Hearing m/s (a) = (44 m/s)/ = 7 m/s, = 0 = = (000 Hz) = 000 Hz + 44 m/s 7 m/s + 44 m/s + 7 m/s (b) = 0, =+ 7 m/s = = (000 Hz) = 00 Hz + 44 m/s EVAUATE: (c) The answer in (b) is much less than the answer in (a) It is the elocity o the source and listener relatie to the air that determines the eect, not the relatie elocity o the source and listener relatie to each other IDENTIFY: Apply = + ET UP: The positie direction is rom the motorcycle toward the car The car is stationary, so = 0 which gies ( 490 Hz )( ) + EXECUTE: = = ( + ), = = 44 m s = 98 m/s + 0 Hz You must be traeling at 98 m/s EVAUATE: < 0 means that the listener is moing away rom the source 648 IDENTIFY: Apply the Doppler eect ormula, Eq(69) (a) ET UP: The positie direction is rom the listener toward the source, as shown in Figure 648a Figure 648a + 44 m/s + 80 m/s EXECUTE: = = (6 Hz) = 0 Hz + 44 m/s 00 m/s EVAUATE: istener and source are approaching and > (b) ET UP: ee Figure 648b Figure 648b + 44 m/s 80 m/s EXECUTE: = = (6 Hz) = 8 Hz + 44 m/s + 00 m/s EVAUATE: istener and source are moing away rom each other and < 649 IDENTIFY: The radar beam consists o electromagnetic waes and Eq(60) applies Apply the Doppler ormula twice, once with the storm as a receier and then again with the storm as a source 8 ET UP: c = 00 0 m/s When the source and receier are moing toward each other, as is the case here, then is negatie c+ EXECUTE: et be the requency receied by the storm; = Then seres as the source c c+ c+ c+ requency when the waes are relected and R = = c c c c+ (0 m/s) Δ = = = = = c c 00 0 m/s 0 m/s 6 R (000 0 Hz) 68 Hz 8

12 6- Chapter 6 EVAUATE: ince! c, in the expression Δ = it is a ery good approximation to replace c c Δ Δ by c and then = is ery small, so c c is ery small ince the storm is approaching the station the inal receied requency is larger than the original transmitted requency 60 IDENTIFY: Apply Eq(60) The source is moing away, so is positie ET UP: 8 c = 00 0 m/s = m/s EXECUTE: 8 c 00 0 m/s 00 0 m/s 4 4 R = = 8 (0 0 Hz) = 9 0 Hz c m/s m/s EVAUATE: R < since the source is moing away The dierence between R is ery small since! c 6 IDENTIFY: Apply Eq(60) ET UP: Require R = 00 ince R > the star would be moing toward us and < 0, so = 8 c = 00 0 m/s EXECUTE: c+ c+ R = R = 00 gies = (00) oling or gies c c (00) c 7 = = 0090c= 8 0 m/s + (00) EVAUATE: 9% c = Δ R 00% = = c and Δ are approximately equal 6 IDENTIFY: Apply Eq(6) ET UP: The Mach number is the alue o /, where is the speed o the shuttle and is the speed o sound at the altitude o the shuttle EXECUTE: (a) = sinα = sin 80 = 0848 The Mach number is 8 = 0848 = m/s (b) = = = 90 m/s sinα sin m/s (c) = 44 m/s = The Mach number would be 44 m/s sinα = = and α = m/s EVAUATE: The smaller the Mach number, the larger the angle o the shock-wae cone 6 IDENTIFY: Apply Eq(6) to calculate α Use the method o Example 60 to calculate t ET UP: Mach 70 means / = 70 EXECUTE: (a) In Eq(6), = 70 = 088 and α = arcsin(088) = 60 ( 90 m) (b) As in Example 60, t = = s (70)(44 m s)(tan(60 )) EVAUATE: The angle α decreases when the speed o the plane increases 64 IDENTIFY: The displacement yxt (, ) is gien in Eq(6) and the pressure ariation is gien in Eq(64) The pressure ariation is related to the displacement by Eq(6) ET UP: k = π / λ EXECUTE: (a) Mathematically, the waes gien by Eq(6) and Eq(64) are out o phase Physically, at a displacement node, the air is most compressed or rareied on either side o the node, and the pressure gradient is zero Thus, displacement nodes are pressure antinodes (b) The graphs hae the same orm as in Figure 6 in the textbook yxt (, ) (c) pxt (, ) = B When yxt (, ) ersus x is a straight line with positie slope, p( xt, ) is constant and x negatie When yxt (, ) ersus x is a straight line with negatie slope, p( xt, ) is constant and positie The graph o 4 p( x,0) is gien in Figure 64 The slope o the straightline segments or yx (,0) is 6 0, so or the wae in 4 Figure 64 in the textbook, p-non = (6 0 ) B The sinusoidal wae has amplitude 4 p = BkA= ( 0 ) B The dierence in the pressure amplitudes is because the two yx (,0) unctions hae dierent slopes

