Chapter 17 Solutions

Transcription

1 Chapter 17 Solutions 17.1 Since v light >> v sound, d (343 m/s)(16. s) 5.56 km Goal Solution G: There is a common rule of thumb that lightning is about a mile away for every 5 seconds of delay between the flash and thunder (or ~3 s/km). Therefore, this lightning strike is about 3 miles (~5 km) away. O: The distance can be found from the speed of sound and the elapsed time. The time for the light to travel to the observer will be much less than the sound delay, so the speed of light can be ignored. A: Assuming that the speed of sound is constant through the air between the lightning strike and the observer, v s d t or d v s t (343 m/s)(16. s) 5.56 km L: Our calculated answer is consistent with our initial estimate, but we should check the validity of our assumption that the speed of light could be ignored. The time delay for the light is t light d c 5560 m m/s s and t t sound t light 16. s s 16. s (when properly rounded) Since the travel time for the light is much smaller than the uncertainty in the time of 16. s, t light can be ignored without affecting the distance calculation. However, our assumption of a constant speed of sound in air is probably not valid due to local variations in air temperature during a storm. We must assume that the given speed of sound in air is an accurate average value for the conditions described. 17. v B ρ km/s 17.3 Sound takes this time to reach the man: (0.0 m 1.75 m) 343 m/s s so the warning should be shouted no later than s s s before the pot strikes.

2 Chapter 17 Solutions Since the whole time of fall is given by y 1 gt 18.5 m 1 (9.80 m/s ) t t 1.93 s the warning needs to come 1.93 s s 1.58 s into the fall, when the pot has fallen 1 (9.80 m/s )(1.58 s) 1. m to be above the ground by 0.0 m 1. m 7.8 m 17.4 v(air) 343 m/s and v(salt water) 1533 m/s Let d width of inlet t d v w ; t d v a, so d v w d v a d 4.50v wv a (4.50)(1533)(343) 1.99 km v w v a (a) At 9000 m, T 9000 ( 1.00 C) 60.0 C so T 30.0 C 150 Using the chain rule: dv dt dv dt dx dtdx dt v dv dt dtdx v(0.607) v 47, so dt (47 s) dv v t dt (47 s) v f dv 0 vi v t (47 s) ln v f v i (47 s) ln (30.0) ( 30.0) t 7. s for sound to reach ground t h v 9000 [ (30.0)] 5.7 s It takes longer when the air cools off than if it were at a uniform temperature From λ v f 340 m/s, we get: λ s mm

3 Chapter 17 Solutions It is easiest to solve part first: The distance the sound travels to the plane is d s h + (h/) h 5 / The sound travels this distance in.00 s, so d s h 5 (343 m/s)(.00 s) 686 m (686 m) giving the altitude of the plane as h 614 m 5 (a) The distance the plane has traveled in.00 s is v(.00 s) h/ 307 m Thus, the speed of the plane is: v 307 m.00 s 153 m/s 17.8 P max ρvωs max s max P max ρvω ( N/m ) (1.0 kg/m 3 )(343 m/s)(π)( s 1 ) m 17.9 (a) A.00 µm λ π m 40.0 cm 15.7 v ω k m/s 157 s.00 cos [(15.7)(0.0500) (858)( )] 0.433µm (c) v max Aω (.00 µm)(858 s 1 ) 1.7 mm/s (a) P (1.7 Pa)sin(πx/m 340πt/s) (SI units) The pressure amplitude is: P max 1.7 Pa (c) (d) ω πf 340π/s, so f 170 Hz k π λ π/m, giving λ.00 m v λf (.00 m)(170 Hz) 340 m/s

