TRAVELING WAVES. Conceptual Questions v a v b v c. Wave speed is independent of wave amplitude (a)

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1 TRAVELING WAVES 0 Conceptual Questions 0.1. a b c. Wae speed is independent o wae amplitude. 0.. (a) T T T 0 cm/s (b) 100 cm/s 4 T 4m (c) cm/s (d) so the speed is unchanged: 00 cm/s. /4 4L 0.3. The constant 1 mm displacement interal appears at later times. Thereore, the displacement at this point reached mm beore it settled at 1 mm (a) The wae is traeling to the right. It reaches the cm point at later times, 7 starting rom the let. (b) 00 cm/s. The leading edge reaches the cm point at t 0.03s. At t 0.01s it was 4 cm to the let, at cm. x 4 cm 00 cm/s t (0.03 s 0.01s) 0.5. a b c because constant or these requencies, so large implies small The requency is unchanged because that is the requency o the driing orce that moes successie oscillators. But, so i the speed is doubled, so is the waelength The amplitude o the wae is the maximum displacement, which is 4.0 cm. The waelength is the distance between two consecutie peaks, which gies 14 m m 1 m. The requency o the wae is We sole or 0 rom the initial conditions at 4 m/s.0 Hz 1 m 1 x 0m and t 0 s: 4sin(0 0) sin( 0) The amplitude o the wae is the maximum displacement, which is 1.0 cm. The waelength is the distance between two consecutie peaks, which gies.0 m. The requency o the wae is 1.0 m/s 0.50 Hz.0 m Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist. 0-1

2 0- Chapter 0 1 We sole or 0 rom the initial conditions at x 0m and t 1s: 11sin (a) 0; they are on the same wae ront. (b) 4 rad; because they are two wae crests apart. (c) rad; because F is on a crest and E on an adjacent trough P C P B P A. Use E P t For one proessor in each case W/m. The new sound intensity leel is J 10 J J PA 1 W PB W PC W s 5 s 1s I ( W/m ) W/m ; or 100 proessors I W/m (10 db)log10 7 db W/m Notice that or 10 times the intensity, the sound intensity leel goes up by 10 db; or 100 times the intensity, the sound intensity leel goes up by 0 db The correct answer is D because the initial shit is to a lower requency, which means the source is initially moing away rom you until t s. Then the shit is to a higher requency, which means the source is moing toward you. Exercises and Problems Section 0.1 The Wae Model 0.1. Model: The wae is a traeling wae on a stretched string. Sole: The wae speed on a stretched string with linear density is string For a wae speed o 10 m/s, the required tension will be TS 75 N m/s kg/m 3 TS string ( kg/m)(10 m/s) 110 N 0.. Model: The wae is a traeling wae on a stretched string. Sole: The wae speed on a stretched string with linear density is string T S /. The wae speed i the tension is doubled will be 0.3. Sole: string TS 1 1 (00 m/s) 141m/s string TS TS TS L TS L t t t t L t m/ L m m Assess:.0 m seems like a reasonable length or a string. TS 0 N L ( t) (50 ms).0 m m 0.05 kg Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

3 Traeling Waes 0-3 Section 0. One-Dimensional Waes 0.4. Model: This is a wae traeling at constant speed. The pulse moes 1 m to the right eery second. Visualize: The snapshot graph shows the wae at all points on the x-axis at t 0 s. The leading edge o the wae will reach x 5.0 m1s later, at t 1s. The irst part o the wae causes a sudden upward displacement at t 1s. The lat top o the wae is 3 m wide, so it will take 3 s to pass the x 5.0 m point, keeping this point high until t 4 s. The alling slope is 1 m wide, so it will take 1 s or the the displacement at x 5.0 m to drop back to zero. The trailing edge o the pulse arries at displacement now becomes zero and stays zero or all later times. x 5.0 m at t 1s, which is 5 s ater the igure gien in the problem. The 0.5. Model: This is a wae traeling at constant speed. The pulse moes 1 m to the let eery second. Visualize: The snapshot graph shows the wae at all points on the x-axis at t s The leading edge o the wae moing let will reach x 0 m1s later, at t 3 s. The irst part o the wae causes a sudden upward displacement at t 3 s. The alling slope o the wae is 4 m wide, so it will take 4 s or the the displacement at x 0m to decrease rom 1cm to 1cm. The trailing edge o the pulse arries at x 0m at t 7 s, which is 5 s ater the igure gien in the problem. The displacement now becomes zero and stays zero or all later times Model: This is a wae traeling to the right at a constant speed o 1 m/s. Visualize: This is the history graph o a wae at x 0 m. At t 1s, the time or which we re to draw the snapshot graph, the displacement at x 0 m is 1cm. The graph shows that the x 0m point irst went high at t 0 s, and this leading edge will hae moed 1 m to the right by the time o our snapshot graph at t 1s. The trailing edge reaches x 0m at t 6 s. That s 5 s ater our snapshot graph, so at t 1s the trailing edge must still be 5 m let o x 0 m. Thus at t 1s the wae will stretch rom x 5 to x 1m. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