13 ound and Hearing 6- EVAUATE: (d) p( xt, ) has its largest magnitude where yxt (, ) has the greatest slope This is where yxt (, ) = 0 or a sinusoidal wae but it is not true in general Figure 64 6 IDENTIFY: The sound intensity leel is β = (0 db) log ( I/ I0), so the same sound intensity leel β means the same intensity I The intensity is related to pressure amplitude by Eq(6) and to the displacement amplitude by Eq(6) ET UP: = 44 m/s ω = π Each octae higher corresponds to a doubling o requency, so the note sung by the bass has requency (9 Hz)/ 8 = 6 Hz et reer to the note sung by the soprano and reer to the note sung by the bass I 0 = 0 W/m p EXECUTE: (a) I = and I = Igies,, B p = p ; the ratio is 00 (b) I = ρbω A = ρb4 π A I = Igies A = A A = = 800 A I 7 (c) β = 70 db gies log( I/ I 0) = 7 = 0 and I = 8 0 W/m I = ρb4 π A I 0 I A = = = = (8 0 W/m ) π ρb π(9 Hz) (0 kg/m )(4 0 Pa) m 47 nm EVAUATE: Een or this loud note the displacement amplitude is ery small For a gien intensity, the displacement amplitude depends on the requency o the sound wae but the pressure amplitude does not 66 IDENTIFY: Use the equations that relate intensity leel and intensity, intensity and pressure amplitude, pressure amplitude and displacement amplitude, and intensity and distance (a) ET UP: Use the intensity leel β to calculate I at this distance β = (0 db) log( I/ I0) EXECUTE: 0 db = (0 db)log( I /(0 W/m )) log( I /(0 W/m )) 0 = implies I = 8 0 W/m ET UP: Then use Eq(64) to calculate p : p I = so p = ρi ρ From Example 66, EXECUTE: p ρ = 0 kg/m or air at 0 C 7 7 = ρi = (0 kg/m )(44 m/s)(8 0 W/m ) = 004 Pa p (b) ET UP: Eq(6): p = BkA so A = Bk For air B = 4 0 Pa (Example 6) π π ( π rad)(87 Hz) EXECUTE: k = = = = 07 rad/m λ 44 m/s p 004 Pa 9 A = = = m Bk (4 0 Pa)(07 rad/m) (c) ET UP: β β = (0 db)log( I/ I) (Example 6) Eq(6): I / I = r / r so I / I = r / r EXECUTE: β β = (0 db)log( r / r ) = (0 db)log( r / r ) β = 0 db and r = 00 m Then β = 00 db and we need to calculate r 0 db 00 db = (0 db) log( r / r ) 0 db = (0 db) log( r / r ) log( r/ r ) = 0 so r = 6r = 60 m

14 6-4 Chapter 6 EVAUATE: The decrease in intensity leel corresponds to a decrease in intensity, and this means an increase in distance The intensity leel uses a logarithmic scale, so simple proportionality between r and β doesn t apply 67 IDENTIFY: The sound is irst loud when the requency 0 o the speaker equals the requency o the γ RT undamental standing wae or the gas in the tube The tube is a stopped pipe, and = = The sound 4 M is next loud when the speaker requency equals the irst oertone requency or the tube ET UP: A stopped pipe has only odd harmonics, so the requency o the irst oertone is = EXECUTE: (a) 0 γ RT 6 M0 = = = This gies T = 4 4 M γ R (b) 0 EVAUATE: (c) Measure 0 and Then 0 = gies = F F 68 IDENTIFY: beat = A B = and = gies = Apply τ z = 0 to the bar to ind the m m tension in each wire ET UP: For τ = 0 take the piot at wire A and let counterclockwise torques be positie The ree-body z diagram or the bar is gien in Figure 68 et be the length o the bar EXECUTE: τ z = 0 gies F B wlead( / 4) wbar ( / ) = 0 F = wlead / 4 + wbar / = (8 N)/ 4 + (6 N)/ = N F + F = wbar + wlead so B 9 N FA = wbar + wlead FB = 6 N + 8 N N = 9 N A = = 884 Hz (0 0 kg)(070 m) N B = A = 7 Hz beat = B A = 7 Hz 9 N EVAUATE: The requency increases when the tension in the wire increases A B Figure IDENTIFY: The lute acts as a stopped pipe and its harmonic requencies are gien by Eq(6) The resonant requencies o the string are n = n, n=,,, The string resonates when the string requency equals the lute requency ET UP: For the string s = 6000 Hz For the lute, the undamental requency is 440 m s = = = 8000 Hz et n label the harmonics o the lute and let ns label the harmonics o the 4 4(007 m) string EXECUTE: For the lute and string to be in resonance, n = n s s, where s = 6000 Hz is the undamental 4 requency or the string ns = n( s) = n n s is an integer when n = N, N =,,, (the lute has only odd harmonics) n = N gies ns = 4N Flute harmonic N resonates with string harmonic 4 N, N =,,, EVAUATE: We can check our results or some speciic alues o N For N =, n = and = 400 Hz For this N, s 4 n = and 4s 400 Hz = For, N = 9 n = and Hz, = and s, n = s 700 Hz = Our general results do gie equal requencies or the two objects