4 4 Chapter 17 Solutions k π λ π (0.100 m) 6.8 m 1 ω πv λ π(343 m/s) (0.100 m) s 1 Therefore, P (0.00 Pa) sin[6.8x/m t/s] 17.1 ω πf πv λ π(343 m/s) (0.100 m) rad/s s max P max ρvω (0.00 Pa) (1.0 kg/m 3 )(343 m/s)( s 1 ) m k π λ π (0.100 m) 6.8 m 1 Therefore, s s max cos(kx ωt) ( m) cos(6.8x/m t/s) (a) The sound "pressure" is extra tensile stress for one-half of each cycle. When it becomes (0.500%)( Pa) Pa, the rod will break. Then, P max ρvωs max s max P max ρvω N/m ( kg/m mm )(3560 m/s)(π500/s) From s s max cos(kx ωt) v s t ωs max sin(kx ωt) v max ωs max (π 500/s)(6.5 mm) 0.5 m/s P max ρωvs max (1.0 kg/m 3 )[π(000 s 1 )](343 m/s)( m) P max Pa P max ρvω s max ρv π v λ s max λ π ρv s max P max λ π (1.0)(343) ( ) m

5 Chapter 17 Solutions (a) P P max sin [kx ωt + φ] with P max 4.00 Pa P(0, 0) P max sin φ 0 φ 0 ω πf π(5000 s 1 ) π 10 4 s 1 Therefore, P (4.00 Pa) sin (kx π 10 4 t/s) At x 0, t s, P (4.00 Pa) sin (0.00π) 0 k π λ ω v π 104 s m/s 91.5 m 1 At x m, t 0, P (4.00 Pa) sin [(91.5 m 1 )(0.000 m) 0] P 3.87 Pa β 10 log I I 0 10 log db (a) 70.0 db 10 log I W/m Therefore, I ( W/m )10 (70.0/10) W/m I P max /ρv, so P max ρvi (1.0 kg/m 3 )(343 m/s)( W/m ) P max 90.7 mpa I 1 ρω s max v (a) At f 500 Hz, the frequency is increased by a factor of.50, so the intensity (at constant s max ) increases by (.50) 6.5. Therefore, 6.5(0.600) 3.75 W/m W/m

6 6 Chapter 17 Solutions 17.0 The original intensity is I 1 1 ρω s max v π ρvf s max (a) If the frequency is increased to f ' while a constant displacement amplitude is maintained, the new intensity is I π ρv(f ') s max so I I 1 π ρv(f ')s max π ρvf s max f ' f or I f ' f I1 If the frequency is reduced to f ' f/ while the displacement amplitude is doubled, the new intensity is I π ρv f (smax ) π ρvf s max I 1 or the intensity is unchanged (a) I 1 ( W/m )10 (β 1/10) ( W/m ) /10 or I W/m I ( W/m )10 (β /10) ( W/m ) /10 or I W/m W/m When both sounds are present, the total intensity is I I 1 + I W/m W/m W/m The decibel level for the combined sounds is β 10 log W/m W/m 10 log( ) 81. db 17. We begin with β 10 log I I 0, and β 1 10 log I 1 I 0, so β β 1 10 log I I 1 Also, I 4πr, and I 1 4πr 1 r 1 r, giving I I 1 Then, β β 1 10 log r 1 r 0 log r 1 r

7 Chapter 17 Solutions Since intensity is inversely proportional to the square of the distance, I I 0.4 and I 0.4 P max ρv (10.0) (1.0)(343) 0.11 W/m The difference in sound intensity level is β 10 log At km, I 4 km I 0.4 km 10( 00) 0.0 db β log 0.11 W/m 10 1 W/m db At 4.00 km, β 4 β β ( ) db 90.8 db Allowing for absorption of the wave over the distance traveled, β' 4 β 4 (7.00 db/km)(3.60 km) 65.6 db This is equivalent to the sound intensity level of heavy traffic. Goal Solution G: At a distance of 4 km, an explosion should be audible, but probably not extremely loud. So based on the data in Table 17., we might expect the sound level to be somewhere between 50 and 100 db. O: From the sound pressure data given in the problem, we can find the intensity, which is used to find the sound level in db. The sound intensity will decrease with increased distance from the source and from the absorption of the sound by the air. A: At a distance of 400 m from the explosion, P max 10 Pa. At this point, I max (10 N/m ) (1.0 kg/m 3 )(343 m/s) Therefore, the maximum sound level is 0.11 W/m β max 10 log I max I 0 10 log 0.11 W/m W/m 111 db