4 0-4 Chapter Model: This is a wae traeling to the let at a constant speed o 1 m/s. Visualize: This is the history graph o a wae at x m. At t 0 s, the time or which we re to draw the snapshot graph, the displacement at pulse is just reaching graph, so at x m to x 6 m. x m is 0 cm but is just beginning to rise. That is, the leading edge o a let-moing x m at t 0 s. The trailing edge reaches x m at t 4 s. That s 4 s ater our snapshot t 0 s the trailing edge must still be 4 m right o x m. Thus at t 0s the wae will stretch rom 0.. Visualize: Figure EX0. shows a snapshot graph at t 0 s o a longitudinal wae. This diagram shows a row o particles with an inter-particle separation o 1.0 cm at equilibrium. Because the longitudinal wae has a positie amplitude o 0.5 cm between x3 cm and x cm, the particles at x 3, 4, 5, 6, 7, and cm are displaced to the right by 0.5 cm Visualize: We irst draw the particles o the medium in the equilibrium positions, with an inter-particle spacing o 1.0 cm. Just underneath, the positions o the particles as a longitudinal wae is passing through are shown at time t 0 s. It is clear that relatie to the equilibrium the particle positions are displaced negatiely on the let side and positiely on the right side. For example, the particles at x0 cm and x 1cm are at equilibrium, the particle at x cm is displaced let by 0.5 cm, the particle at x 3 cm is displaced let by 1.0 cm, the particle at x 4 cm is displaced let by 0.5 cm, and the particle at x 5 cm is undisplaced. The behaior o particles or x 5 cm is opposite o that or x 5 cm. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

5 Traeling Waes 0-5 Section 0.3 Sinusoidal Waes Sole: (a) The waelength is (b) The requency is 4. m k 1.5 rad/m 00 m/s 4 Hz 4.19 m Sole: (a) The wae number is (b) The wae speed is k 3.1 rad/m.0 m 30 rad/s (.0 m) 9.5 m/s 0.1. Model: The wae is a traeling wae. Sole: (a) A comparison o the wae equation with Equation 0.14 yields: A 3.5 cm, k. rad/m, rad/s, and 0 0 rad. The requency is (b) The waelength is (c) The wae speed 46 m/s. 14 rad/s 19.7 Hz 0 Hz.33 m.3 m k.7 rad/m Model: The wae is a traeling wae. Sole: (a) A comparison o the wae equation with Equation 0.14 yields: A 5. cm, k rad/m, rad/s, and 0 0 rad The requency is (b) The waelength is (c) The wae speed 13 m/s. 7 rad/s 11 5 Hz 11 Hz 114 m 11 m k 55 rad/m Sole: The amplitude o the wae is the maximum displacement, which is 6.0 cm. The period o the wae is 0.60 s, so the requency 1/ T 1/0.60 s 1.67 Hz. The waelength is m/s 1. m Hz Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

6 0-6 Chapter 0 Section 0.4 Waes in Two and Three Dimensions Sole: According to Equation 0., the phase dierence between two points on a wae is ( ) 1 r r 1 I 1 rad at r1 4.0 m, we can determine at any r alue at the same instant using this equation. At r 3.5 m, 3 At r 4.5 m, rad. 1 ( r r1 ) rad (35 m 40 m) rad 0 m Sole: According to Equation 0., the phase dierence between two points on a wae is r 1 ( r r1 ) (3 rad 0 rad) (0 cm 0 cm) 40 cm Visualize: A phase dierence o rad corresponds to a distance o. Set up a ratio. Sole: x 5.5 rad 5.5 rad 5.5 rad 5.5 rad 340 m/s = x = = = =.5 m rad rad rad rad 10 Hz Assess:.5 m seems like a reasonable distance Visualize: Sole: (a) Because the same waeront simultaneously reaches listeners at 0 rad ( r r1 ) r r1 Thus, the source is at x.0 m, so that it is equidistant rom the two listeners. (b) The third person is also 5.0 m away rom the source. Her y-coordinate is thus Section 0.5 Sound and Light x 7.0 m and x 3.0 m, y (5 m) ( m) 4 6 m Sole: Two pulses o sound are detected because one pulse traels through the metal to the microphone while the other traels through the air to the microphone. The time interal or the sound pulse traeling through the air is x 400 m tair s ms 343 m/s air Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