15 ound and Hearing IDENTIFY: The harmonics o the string are n = n = n, where l is the length o the string The tube is a l stopped pipe and its standing wae requencies are gien by Eq(6) For the string, = F/ μ, where F is the tension in the string string ET UP: The length o the string is d = 0, so its third harmonic has requency = d F μ The pipe s stopped pipe has length, so its irst harmonic has requency = 4 string pipe EXECUTE: (a) Equating and and using d = 0 gies F = μs 600 (b) I the tension is doubled, all the requencies o the string will increase by a actor o In particular, the third harmonic o the string will no longer be in resonance with the irst harmonic o the pipe because the requencies will no longer match, so the sound produced by the instrument will be diminished (c) The string will be in resonance with a standing wae in the pipe when their requencies are equal Using pipe string pipe string =, the requencies o the pipe are n = n, (where n =,,, ) etting this equal to the string requencies o the string n, the harmonics o the string are n = n=, 9,, The nth harmonic o the pipe is in resonance with the nth harmonic o the string EVAUATE: Each standing wae or the air column is in resonance with a standing wae on the string But the reerse is not true; not all standing waes o the string are in resonance with a harmonic o the pipe 66 IDENTIFY and ET UP: The requency o any harmonic is an integer multiple o the undamental For a stopped pipe only odd harmonics are present For an open pipe, all harmonics are present ee which pattern o harmonics its to the obsered alues in order to determine which type o pipe it is Then sole or the undamental requency and relate that to the length o the pipe EXECUTE: (a) For an open pipe the successie harmonics are = n, n =,,, For a stopped pipe the successie harmonics are = n, n =,,, I the pipe is open and these harmonics are successie, then n n = n = 7 Hz and n+ = ( n+ ) = 764 Hz ubtract the irst equation rom the second: 7 Hz ( n+ ) n = 764 Hz 7 Hz This gies = 9 Hz Then n = = But n must be an integer, so 9 Hz the pipe can t be open I the pipe is stopped and these harmonics are successie, then = n = 7 Hz and n+ = ( n+ ) = 764 Hz (in this case successie harmonics dier in n by ) ubtracting one equation rom the other gies = 9 Hz and = 96 Hz Then n= 7 Hz / = 7 so 7 Hz = 7 and 764 Hz = 9 The solution gies integer n as it should; the pipe is stopped (b) From part (a) these are the 7th and 9th harmonics (c) From part (a) = 96 Hz 44 m/s For a stopped pipe = and = = = 049 m 4 4 4(96 Hz) EVAUATE: It is essential to know that these are successie harmonics and to realize that 7 Hz is not the undamental There are other lower requency standing waes; these are just two successie ones 66 IDENTIFY: The steel rod has standing waes much like a pipe open at both ends, since the ends are both displacement n antinodes An integral number o hal waelengths must it on the rod, that is, n =, with n =,,, ET UP: Table 6 gies = 94 m/s or longitudinal waes in steel EXECUTE: (a) The ends o the rod are antinodes because the ends o the rod are ree to oscillate (b) The undamental can be produced when the rod is held at the middle because a node is located there ()( 94 m s) (c) = = 980 Hz 0 m ( ) (d) The next harmonic is n=, or = 96 Hz We would need to hold the rod at an n = node, which is located at 4 = 07 m rom either end EVAUATE: For the 0 m long rod the waelength o the undamental is x= = 00 m The node to antinode distance is λ / 4 = 07 m For the second harmonic λ = = 0 m and the node to antinode distance is 07 m There is a node at the middle o the rod, but orcing a node at 07 m rom one end, by holding the rod there, preents rod rom ibrating in the undamental n n