8 8 Chapter 17 Solutions From equations 17.8 and 17.7, we can calculate the intensity and decibel level (due to distance alone) 4 km away: I(400 m) I' (4000 m) W/m and β 10 log I I 0 10 log db At a distance of 4 km from the explosion, absorption from the air will have decreased the sound level by an additional β (7 db/km)(3.6 km) 5. db So at 4 km, the sound level will be β f β β 90.8 db 5. db 65.6 db L: This sound level falls within our expected range. Evidently, this explosion is rather loud (about the same as a vacuum cleaner) even at a distance of 4 km from the source. It is interesting to note that the distance and absorption effects each reduce the sound level by about the same amount (~0 db). If the explosion were at ground level, the sound level would be further reduced by reflection and absorption from obstacles between the source and observer, and the calculation would be much more complicated (if not impossible) Let r 1 and r be the distance from the speaker to the observer that hears 60.0 db and 80.0 db, respectively. Use the result of problem, β β 1 0 log r 1 r, to obtain log r 1 r Thus, log r 1 r 1, so r r. Also: r 1 + r 110 m, so 10.0r + r 110 m giving r 10.0 m, and r m 17.5 We presume the speakers broadcast equally in all directions. (a) r AC m 5.00 m I /4πr W/4π(5.00 m) W/m β 10 db log ( W/m /10 1 W/m ) β 10 db db r BC 4.47 m I W/4π(4.47 m) W/m β 10 db log ( /10 1 ) β 67.8 db

9 Chapter 17 Solutions 9 (c) I 3.18 µw/m µw/m β 10 db log ( /10 1 ) 69.6 db 17.6 I 4πr, where I 1.0 W/m 4πr I 4π(4.00) (1.0) 41 W db 10 db log I 10 1 W/m 4.00 log I 10 1 I 10 1 ( ) W/m 4π r I (4π)(9.00)( ) 1.13 µw *17.8 In I 4πr, intensity I is proportional to 1 r, so between locations 1 and : I I 1 r 1 r In I 1 ρv(ωs max), intensity is proportional to s max, so I I 1 s Then, s s 1 r 1 r or 1 r 1 r, giving r r 1 (50.0 m) 100 m s 1 But, r (50.0 m) + d yields d 86.6 m 17.9 β 10 log I 10 1 I [10(β/10) ](10 1 )W/m I (10 db ) 1.00 W/m ; I (100 db) W/m ; I (10 db) W/m (a) 4πr I so that r 1 I 1 r I r r 1 (I 1 /I ) 1/ (3.00 m) r r 1 (I 1 /I ) 1/ (3.00 m) m m

10 10 Chapter 17 Solutions (a) E t 4πr It 4π(100 m) ( W/m )(0.00 s) 1.76 kj β 10 log db *17.31 (a) The sound intensity inside the church is given by β 10 ln(i/i 0 ) 101 db (10 db) ln(i/10 1 W/m ) I (10 1 W/m ) W/m W/m We suppose that sound comes perpendicularly out through the windows and doors. Then, the radiated power is IA (0.016 W/m )(.0 m ) 0.77 W Are you surprised by how small this is? The energy radiated in 0.0 minutes is E t (0.77 J/s)(0.0 min)(60.0 s/1.00 min) 33 J If the ground reflects all sound energy headed downward, the sound power, 0.77 W, covers the area of a hemisphere. One kilometer away, this area is A πr π(1000 m) π 10 6 m. The intensity at this distance is I A 0.77 W π 10 6 m W/m and the sound intensity level is β (10 db) ln W/m W/m 46.4 db 17.3 (a) P max 5.0 r Pa 4.00 v ω k m/s so the material is water (c) I P max ρv (6.5) ()(1000)(1496) W/m β 10 log db

11 Chapter 17 Solutions 11 (d) P sin [(1.5)(5.00) (1870)(0.0800)] 4.59 Pa *17.33 (a) f ' f v (v ± v S ) Approach: f ' 30 (343) ( ) (343) Receding: f ' 30 ( ) 36. Hz 86.5 Hz The change in frequency observed Hz λ v f 343 m/s 36 Hz m (a) f ' f(v + v O) (v + v S ) where from observer to source is positive. ( ) f ' 500 ( ) ( ) f ' 500 ( ) ( ) (c) f ' 500 ( ) ( ) f ' 500 ( ) 3.04 khz.08 khz.6 khz while police car overtakes.40 khz after police car passes Approaching car f f ' 1 v S (Equation 17.14) Departing car f f '' 1 + v S (Equation 17.15)