7 Traeling Waes 0-7 Sound traels aster through solids than gases, so the pulse traeling through the metal will reach the microphone beore the pulse traeling through the air. Because the pulses are separated in time by 9.00 ms, the pulse traeling through the metal takes tmetal.66 ms to trael the 4.00 m to the microphone. Thus, the speed o sound in the metal is metal x 400 m 1504 m/s 1500 m/s t s metal 0.0. Sole: (a) In aluminum, the speed o sound is 640 m/s. The waelength is thus equal to 640 m/s m 3.1 mm 3. mm Hz (b) The speed o an electromagnetic wae is c. The requency would be 0.1. Sole: (a) The requency is c m/s Hz m c m/s Hz 1.5 GHz 0.0 m (b) The speed o a sound wae in water is water 140 m/s. The waelength o the sound wae would be water 140 m/s m 990 nm Hz 0.. Model: Light is an electromagnetic wae that traels with a speed o Sole: (a) The requency o the blue light is (b) The requency o the red light is (c) Calculate the index o reraction, material blue 9 c m/s Hz m m/s Hz m red 9 acuum acuum 650 nm n 1.44 n 450 nm material 0.3. Model: Radio waes are electromagnetic waes that trael with speed c. Sole: (a) The waelength is c 3010 m/s 96 m 1013 MHz (b) The speed o sound in air at 0 C is 343 m/s. The requency is sound 343 m/s 116 Hz 96 m 0.4. Model: Microwaes are electromagnetic waes that trael with a speed o Sole: (a) The requency o the microwae is 3 10 m/s m/s. c 3010 m/s 10 microwaes 1010 Hz 10 GHz 3010 m (b) The reractie index o air is , so the speed o microwaes in air is air c/1.00 c. The time or the microwae signal to trael is 3 50 km 5010 m t 0167 ms 017 ms air (30 10 m/100) Assess: A small time o 0.17 ms or the microwaes to coer a distance o 50 km shows that the electromagnetic waes trael ery ast. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

8 0- Chapter Model: Light is an electromagnetic wae. Sole: (a) The time light takes is (b) The thickness o water is glass mm 3010 m 3010 m 11 t 1510 s c/ n (30 10 m/s)/150 water 3010 m/s ( s) 3 4 mm c d watert t n Model: Assume that the glass has index o reraction n 15 This means that glass c/ n 10 m/s Visualize: We apply twice, once in air and then in the glass. The requency will be the same in both cases. Sole: (a) In the air The requency is the same in both media, so air air air 3010 m/s Hz 6 10 Hz 035 m 6 10 Hz. glass (b) Now that we know glass and glass, we can ind glass glass glass glass 0 10 m/s 3 cm Hz Assess: We get the same answer rom glass air / n glass 35 cm/1 5 3 cm Sole: (a) The speed o light in a material is gien by Equation 0.9: c c n mat n The reractie index is (b) The requency is mat 40 nm n c ( m/s) 1 10 m/s ac solid solid mat ac 670 nm solid Section 0.6 Power, Intensity, and Decibels solid 1 10 m/s Hz 40 nm 0.. Sole: The energy deliered to the eardrum in time t is E Pt, where P is the power o the wae. The intensity o the wae is I P/ a where a is the area o the ear drum. Putting the aboe inormation together, we hae E Pt ( Ia) t Ir t (010 W/m ) (30 10 m) (60 s) 3410 J 0.9. Sole: The energy deliered to an area a in time t is E Pt, where the power P is related to the intensity I as I P/. a Thus, the energy receied by your back is 5 E Pt Iat (0.0)(1400 W/m )( m )(3600 s) J Sole: (a) The intensity o a uniorm spherical source o power P source a distance r away is I P source /4 r. Thus, the intensity at the position o the microphone is 35 W 3 I50 m 1110 W/m 4 (50 m) Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

9 Traeling Waes 0-9 (b) The sound energy impinging on the microphone per second is P Ia (1 110 W/m )(1 010 m ) 1110 W 1110 J/s 7 Energy impinging on the microphone in 1 second 1110 J Model: Assume the intensity scales inersely with the square o the distance. I Visualize: = (10 db) log10. I0 30 m Sole: The intensity at 1.0 km is I I m 30 m 1000 m (10 db) I I log10 (10 db) log10 I0 I0 30 m I (10 db) log (10 db) log 30 db 140 db =110 db Assess: 110 db is still loud, but not as damaging m I Sole: Because the sun radiates waes uniormly in all directions, the intensity I o the sun s rays when they impinge upon the earth is r With sun-venus planets are I enus 6 Psun Psun 410 W I earth 11 4 r 4 rearth 4 ( m) m and rsun-mars m, IMars I 1400 W/m the intensities o electromagnetic waes at these 700 W/m and 610 W/m Visualize: Equation 0.35 gies the sound intensity leel as I (10 db)log10 I0 where 1 I W/m. Sole: (a) 6 I 3010 W/m (10 db)log 10 (10 db)log10 65 db I W/m (b) I W/m (10 db)log 10 (10 db)log db I W/m Assess: As mentioned in the chapter, each actor o 10 in intensity changes the sound intensity leel by 10 db; 4 between the irst and second parts o this problem the intensity changed by a actor o 10, so we expect the sound intensity leel to change by 40 db. /10 db Visualize: We can sole Equation 0.35 or the sound intensity, inding I I Sole: (a) I I /10 db (1 010 W/m ) W/m (b) 0 0 /10 db I I 10 (1 010 W/m ) W/m Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