16 6-6 Chapter 6 66 IDENTIFY: The shower stall can be modeled as a pipe closed at both ends, and hence there are nodes at the two end walls Figure in the textbook shows standing waes on a string ixed at both ends but the sequence o harmonics is the same, namely that an integral number o hal waelengths must it in the stall ET UP: The irst three normal modes correspond to one hal, two hales or three hales o a waelength along the length o the air column EXECUTE: (a) The condition or standing waes is n = n, so the irst three harmonics are or n =,, (b) A particular physics proessor s shower has a length o = 48 m Using = n and = 44 m s gies n resonant requencies 6 Hz, Hz and 49 Hz Note that the undamental and second harmonic, which would hae the greatest amplitude, are requencies typically in the normal range o male singers Hence, men do sing better in the shower! (For a urther discussion o resonance and the human oice, see Thomas D Rossing, The cience o ound, econd Edition, Addison-Wesley, 990, especially Chapters 4 and 7) EVAUATE: The standing wae requencies or a pipe closed at both ends are the same as or an open pipe o the same length, een though the nodal patterns are dierent 664 IDENTIFY: tress is F/ A, where F is the tension in the string and A is its cross sectional area ET UP: A= π r For a string ixed at each end, F F = = = μ m 8 6 EXECUTE: (a) The cross-section area o the string would be A = (900 N) (70 0 Pa) = 9 0 m, corresponding to a radius o 0640 mm The length is the olume diided by the area, and the olume is V = m/ ρ, so V m ρ (400 0 kg) = = = = 040 m 6 A A (78 0 kg m )(9 0 m ) 900 N (b) For the imum tension o 900 N, = = 7 Hz, or 80 Hz to two igures (400 0 kg)(040 m) EVAUATE: The string could be shorter and thicker A shorter string o the same mass would hae a higher undamental requency 66 IDENTIFY and ET UP: There is a node at the piston, so the distance the piston moes is the node to node distance, λ / Use Eq() to calculate and Eq(60) to calculate γ rom EXECUTE: (a) λ /= 7 cm, so λ = (7 cm) = 70 cm = 070 m = λ = (00 Hz)(070 m) = 7 m/s (b) = γ RT/ M (Eq60) M (88 0 kg/mol)(7 m/s) γ = = = 9 RT (84 J/mol K)(0 K) (c) EVAUATE: There is a node at the piston so when the piston is 80 cm rom the open end the node is inside the pipe, 80 cm rom the open end The node to antinode distance is λ / 4 = 88 cm, so the antinode is 08 cm beyond the open end o the pipe The alue o γ we calculated agrees with the alue gien or air in Example IDENTIFY: Model the auditory canal as a stopped pipe with length cm ET UP: The requencies o a stopped pipe are gien by Eq(6) EXECUTE: (a) The requency o the undamental is = /4 = (44 m/s)/[4(00 m)] = 440 Hz 00 Hz is near the resonant requency, and the ear will be sensitie to this requency (b) The next resonant requency would be = 0,00 Hz and the ear would be sensitie to sounds with requencies close to this alue But 7000 Hz is not a resonant requency or a stopped pipe and the ear is not sensitie at this requency EVAUATE: For a stopped pipe only odd harmonics are present 667 IDENTIFY: The tuning ork requencies that will cause this to happen are the standing wae requencies o the F wire For a wire o mass m, length and with tension F the undamental requency is = = 4m The standing wae requencies are, n = n n =,,, ET UP: F = Mg, where M = 040 kg The mass o the wire is m= ρv = ρπ d /4, where d is the diameter F Mg (400 0 kg)(980 m s ) 6 EXECUTE: (a) = = = = 77 Hz 4 m πd ρ π( 0 m) (04 m) (4 0 kg m ) The tuning ork requencies or which the ork would ibrate are integer multiples o 77 Hz