12 1 Chapter 17 Solutions Since f ' 560 Hz and f '' 480 Hz, v S v S 1040 v S v 80.0 v S 80.0(343) 1040 m/s 6.4 m/s Goal Solution G: We can assume that a police car with its siren on is in a hurry to get somewhere, and is probably traveling between 0 and 100 mph (~10 to 50 m/s), depending on the driving conditions. O: We can use the equation for the Doppler effect to find the speed of the car. f A: Approaching car: f ' v s f Departing car: f '' 1 + v s (Eq ) (Eq ) Where f ' 560 Hz and f '' 480 Hz. Solving the two equations above for f and setting them equal gives: f ' 1 v s f '' 1 + v s or f ' f '' v s v (f ' + f '') v(f ' f '') so the speed of the source is v s f ' + f '' (343 m/s)(560 Hz 480 Hz) (560 Hz Hz) 6.4 m/s L: This seems like a reasonable speed (about 50 mph) for a police car, unless the street is crowded or the car is traveling on an open highway. *17.36 (a) ω πf π 115/min 60.0 s/min 1.0 rad/s v max ωa (1.0 rad/s)( m) m/s The heart wall is a moving observer. f ' f v + v O (,000,000 Hz) Hz 1500

13 Chapter 17 Solutions 13 (c) Now the heart wall is a moving source. f '' f ' v v v s f f v v + v S t fall (,000,09 Hz) Hz 485(340) + (485)(9.80t f ) (51)(340) t f s d 1 1 gt f 18.3 m t return s The fork continues to fall while the sound returns. t total fall t f + t return 1.93 s s s d total 1 gt total fall 19.3 m (a) The maximum speed of the speaker is described by 1 mv max 1 ka v max k/m A 0.0 N/m (0.500 m) 1.00 m/s 5.00 kg The frequencies heard by the stationary observer range from f' min f v to f ' v + v max max f v v v max where v is the speed of sound. f min ' 440 Hz 343 m/s 343 m/s m/s 439 Hz f' max 440 Hz 343 m/s 441 Hz 343 m/s 1.00 m/s

14 14 Chapter 17 Solutions β 10 db log (I/I 0 ) 10 db log /4πr I 0 The maximum intensity level (of 60.0 db) occurs at r r min 1.00 m. The minimum intensity level occurs when the speaker is farthest from the listener (i.e., when r r max r min + A.00 m). Thus, β max β min 10 db log 4πI 0 r min 10 db log 4πI 0 r max or β max β min 10 db log 4πI 0 r max 4πI 0 r min 10 db log r max r min This gives: 60.0 db β min 10 db log(4.00) 6.0 db, and β min 54.0 db f ' f (v ± v O) (v ± v S ) ( ) (a) f ' Hz ( ) ( ) f ' Hz ( ) *17.40 (a) v (331 m/s) 1 + T 10.0 C (331 m/s) m/s 73 C 73 C Approaching the bell, the athlete hear a frequency of f ' f v + v O After passing the bell, she hears a lower frequency of f '' f v v O The ratio is f '' f ' v v O v + v O 5 6, which gives 6v 6v O 5v + 5v O or v O v m/s m/s

15 Chapter 17 Solutions 15 v sound sin θ v sound 1 v jet 1.0v sound 1.0 θ 56.4 θ θ 17.4 The half angle of the shock wave cone is given by sin θ v light v S v S v light sin θ m/s.8 10 sin (53.0 ) 8 m/s sin θ v 1 v S 3.00 tanθ h x θ 19.5 h x tan θ x 0000 m tan m 56.6 km (a) It takes the plane t x m 56.3 s to travel this distance. v S 3.00(335 m/s) x t 0 θ h h θ Observer Observer hears the boom a. b θ sin 1 v sin v 1 1 S