10 0-10 Chapter 0 Assess: Since the sound intensity leels in the two parts o this problem dier by 57 db, we expect the sound 57 intensities to dier by a actor o Model: Assume the pole is tall enough that we don t hae to worry about the ground absorbing or relecting sound. Visualize: The area o a sphere o radius R is A 4 R. Also, I = P/ A. We seek P when R 0 m. Sole: /10 db 90 db/10 db 0 0 P = IA ( I 10 )(4 R ) ( I 10 )(4 (0 m) ) = 5.0 W Assess: 5.0 W is a reasonable power output or a speaker. Section 0.7 The Doppler Eect Model: Your riend s requency is altered by the Doppler eect. The requency o your riend s note increases as he races towards you (moing source and a stationary obserer). The requency o your note or your approaching riend is also higher (stationary source and a moing obserer). Sole: (a) The requency o your riend s note as heard by you is Hz 43 Hz S 50 m/s m/s (b) The requency heard by your riend o your note is 0 50 m/s 0 1 (400 Hz) 1 49 Hz 340 m/s Model: The requency o the opera singer s note is altered by the Doppler eect. Sole: (a) Using 90 km/h m/s, the requency as her conertible approaches the stationary person is Hz 650 Hz 1 5 m/s S/ m/s (b) The requency as her conertible recedes rom the stationary person is Hz 560 Hz 1 5 m/s S/ m/s 0.3. Model: The bat s chirping requency is altered by the Doppler eect. The requency is increased as the bat approaches and it decreases as the bat recedes away. Sole: The bat must ly away rom you, so that the chirp requency obsered by you is less than 5 khz. From Equation 0.3, 0 5,000 Hz 0,000 Hz S 5 m/s 6 m/s 1 S/ 1 S 343 m/s Assess: This is a rather large speed: 5. m/s 10 mph. This is not possible or a bat Model: The mother hawk s requency is altered by the Doppler eect. Sole: The requency is as the hawk approaches you is 900 Hz 00 Hz 3 1 m/s m/s 0 S 1 S / S Assess: The mother hawk s speed o 3.1 m/s 0 mph is reasonable. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

11 Traeling Waes Visualize: The unction D(x, t) represents a pulse that traels in the positie x-direction without changing shape. Sole: (a) (b) The leading edge o the pulse moes orward 3 m each second. Thus, the wae speed is 3 m/s. (c) x 3t is a unction o the orm D( x t), so the pulse moes to the right at 3 m/s Sole: (a) We see rom the history graph that the period T 0.0 s and the wae speed 4.0 m/s. Thus, the waelength is T (40 m/s)(00 s) 0 0 m (b) The phase constant 0 is obtained as ollows: D(0 m, 0 s) Asin mm ( mm)sin sin 1 rad (c) The displacement equation or the wae is x x t D( x, t) Asin 1 t 0 (0 mm)sin (0 mm)sin(5 x 10 t ) 00 m 00 s where x and t are in m and s, respectiely Sole: (a) We can see rom the graph that the waelength is.0 m. We are gien that the wae s requency is 5.0 Hz. Thus, the wae speed is 10 m/s. (b) The snapshot graph was made at t 0 s. Reading the graph at x 0 m, we see that the displacement is D( x 0 m, t 0 s) D(0 m, 0 s) 05 mm 1 A Thus D(0 m, 0 s) A Asin 00 sin rad or rad 6 6 Note that the alue o D(0 m, 0 s) alone gies two possible alues o the phase constant. One o the alues will cause the displacement to start at 0.5 mm and increase with distance as the graph shows while the other will cause the displacement to start at 0.5 mm but decrease with distance. Which is which? The wae equation or t 0 s is x D( x, t 0) Asin 0 Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

12 0-1 Chapter 0 I x is a point just to the right o the origin and is ery small, the angle ( x/ 0) is just slightly bigger than the angle 0 Now sin31 sin30, but sin151 sin150, so the alue 1 rad is the phase constant or which the 0 6 displacement increases as x increases. (c) The equation or a sinusoidal traeling wae can be written as x x D( x, t) Asin t 0 Asin t 0 Substituting in the alues ound aboe, x 1 D( x, t) (10 mm)sin 50 s t 0 m Model: The wae is a traeling wae on a stretched string. Sole: The wae speed on a string whose radius is R, length is L, and mass density is is string T S / with I the string radius doubles, then m V R L R L L L T S 0 string ( R) Model: The wae pulse is a traeling wae on a stretched string. Sole: While the tension T S is the same in both the strings, the wae speeds in the two strings are not. We hae TS 1 and 1 T T S 1 1 S Because 1 L1 / t1 and L / t, and because the pulses are to reach the ends o the string simultaneously, the aboe equation can be simpliied to Since L1L 4 m, 11 1 t t L 1 L L L 40 g/m L L 0 g/m 1 L L 4 m L 166 m 1 7 m and L1 (166 m) 34 m 3 m Sole: t is the time the sound wae takes to trael down to the bottom o the ocean and then up to the ocean surace. The depth o the ocean is d ( ) t d (750 m/s) t sound in water Using this relation and the data rom Figure P0.45, we can generate the ollowing table or the ocean depth (d ) at arious positions (x) o the ship. x (km) t (s) d (km) Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