17 ound and Hearing EVAUATE: (b) The ratio mm 9 0, so the tension does not ary appreciably along the string due to the mass o the wire Also, the suspended mass has a large inertia compared to the mass o the wire and assuming that it is stationary is an excellent approximation 668 IDENTIFY: For a stopped pipe the requency o the undamental is = The speed o sound in air depends on 4 temperature, as shown by Eq(60) ET UP: Example 6 shows that the speed o sound in air at 0 C is 44 m/s 44 m/s EXECUTE: (a) = = = 046 m 4 4(49 Hz) (b) The requency will be proportional to the speed, and hence to the square root o the Kelin temperature The temperature necessary to hae the requency be higher is (9 K)([70 Hz]/[49 Hz]) = 9 K, which is 6 C EVAUATE: 6 C = F, so this extreme rise in pitch won't occur in practical situations But changes in temperature can hae noticeable eects on the pitch o the organ notes γ RT 669 IDENTIFY: = λ = ole or γ M ET UP: The waelength is twice the separation o the nodes, so λ =, where = 000 m γ RT EXECUTE: = λ = = oling or γ, M M (60 0 kg/mol) γ = ( ) = ( (000 m)(00 Hz) ) = 7 RT (84 J mol K) (9 K) EVAUATE: This alue o γ is smaller than that o air We will see in Chapter 9 that this alue o γ is a typical alue or polyatomic gases 670 IDENTIFY: Destructie intererence occurs when the path dierence is a hal-integer number o waelengths Constructie intererence occurs when the path dierence is an integer number o waelengths 44 m/s ET UP: λ = = = 049 m 784 Hz EXECUTE: (a) I the separation o the speakers is denoted h, the condition or destructie intererence is x + h x= βλ, where β is an odd multiple o one-hal Adding x to both sides, squaring, canceling the h β term rom both sides and soling or x gies x = λ Using λ = 049 m and h = 00 m yields 90 m βλ or β =, 7 m or β =, 7 m or β =, 0 m or β = 7, and 006 m or 9 β = These are the only allowable alues o β that gie positie solutions or x (b) Repeating the aboe or integral alues o β, constructie intererence occurs at 44 m, 84 m, 086 m, 06 m Note that these are between, but not midway between, the answers to part (a) (c) I h = λ, there will be destructie intererence at speaker B I λ > h, the path dierence can neer be as large as λ (This is also obtained rom the aboe expression or x, with x = 0 and β = ) The minimum requency is then h= (44 m s) (40 m) = 86 Hz EVAUATE: When increases, λ is smaller and there are more occurrences o points o constructie and destructie intererence + 67 IDENTIFY: Apply = + ET UP: The positie direction is rom the listener to the source (a) The wall is the listener = 0 m/s = 0 = 600 Hz (b) The wall is the source and the car is the listener = 0 =+ 0 m/s = 600 Hz m/s 0 m/s EXECUTE: (a) = = = (600 Hz) = 48 Hz m/s + 44 m/s + 0 m/s (b) = = (600 Hz) = 6 Hz + 44 m/s EVAUATE: ince the singer and wall are moing toward each other the requency receied by the wall is greater than the requency sung by the soprano, and the requency she hears rom the relected sound is larger still x

18 6-8 Chapter IDENTIFY: Apply = The wall irst acts as a listener and then as a source + ET UP: The positie direction is rom listener to source The bat is moing toward the wall so the Doppler eect increases the requency and the inal requency receied,, is greater than the original source requency, = 000 Hz = 00 Hz EXECUTE: The wall receies the sound: = = bat = bat The wall receies the sound: = = 0 and =+ bat = = = bat bat bat bat bat + = and = 0 = + + bat bat Δ (44 m/s)(00 Hz) =Δ = = bat = = = 088 m/s bat bat +Δ (000 Hz) + 00 Hz bat EVAUATE: <Δ, so we can write our result as the approximate but accurate expression Δ = 67 IDENTIFY and ET UP: Use Eq(6) or the intensity and Eq(64) to relate the intensity and pressure amplitude EXECUTE: (a) The amplitude o the oscillations is Δ R I = ρb π A = ρbπ Δ R (b) (c) ( ) ( ) P= I(4 πr ) = 8 π ρb R ( Δ R) I / I = d / R R d I = ( R/ d) I = π ρb( R / d) ( Δ R) d I = p ρb so / R ( ) p = ρbi = π ρb( R / d) Δ R p p p A = / B( R/ d) R Bk = λ ρ Bπ = Bπ = Δ But = B/ ρ so ρ / B = so A= ( R/ d) Δ R EVAUATE: The pressure amplitude and displacement amplitude all o like / d and the intensity like gies / d IDENTIFY: Apply = The heart wall irst acts as the listener and then as the source + ET UP: The positie direction is rom listener to source The heart wall is moing toward the receier so the Doppler eect increases the requency and the inal requency receied,, is greater than the source requency, = 8 Hz EXECUTE: Heart wall receies the sound: = = = 0 = + wall = + = wall Heart wall emits the sound: = =+ wall = 0 wall wall wall wall = = = = = + wall + wall + wall + wall + wall ( ) ( ) (8 Hz)(00 m/s) wall = " and wall = = = 009 m/s = 9 cm/s 6 ( ) (00 0 Hz) 6 EVAUATE: = 00 0 Hz and 8 Hz, gies = so the approximation we made is ery accurate Within this approximation, the requency dierence between the relected and transmitted waes is directly proportional to the speed o the heart wall