16 16 Chapter 17 Solutions Let d be the distance the stone drops. t d v S + d g d + g v S d v S t 0 d 1 g v S + v S g + 4v St d 1 ( ± 38000) Choose the positive root so that d > 0 d 0.0 d 400 m If the speed of sound is ignored, t d g d 1 gt 510 m The percentage error is given by d' d d % *17.46 Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versus time. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in all directions and some of it is incident on each one of the solid vertical risers of the bleachers. Suppose that, at the ambient temperature, sound moves at 340 m/s; and suppose that the horizontal width of each row of seats is 60 cm. Then there is a time delay of 0.6 m (340 m/s) 0.00 s between your sound impulse reaching each riser and the next. Whatever its material, each will reflect much of the sound that reaches it. The reflected wave sounds very different from the sharp pop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twenty crests, each separated from the next in time by (0.6 m) (340 m/s) s

17 Chapter 17 Solutions 17 This is the extra time for it to cross the width of one seat twice, once as an incident pulse and once again after its reflection. Thus, you hear a sound of definite pitch, with period about s, frequency s wavelength ~ 300 Hz λ v f (340 m/s) (300/s) 1. m ~ 10 0 m and duration 0(0.004 s) ~ 10 1 s *17.47 (a) λ v f 343 m/s 1480 s m β 81.0 db 10 db log [I/(10 1 W/m )] I (10 1 W/m ) W/m W/m 1 ρvω s max (c) s max λ' v f' I ρvω ( W/m ) (1.0 kg/m 3 )(343 m/s)4π (1480 s 1 ) m 343 m/s 1397 s m λ λ' λ 13.8 mm Since cos θ + sin θ 1 sin θ ±(1 cos θ) 1/ each sign applying half the time. P P max sin (kx ωt) ±ρvωs max [1 cos (kx ωt)] 1/ ±ρvω [s max s max cos (kx ωt)] 1/ ±ρvω(s max s ) 1/ *17.49 The trucks form a train analogous to a wave train of crests with speed v 19.7 m/s and unshifted frequency f 3.00 min 0.667/min. (a) The cyclist as observer measures a lower Doppler-shifted frequency: v v f ' f O (0.667/min) /min 19.7

18 18 Chapter 17 Solutions f '' f v v O ' (0.667/min) /min *17.50 v d t The cyclist s speed has decreased very significantly, but there is only a modest increase in the frequency of trucks passing him. d vt 1 ( m/s)(1.85 s) 6.01 km *17.51 Call d the distance to the reflection point. We have d (6.0 km/s)t and d (3.0 km/s)(t +.40 s) To solve for d we eliminate t by substitution: d t 6.0 km/s d (3.0 km/s) d s 6.0 km +.40 s d 1.03d km d 7.68 km km *17.5 (a) From the equation given in Example 17.1, the speed of a compression wave in a bar is v Y/ρ ( N/m )/(7860 kg/m 3 ) m/s The signal to stop passes between layers of atoms as a sound wave, reaching the back end of the bar in time t L/v (0.800 m)/( m/s) s (c) As described by Newton s first law, the rearmost layer of steel has continued to move forward with its original speed v i for this time, compressing the bar by L v i t (1.0 m/s)( s) m 1.90 mm (d) The strain in the rod is: L/L ( m)/(0.800 m) (e) The stress in the rod is: σ Y( L/L) ( N/m )( ) 476 MPa Since σ > 400 MPa, the rod will be permanently distorted.

19 Chapter 17 Solutions 19 (f) We go through the same steps as in parts (a) through (e), but use algebraic expressions rather than numbers: The speed of sound in the rod is v Y/ρ. The back end of the rod continues to move forward at speed v i for a time of t L/v L ρ/y, traveling distance L v i t after the front end hits the wall. The strain in the rod is: L/L v i t/l v i ρ/y The stress is then: σ Y( L/L) Yv i ρ/y v i ρy For this to be less than the yield stress, σ y, it is necessary that v i ρy < σ y or v i < σ y ρy With the given numbers, this speed is 10.1 m/s. The fact that the length of the rod divides out means that the steel will start to bend right away at the front end of the rod. There it will yield enough so that eventually the remainder of the rod will experience only stress within the elastic range. You can see this effect when sledgehammer blows give a mushroom top to a rod used as a tent stake. v (a) f f (v v diver ) so 1 v diver v f f v diver v 1 f f with v 343 m/s, f ' 1800 Hz and f 150 Hz we find v diver m/s If the waves are reflected, and the skydiver is moving into them, we have f '' f ' (v + v diver) v f '' f v (v v diver ) (v + v diver ) v ( ) so f '' Hz ( )