13 Traeling Waes Visualize: Sole: The explosie s sound traels down the lake and into the granite, and then it is relected by the oil surace. The echo time is thus equal to t t t t t echo water down granite down granite up water up 500 m dgranite dgranite 500 m 094 s dgranite 790 m 140 m/s 6000 m/s 6000 m/s 140 m/s Model: Assume a room temperature o 0 C Visualize: Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

14 0-14 Chapter 0 Sole: The distance between the source and the let ear (E L) is dl x ( y 01 m) [(50 m)cos45 ] [(50 m)sin45 01 m] m Similarly d R 499 m Thus, dl dr d m For the sound wae with a speed o 343 m/s, the dierence in arrial times at your let and right ears is d m t 410 s 343 m/s 343 m/s 0.4. Model: The laser beam is an electromagnetic wae that traels with the speed o light. Sole: The speed o light in the liquid is The liquid s index o reraction is 3010 m liquid m/s 1310 s c 3010 n liquid Thus the waelength o the laser beam in the liquid is ac 633 nm liquid 459 nm n Model: applies. Sole: (a) The requency must remain the same since the harmonic oscillators in one medium excite the oscillators in the second medium. So water 440 Hz. (b) The table in the chapter gies water 140 m/s. 140 m/s 3.4 m 440 Hz Assess: This is a reasonable waelength or a sound wae Sole: The dierence in the arrial times or the P and S waes is d d t ts tp 10 s d d 1310 m 130 km S P 4500 m/s 000 m/s Assess: d is approximately one-ith o the radius o the earth and is reasonable Model: This is a sinusoidal wae. Sole: (a) The equation is o the orm D( y, t) Asin( ky t 0), so the wae is traeling along the y-axis. Because it is t rather than t the wae is traeling in the negatie y-direction. (b) Sound is a longitudinal wae, meaning that the medium is displaced parallel to the direction o trael. So the air molecules are oscillating back and orth along the y-axis. 1 (c) The wae number is k 96 m, so the waelength is m k 1 96 m The angular requency is 3140 s, so the wae s requency is 3140 s 500 Hz Thus, the wae speed (0.70 m)(500 Hz) 350 m/s. The period T 1/ s.00 ms. Assess: The wae is a sound wae with speed 350 m/s. This is greater than the room-temperature speed o 343 m/s, so the air temperature must be greater than 0 1 Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

15 Traeling Waes Model: This is a sinusoidal wae. Sole: (a) The displacement o a wae traeling in the positie x-direction with wae speed must be o the orm D(x, t) D(x t). Since the ariables x and t in the gien wae equation appear together as x t, the wae is traeling toward the let, that is, in the x direction. (b) The speed o the wae is /00 s 1 m/s k rad/4 m The requency is rad/00 s 5 0 Hz The wae number is rad k 6 rad/m 4 m (c) The displacement is 00 m 050 s D(00 m, 050 s) (30 cm)sin 1 15 cm 4 m 00 s Model: This is a sinusoidal wae traeling on a stretched string in the x direction. Sole: (a) From the displacement equation o the wae, A.0 cm, k 1.57 rad/m, and 63 rad/s. Using the equation or the wae speed in a stretched string, TS rad/s string S string T ( kg/m ) 16 N k 157 rad/m (b) The maximum displacement is the amplitude D max ( x, t ) 00 cm. (c) From Equation 0.17, y max A (63 rad/s)(0 10 m) 1 m/s Sole: The wae number and requency are calculated as ollows: rad k 4 rad/m k (40 m/s)(4 rad/m) 16 rad/s 050 m Thus, the displacement equation or the wae is D( y, t) (50 cm)sin[(4 rad/m) y (16 rad/s) t] Assess: The positie sign in the sine unction s argument indicates motion along the y direction Sole: The angular requency and wae number are calculated as ollows: 400 rad/s (00 Hz) 400 rad/s k rad/m 400 m/s The displacement equation or the wae is D( x, t) (0010 mm)sin[( rad/m) x (400 rad/s) t 1 rad] Assess: Note the negatie sign with t in the sine unction s argument. This indicates motion along the x direction Model: We hae a sinusoidal traeling wae on a stretched string. Sole: (a) The wae speed on a string and the waelength are calculated as ollows: TS 0 N 1000 m/s 100 m/s 10 m 000 kg/m 100 Hz (b) The amplitude is determined by the oscillator at the end o the string and is A 1.0 mm. The phase constant can be obtained rom Equation 0.15 as ollows: D(0 m, 0 s) Asin0 10 mm (10 mm)sin 0 0 rad Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