19 ound and Hearing (a) IDENTIFY and ET UP: Use Eq() to calculate λ 48 m/s EXECUTE: λ = = = m 0 0 Hz (b) IDENTIFY: Apply the Doppler eect equation, Eq(69) The Problem-oling trategy in the text (ection 68) describes how to do this problem The requency o the directly radiated waes is =,000 Hz The moing whale irst plays the role o a moing listener, receiing waes with requency The whale then acts as a moing source, emitting waes with the same requency, = with which they are receied et the speed o the whale be W ET UP: whale receies waes (Figure 67a) Figure 67a ET UP: whale re-emits the waes (Figure 67b) Figure 67b EXECUTE: W = W = = + = EXECUTE: W + = = + W But + W + W = so = = W W + Δ = = = = W W W W Then W W W 4 (0 0 Hz)(49 m/s) Δ = = 47 Hz 48 m/s 49 m/s EVAUATE: istener and source are moing toward each other so requency is raised IDENTIFY: Apply the Doppler eect ormula = In the HM the source moes toward and away + rom the listener, with imum speed ω A p p ET UP: The direction rom listener to source is positie EXECUTE: (a) The imum elocity o the siren is ωpap = ωpap You hear a sound with requency = siren ( + ), where aries between + π PAP and π PAP = siren ( π PAP) and ( π ) min= siren + PAP (b) The imum (minimum) requency is heard when the platorm is passing through equilibrium and moing up (down) EVAUATE: When the platorm is moing upward the requency you hear is greater than siren and when it is moing downward the requency you hear is less than siren When the platorm is at its imum displacement rom equilibrium its speed is zero and the requency you hear is siren 677 IDENTIFY: Follow the method o Example 69 and apply the Doppler shit ormula twice, once with the insect as the listener and again with the insect as the source ET UP: et bat be the speed o the bat, insect be the speed o the insect, and i be the requency with which the sound waes both strike and are relected rom the insect The positie direction in each application o the Doppler shit ormula is rom the listener to the source

20 6-0 Chapter 6 EXECUTE: The requencies at which the bat sends and receies the signals are related by + bat + insect + bat = i = oling or insect bat, insect insect = and = gies the result etting rel bat + bat + + bat ( ) ( ) ( ) ( ) bat bat bat insect = = + bat bat + + bat (b) I bat = 807 khz, rel = 8 khz, and bat = 9 m s, then insect = 0 m s EVAUATE: rel > bat because the bat and insect are approaching each other 678 IDENTIFY: Follow the steps speciied in the problem is positie when the source is moing away rom the receier and is negatie when the source is moing toward the receier R is the beat requency ET UP: The source and receier are approaching, so R > and R = 460 Hz EXECUTE: (a) c / c R = = = + c+ + / c c c (b) For small x, the binomial theorem (see Appendix B) gies ( x) x, ( x) + x Thereore, where the binomial theorem has been used to approximate ( x ) x c c (c) For an airplane, the approximation! c is certainly alid oling the expression ound in part (b) or, c c 460 Hz R beat 8 = = = (00 0 ms) = 68 ms Hz The speed o the aircrat is 68 m/s EVAUATE: The approximation! c is seen to be alid is negatie because the source and receier are Δ approaching ince! c, the ractional shit in requency,, is ery small 679 IDENTIFY: Apply the result deried in part (b) o Problem 678 The radius o the nebula is R = t, where t is the time since the supernoa explosion ET UP: When the source and receier are moing toward each other, is negatie