20 0 Chapter 17 Solutions Goal Solution G: Sky divers typically reach a terminal speed of about 150 mph (~75 m/s), so this sky diver should also fall near this rate. Since her friend receives a higher frequency as a result of the Doppler shift, the sky diver should detect a frequency with twice the Doppler shift: f ' 1800 Hz + ( ) Hz 500 Hz. O: We can use the equation for the Doppler effect to answer both parts of this problem. A: Call f e 1800 Hz the emitted frequency; v e, the speed of the sky diver; and f g 150 Hz, the frequency of the wave crests reaching the ground. (a) The sky diver source is moving toward the stationary ground, so we use the equation f g f s v v v s and v e v 1 f e f g (343 m/s) Hz 150 Hz 55.8 m/s The ground now becomes a stationary source, reflecting crests with the 150-Hz frequency at which they reach the ground, and sending them to a moving observer: f e f v + v e g m/s m/s 343 m/s 500 Hz L: The answers appear to be consistent with our predictions, although the sky diver is falling somewhat slower than expected. The Doppler effect can be used to find the speed of many different types of moving objects, like raindrops (Doppler radar) and cars (police radar) (a) f ' fv v u f '' fv v + u f ' f '' fv 1 v u 1 v + u f f fv(v + u v + u) v u (u/v) 1 (u /v ) f uvf v 1 u v 130 km/h 36.1 m/s (36.1)(400) f 85.9 Hz (36.1) 340

21 Chapter 17 Solutions Use the Doppler formula, and remember that the bat is a moving source. If the velocity of the insect is v x, Solving, ( )(340 v x) ( )(340 + v x ) v x 3.31 m/s Therefore, the bat is gaining on its prey at 1.69 m/s sin β v v S 1 N M h v(1.8 s) shock front x v S (10.0 s) tan β h x 1.8 v v S 1.8 N M h x β v s shock front cos β sin β tan β β 38.6 N M 1 sin β 1.60 *17.57 (a) λ v f 343 m/s 1000 s m (c) λ' v f' v v v S f v ( ) m/s 1000 s m (d) λ'' v f'' ( ) m/s 1000 s m (e) f ' f v v O v v S ( ) m/s (1000 Hz) 1.03 khz ( ) m/s

22 Chapter 17 Solutions t L 1 1 v air v cu L v cu v air v air v cu L v airv cu t (331 m/s)( m/s) ( v cu v air ( ) m/s 3 s) L.34 m (a) 10 db 10 db log [I/(10 1 W/m )] I 1.00 W/m /4πr r 4πI 6.00 W 4π(1.00 W/m ) m We have assumed the speaker is an isotropic point source. 0 db 10 db log (I/10 1 W/m ) I W/m r 4πI 6.00 W 4π( W/m 691 km ) We have assumed a uniform medium that absorbs no energy The shock wavefront connects all observers first hearing the plane, including our observer O and the plane P, so here it is vertical. The angle φ that the shock wavefront makes with the direction of the plane s line of travel is given by P φ C θ O sin φ v 340 m/s v S 1963 m/s so φ 9.97 Using the right triangle CPO, the angle θ is seen to be θ 90.0 φ