16 0-16 Chapter 0 (c) The wae (as distinct rom the oscillator) is described by D( x, t) Asin( kx t 0). In this equation the wae number and angular requency are k rad/m k (100 0 m/s)( rad/m) 00 rad/s 10 m Thus, the wae s displacement equation is D( x, t) (10 mm)sin[( rad/m) x (00 rad/s) t 1 rad] (d) The displacement is D(050 m, s) (1.0 mm)sin[( rad/s)(0.50 m) (00 rad/s)(0.015 s) 1] 1.0 mm Model: We hae a wae traeling to the right on a string. Visualize: Sole: The snapshot o the wae as it traels to the right or an ininitesimally small time t shows that the elocity at point 1 is downward, at point 3 is upward, and at point is zero. Furthermore, the speed at points 1 and 3 is the maximum speed gien by Equation 0.17: 1 3 A. The requency o the wae is (45 m/s) 300 rad/s A (300 rad/s)(0 10 m) 19 m/s 030 m Thus, 1 19 m/s, 0 m/s, and 3 19 m/s Model: The wae pulse is a traeling wae on a stretched string. The two masses hanging rom the steel wire are in static equilibrium. Visualize: Sole: The wae speed along the wire is Using Equation 0., wire 40 m m/s 004 s m/s T T 0 4 N (0060 kg/0 m) 1 1 wire T1 Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

17 Traeling Waes 0-17 Because point 1 is in static equilibrium, with Fnet 0, T1 ( Fnet ) x T1 T cos 40 T 71 N cos40 (71 N)sin 40 ( Fnet ) y T sin 40 w 0 N w mg T sin 40 m 17 kg 9 m/s Sole: The wae speeds along the two metal wires are T 50 N T 50 N 500 m/s 300 m/s kg/m 005 kg/m 1 1 The waelengths along the two wires are m/s m/s 1 1 m m 1500 Hz Hz 5 Thus, the number o waelengths oer two sections o the wire are 10 m 10 m 10 m 10 m m 1 m The number o complete cycles o the wae in the.00-m-long wire is Model: The object is in static equilibrium. is the linear density o the string, not a coeicient o riction. Visualize: Use tilted axes with the x-direction along the string. The tension in the string is T S. Sole: From a ree-body diagram we see that F = T Mg sin 0 T Mg sin x S TS S Mg sin Model: The wae is traeling on a stretched string. Sole: The wae speed on the string is TS 50 N 100 m/s 0005 kg/m The speed o the particle on the string, howeer, is gien by Equation The maximum speed is calculated as ollows: 100 m/s y Acos( kx t 0 ) y max A A A (0030 m) 94 m/s 0 m 0.6. Model: A sinusoidal wae is traeling along a stretched string. Sole: From Equation 0.17 and Equation 0.0, y max y max A and a A. These two equations can be combined to gie a y max 00 m/s y max 0 m/s 100 rad/s 159 Hz 16 Hz A 0 cm 0 m/s 100 rad/s y max Sole: (a) At a distance r rom the bulb, the 5 watts o isible light hae spread out to coer the surace o a sphere o radius r. The surace area o a sphere is a a 4 r. Thus, the intensity at a distance o m is P P 50 W I 0095 W/m a 4r 4 (0 m) Note that the presence o the wall has nothing to do with the intensity. The wall allows you to see the light, but the light wae has the same intensity at all points m rom the bulb, whether it is striking a surace or moing through empty space. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

18 0-1 Chapter 0 (b) Unlike the light rom a light bulb, a laser beam does not spread out. We ignore the small diraction spread o the beam. The laser beam creates a dot o light on the wall that is mm in diameter. The ull 5 watts o light is 6 concentrated in this dot o area a r (0001 m) m. The intensity is P 5 W I 16 MW/m a m Although the power o the light source is the same in both cases, the laser produces light on the wall whose intensity is oer 16 million times that o the light bulb Model: The radio wae is an electromagnetic wae. Sole: At a distance r, the 5 kw power station spreads out waes to coer the surace o a sphere o radius r. The surace area o a sphere is 4 r. Thus, the intensity o the radio waes is source W 3 P I 010 W/m 4r 4 (10 10 m) Visualize: To ind the power o a laser pulse, we need the energy it contains, U, and the time duration o the pulse, t Then to ind the intensity, we need the area o the pulse. Its radius is 050 mm Sole: (a) Using P U/Δ t, we ind the ollowing: (b) Then rom I P/, a we obtain P (1 010 J)/(15 10 s) W ( W) I (50 10 m) W/m Model: We hae a traeling wae radiated by the tornado siren. Sole: (a) The power o the source is calculated as ollows: Psource Psource I50m 010 W/m P source (010 W/m )4 (50 m) (1000 ) W 4r 4 (50 m) The intensity at 1000 m is Psource (1000 ) W I1000 m 50 W/m 4 (1000 m) 4 (1000 m) (b) The maximum distance is calculated as ollows: Psource 6 (1000 ) W I 1010 W/m r 16 km 4r 4r Model: Assume the saw is ar enough o the ground that we don t hae to worry about relected sound. Visualize: First note that 1 0dB I1/ I (a change o 10 db corresponds to a change in intensity by a actor o 10). Then use I1A1 Sole: Put all o the aboe together. P and then P IA A P/ I, and inally sole or R A /4. I A P (4 R ) 1 I I I I 1 A R R R 100 (50 m)(10) 50 m I Assess: The scaling laws help and the answer is reasonable. I 0.6. Model: Assume the two loudspeakers broadcast the same power and that the platorms are high enough o the ground that we don t hae to worry about relected sound. Visualize: Call the distance between the loudspeakers d. Call the intensity halway between the speakers (at d/) I 1 and the sound intensity leel there 1 ( 75 db ); call them I and at 1/4 the distance rom one pole and 3/4 the distance rom the other pole on the line between them. We seek. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