and R > The light rom the explosion reached earth 9 years ago, so that is the amount o time the nebula has expanded ly = m 4 R Hz 6 EXECUTE: (a) = c = (00 0 m s) = 0 m s, with the minus sign indicating Hz that the gas is approaching the earth, as is expected since > R (b) The radius is (9 yr)(6 0 s yr)( 0 m s) = 6 0 m = 8 ly (c) The ratio o the width o the nebula to π times the distance rom the earth is the ratio o the angular width (taken as arc minutes) to an entire circle, which is arc minutes The distance to the nebula is then (60)(60) (7 ly) = 0 ly The time it takes light to trael this distance is 00 yr, so the explosion actually took place 00 yr beore 04 CE, or about 400 BCE Δ EVAUATE: = 40 0, so een though is ery large the approximation required or = c is accurate c 680 IDENTIFY and ET UP: Use Eq(60) that describes the Doppler eect or electromagnetic waes! c, so the simpliied orm deried in Problem 678b can be used (a) EXECUTE: From Problem 678b, = R ( / c ) is negatie since the source is approaching: = (40 km/h)(000 m/ km)( h/600 s) = 67 m/s Approaching means that the requency is increased 6 67 m/s Δ = = Hz 09 Hz 8 = c 00 0 m/s

21 ound and Hearing 6- (b) EVAUATE: Approaching, so the requency is increased! c and thereore Δ /! The requency o the waes receied and relected by the water is ery close to 880 MHz, so get an additional shit o 09 Hz and the total shit in requency is (09 Hz) = 8 Hz 68 IDENTIFY: Follow the method o Example 69 and apply the Doppler shit ormula twice, once or the wall as a listener and then again with the wall as a source ET UP: In each application o the Doppler ormula, the positie direction is rom the listener to the source EXECUTE: (a) The wall will receie and relect pulses at a requency 0, and the woman will hear this w w relected wae at a requency w w 0= 0 The beat requency is beat = 0 = 0 w w w w (b) In this case, the sound relected rom the wall will hae a lower requency, and using 0( w)/( + w) as the w w detected requency w is replaced by w in the calculation o part (a) and beat = 0 = 0 + w + w EVAUATE: The beat requency is larger when she runs toward the wall, een though her speed is the same in both cases 68 IDENTIFY and ET UP: Use Fig(67) to relate α and T Use this in Eq(6) to eliminate sin α sinα sinα EXECUTE: Eq(6): sin α = / From Fig67 tan α = h/ T And tan α = = cosα sin α h / h Combining these equations we get = and = T ( / ) T ( / ) T ( / ) = and = h T / h h = as was to be shown h T EVAUATE: For a gien h, the aster the speed o the plane, the greater is the delay time T The imum delay time is h/, and T approaches this alue as T 0 as 68 IDENTIFY: The phase o the wae is determined by the alue o x t, so t increasing is equialent to x decreasing with t constant The pressure luctuation and displacement are related by Eq(6) ET UP: yxt (, ) = pxtdx (, ) B I p( xt, ) ersus x is a straight line, then yxt (, ) ersus x is a parabola For air, B = 4 0 Pa EXECUTE: (a) The graph is sketched in Figure 68a (b) From Eq(64), the unction that has the gien px (, 0) at t= 0 is gien graphically in Figure 68b Each section is a parabola, not a portion o a sine cure The period is λ w 4 = (000 m) (44 m s) = 8 0 s and the amplitude is equal to the area under the p ersus x cure between x= 0 and x= 0000 m diided by B, or m (c) Assuming a wae moing in the + x -direction, y(0, t) is as shown in Figure 68c (d) The imum elocity o a particle occurs when a particle is moing through the origin, and the particle speed y p is y = = The imum elocity is ound rom the imum pressure, and x B = = The imum acceleration is the imum pressure gradient y (40 Pa)(44 m/s)/(4 0 Pa) 969 cm/s diided by the density, (800 Pa) (000 m) a = = (0 kg m ) m s (e) The speaker cone moes with the displacement as ound in part (c ); the speaker cone alternates between moing orward and backward with constant magnitude o acceleration (but changing sign) The acceleration as a unction o time is a square wae with amplitude 667 m/s and requency = / λ = (44 m/s)/(000 m) = 7 khz