23 Chapter 17 Solutions When observer is moving in front of and in the same direction as the source, f ' f v v O where v v S v O and v S are measured relative to the medium in which the sound is propagated. In this case the ocean current is opposite the direction of travel of the ships and v O 45.0 km/h ( 10.0 km/h) 55.0 km/h 15.3 m/s, and v S 64.0 km/h ( 10.0 km/h) 74.0 km/h 0.55 m/s Therefore, f ' (100.0 Hz) 150 m/s 15.3 m/s 104. Hz 150 m/s 0.55 m/s 17.6 We suppose the sound level is uniform over the outer surface of area (0.400 m)(0.400 m) + 4(0.400 m)(0.500 m) 1.1 m The intensity is given by 40.0 db 10 db log(i/10 1 W/m ) I W/m 10 8 W/m The sound power is IA W So the oven's energy efficiency as a sound source is / input ( W)/( W) (a) θ sin 1 v sound v obj sin θ ' sin P max ρωvs max ρ πv λ vs max Also, P and s are 90 out of phase. Therefore, P πρv s max cos (kx ωt) λ

24 4 Chapter 17 Solutions For the longitudinal wave v L (Y/ρ) 1/ For the transverse wave v T (T/µ) 1/ If we require v L v T 8.00, we have T µy 64.0ρ where µ m L and ρ mass volume m πr L This gives T πr Y 64.0 π( m) ( N/m ) N β 1 β 10 log β 10 log β 67.0 db t 3 J 4π(500 m) (10 1 W/m )(10 6 ) 95.5 s The total output sound energy is ee t, where is the power radiated. Thus, t ee ee IA ee (4πr )I But, β 10 log I. Therefore, I I I 0 0 (10 β/10 ) and t ee 4πr I 0 (10 β/10 ) (a) If the source and the observer are moving away from each other, we have: θ S θ O 180, and since cos 180 1, we get Equation with the lower signs. v If v O 0 m/s, then f ' v v S cos θ S f Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection, cos θ S 4 5 so f ' 343 m/s (500 Hz) 343 m/s 0.800(5.0 m/s) or f ' 531 Hz

25 Chapter 17 Solutions 5 Note that as the train approaches, passes, and departs from the intersection, θ S varies from 0 to 180 and the frequency heard by the observer varies from: v f max ' v v S cos m/s f (500 Hz) 539 Hz 343 m/s 5.0 m/s v f min ' v v S cos 180 f 343 m/s (500 Hz) 466 Hz 343 m/s m/s *17.70 Let T represent the period of the source vibration, and E be the energy put into each wavefront. Then av E/T. When the observer is at distance r in front of the source, he is receiving a spherical wavefront of radius vt, where t is the time since this energy was radiated, given by vt v S t r. Then, r t v v S The area of the sphere is 4π(vt) 4πv r (v v S ). The energy per unit area over the spherical wavefront is uniform with the value E A avt(v v S ) 4πv r. The observer receives parcels of energy with the Doppler shifted frequency f ' f v v, so the observer receives a wave with intensity v v S T(v v S ) I E A f ' av T(v v S ) v 4πv r T(v v S ) av 4πr v v S v (a) The time required for a sound pulse to travel distance L at speed v is given by t L v L. Using this expression we find Y/ρ t 1 L 1 Y 1 /ρ 1 L 1 ( N/m )/(700 kg/m 3 ) ( L 1 ) s t 1.50 m L 1 Y /ρ 1.50 m L 1 ( N/m )/( kg/m 3 ) L 1 L or t ( L 1 ) s L 3 t m ( N/m 3 )/(8800 kg/m 3 ) t s

26 6 Chapter 17 Solutions We require t 1 + t t 3, or L L This gives L m and L m The ratio of lengths is then L 1 L 6.45 The ratio of lengths L 1 /L is adjusted in part (a) so that t 1 + t t 3. Sound travels the two paths in equal time and the phase difference, φ Let θ θ 0 log R, I kr: β 10 log I 10 log I 10 log I I log kr 10 log I 0 or β 10 log R + 10(log k log I 0 ) 10 θ + 10(log k log I θ 0 ) 0 β 10 θ 0 θ + 10 log k the equation of a straight line. [y mx + b] I 0 *17.73 To find the separation of adjacent molecules, use a model where each molecule occupies a sphere of radius r given by ρ air average mass per molecule 4 3 πr3 or 1.0 kg/m kg 4, r 3( kg) 1/ πr3 4π(1.0 kg/m 3 9 m ) Intermolecular separation is r m, so the highest possible frequency sound wave is v f max v λ min r 343 m/s m Hz ~ Hz