19 Traeling Waes 0-19 We irst apply a general approach or dierent sound intensity leels: I I1 I/ I0 I 1 (10dB) log10 log 10 (10 db)log 10 (10 db)log10 I0 I0 I1/ I0 I1 Sole: Recall that or the general case o spherical symmetry I P/ A, where P is the power emitted by the source and A 4 R is the area o the sphere. Now we ind the ratio o the intensities I/ I 1 and then plug it in the ormula aboe and add it to 75 db. P P P I1 4 ( d/ ) 4 ( d/) d I P P 4P 4 P (36 4) P 40P 0 I 4 ( d/4) 4 (3 d/4) d 9 d 9 d 9 d 9 1 I 0 (10 db)log 10 (10 db)log10 34 db I db 34 db 7 db Assess: An increase o about 3dB corresponds to a doubling o the intensity. 0/9 is close to double Sole: I we sole the equation or I, we hae: 6 I I 0 10 ( /10 db) Now plugging in 60 db or, we get I 10 W/m and plugging in 61 db or, we get ratio o the latter to the ormer is 13 6 I 1310 W/m The Visualize: Take the log 10 o both sides o I cpsource r x. log I = log( cp ) log( r ) = log( cp ) xlog r source x source Sole: Now use the equation or sound leel intensity. I (10dB)log (10dB)[log I log I0] I0 Insert the expression or log I rom aboe. 10dB log( cp ) xlog r log I 1 Diide both sides by 10 db and note log10 1. Group constant terms. This equation leads us to beliee that a graph o 10 db whose intercept is log( cpsource ) 1. source 0 = log( cpsource ) xlog r ( 1) 10 db = xlog r (log( cpsource ) 1) 10 db s. log r would produce a straight line whose slope is x and Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

20 0-0 Chapter 0 The spreadsheet shows that the linear it is excellent and that the slope is 1.5. So we conclude that x 1.5. Assess: We expected a number less than. Also, the intercept o 10 tells us that log( cp ) Model: As suggested, model the bald head as a hemisphere with radius R 0 00m. This means the surace area o the bald head (hemisphere) is A R 0 040m. source /10dB Visualize: We are gien 93 db and E 0 10 J. We also know that I I 0 10 and P IA. Also recall P E t. Sole: Put all o the aboe together to ind t. E E E 010 J t 150 s 1 min P IA /10dB 1 93 ( I 10 )( R ) (10 W/m 10 )(0040 m ) 0 Assess: 1 min seems like quite a while to delier 0.10 J o energy, but sounds waes don t carry a lot o energy unless the intensity is high Model: The sound generator s requency is altered by the Doppler eect. The requency increases as the generator approaches the student, and it decreases as the generator recedes rom the student. Sole: The generator s speed is 100 S r r( ) (10 m) re/s 1047 m/s 60 The requency o the approaching generator is Hz Hz 60 Hz m/s S/ m/s Doppler eect or the receding generator, on the other hand, is Hz 5 Hz 50 Hz m/s S/ m/s Thus, the highest and the lowest requencies heard by the student are 60 Hz and 50 Hz Sole: We will closely ollow the details o section 0.7 in the textbook. Figure 0.9 shows that the wae 1 crests are stretched out behind the source. The waelength detected by Pablo is d where d is the distance the wae has moed plus the distance the source has moed at time t 3T. These distances are xwae t 3T and x t 3 T. The waelength o the wae emitted by a receding source is thus source S S d xwae xsource 3T 3ST ( S) T The requency detected in Pablo s direction is thus 0 ( ) T 1 / Model: We are looking at the Doppler eect or the light o an approaching source. Sole: (a) The time is 5410 km t 5 10 s 3 0 min 310 km/s (b) Using Equation 0.40, the obsered waelength is s 6 1 s/ c 10.1 c/ c 0 (540 nm) (0.9045)(540 nm) 4 nm 490 nm 1 / c 10.1 c/ c Assess: 490 nm is slightly blue shited rom green. s S 3, Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

21 Traeling Waes Model: The Doppler eect or light o a receding source yields an increased waelength. Sole: Because the measured waelengths are 0.5% longer, that is, , the distant galaxy is receding away rom the earth. Using Equation 0.40, 1 / c 1 / c 1005 (1005) m/s s s s c 1 s/ c 1 s/ c Model: The Doppler eect or light o an approaching source leads to a decreased waelength. Sole: The red waelength ( nm) is Doppler shited to green ( 540 nm) due to the approaching light source. In relatiity theory, the distinction between the motion o the source and the motion o the obserer disappears. What matters is the relatie approaching or receding motion between the source and the obserer. Thus, we can use Equation 0.40 as ollows: 1 s/ c 1 s/ c nm (650 nm) 1 / c 1 / c The ine will be s s km/s 010 km/h 1 $ (010 km/h 50 km/h) $00 million 1 km/h Assess: The police oicer knew his physics Sole: The time or the wae to trael rom Caliornia to the South Paciic is d m t s 140 m/s d A time decrease to s implies the speed has changed to 140 m/s t Since the 4.0 m/s increase in elocity is due to an increase o 1C, an increase o 0. m/s occurs due to a temperature increase o 1C (0 m/s) 007C 40 m/s Thus, a temperature increase o approximately 0.07C can be detected by the researchers Model: As the guitar string is stretched its linear density will decrease, but only slightly, so we will assume that 1.3g/m will apply through the problem. Visualize: The act that we hae a length L, a tension orce, and are looking or a change in length suggests this is a Young s modulus problem. T S Sole: First sole or the tension orce rom =. This tension orce becomes the F in the Young s modulus equation. From the chapter on elasticity we hae L = LF YA 10 where Y 0 10 N/m is Young s modulus or steel rom Table 15.3, F T and A R is the crosssectional area o the string. 6 s S, LF L( ) (0.75 m)(50 m/s) ( kg/m) L = = = =1. mm YA 10 Y( R ) (0 10 N/m ) ( m) Assess: That 196 Hz when the string is properly tuned is correct but irreleant inormation. That requency, howeer, illustrates that only 65 cm o the string is really ibrating, the other 10 cm are wrapped around the tuning screw. Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

22 0- Chapter Model: The wae pulse is a traeling wae on a stretched wire. Visualize: Sole: (a) At a distance y aboe the lower end o the rope, the point P is in static equilibrium. The upward tension in the rope must balance the weight o the rope that hangs below this point. Thus, at this point T w Mg ( y) g where m/l is the linear density o the entire rope. Using Equation 0., we get T yg gy (b) The time to trael a distance dy at y, where the wae speed is gy, is dy dy dt gy Finding the time or a pulse to trael the length o the rope requires integrating rom one end o the rope to the other: 0.0. Visualize: T L dy 1 L t dt y L t Ñ Ñ 0 0 gy g 0 g L g Sole: (a) Using the graph, the reractie index n as a unction o distance x can be mathematically expressed as n n1 n n1 x L At position x, the light speed is c/. n The time or the light to trael a distance dx at x is dx n 1 n n1 dt dx n1 xdx c c L Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

23 Traeling Waes 0-3 To ind the total time or the light to coer a thickness L o a glass we integrate as ollows: T L L L 1 n n1 n1 ( n n1 ) n1 n n1 L n1 n Ñ 1 c Ñ L c Ñ cl Ñ c cl c T dt n xdx dx x dx L L (b) Substituting the gien alues into this equation, ( ) 11 T 0010 m s (30 10 m/s) 0.1. Model: Use the shallow wae equation gd. Visualize: Say the wae is moing to the right rom an initial depth o 5.0 m at 100 m rom shore to a depth o 0.0 m at the shore. Sole: We want T = dt where dt dx/ dx/ gd. 100 dx 100 dx dx T dt 0 gd 0 5 5g x g (100 x ) x s 5g 0 5g Assess: 9 s seems like a reasonable time or a shallow wae to trael 100 m. 0.. Model: Assume 0. Visualize: Apply the Doppler eect twice, irst with the obserer (the object) approaching the source, and second with the source (the object now relecting the sound) approaching the obserer. Sole: (a) With the object (obserer) approaching we hae Plug that as the new 0 into the equation or the receding source (now the object) = The shit is now = 0 = (b) Bernoulli s equation or luid dynamics is p1 1 gy1 = p gy. In this case 1 y y, and the speed o the blood in the heart is 1 0, so 1 p= Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.

24 0-4 Chapter 0 The blood in this part is the object in part (a), so 0 there is now =, where 0 =.5 MHz, = 6000 Hz, and the speed o sound in blood is the same as the speed o sound in water: =140 m/s Hz p = = (1060 kg/m ) (140 m/s) = 167 Pa 0 (.5 MHz) Conert to mm o Hg. 167 Pa =1.54 mm o Hg 13 mm o Hg Assess: 13 mm o Hg seems like a reasonable pressure dierence. 0 Copyright 013 Pearson Education, Inc. All rights resered. This material is protected under all copyright laws as they currently exist.