Chapter 16 Solutions

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1 Chapter 16 Solutions 16.1 Replace x by x vt = x 4.5t to get y = 6 [(x 4.5t) + 3] 16. y (c) y (c) y (c) t = s t = 1 s t = 1.5 s x x x y (c) y (c) 4 t =.5 s 4 t = 3 s x x e (x + 5t) is of the for f(x + vt) so it describes a wave oving to the left at v = 5.00 /s 16.4 y () x () y () x ()

2 Chapter 16 Solutions 16.5 (a) he longitudinal wave travels a shorter distance and is oving faster, so it will arrive at point B first. he wave that travels through the Earth ust travel a distance of R sin 30.0 = ( ) sin 30.0 = at a speed of 7800 /s. herefore, it takes = 817 s 7800 /s he wave that travels along the Earth's surface ust travel a distance of S = Rθ = R π 3 rad = at a speed of 4500 /s herefore, it takes = 148 s 4500 he tie difference is 665 s = 11.1 in. *16.6 he distance the waves have traveled is d = (7.80 k/s)t = (4.50 k/s)(t s) where t is the travel tie for the faster wave. hen, ( )(k/s)t = (4.50 k/s)(17.3 s) (4.50 k/s)(17.3 s) or t = ( )(k/s) = 3.6 s, and the distance is d = (7.80 k/s)(3.6 s) = 184 k 16.7 (a) φ 1 = (0.0 rad/c)(5.00 c) (3.0 rad/s)(.00 s) = 36.0 rad φ 1 = (5.0 rad/c)(5.00 c) (40.0 rad/s)(.00 s) = 45.0 rad φ = 9.00 radians = 516 = 156 φ = 0.0x 3.0t [5.0x 40.0t] = 5.00x t At t =.00 s, the requireent is φ = 5.00x (.00) = (n + 1)π for any integer n. For x < 3.0, 5.00x is positive, so we have 5.00x = (n + 1)π, or x = 3.0 (n + 1)π 5.00 he sallest positive value of x occurs for n = and is x = 3.0 (4 + 1)π 5.00 = 3.0 π = c

3 Chapter 16 Solutions y = y 1 + y = 3.00 cos (4.00x 1.60t) sin (5.00x.00t) evaluated at the given x values. (a) x = 1.00, t = 1.00 y = 3.00 cos (.40 rad) sin (+3.00 rad) = 1.65 x = 1.00, t = y = 3.00 cos (+3.0 rad) sin (+4.00 rad) = 6.0 (c) x = 0.500, t = 0 y = 3.00 cos (+.00 rad) sin (+.50 rad) = (a) y 1 = f(x vt), so wave 1 travels in the +x direction. y = f(x + vt), so wave travels in the x direction. o cancel, y 1 + y = 0: 5 (3x 4t) + = +5 (3x + 4t 6) + (3x 4t) = (3x + 4t 6) 3x 4t = ±(3x + 4t 6) +root 8t = 6 t = s (at t = s, the waves cancel everywhere) (c) root 6x = 6 x = 1.00 (at x = 1.00, the waves cancel always) *16.10 he down and back distance is = he speed is then v = d total t 4(8.00 ) = = 40.0 /s = /µ s 0.00 kg Now, µ = 4.00 = kg/, so = µv = ( kg/)(40.0 /s) = 80.0 N kg he ass per unit length is: µ = 5.00 = kg/ he required tension is: = µv = (0.010 kg/)(50.0 /s) = 30.0 N

4 4 Chapter 16 Solutions 16.1 v = µ = 1350 kg /s kg/ = 50 /s = Mg is the tension v = µ = Mg (/) = Mg = t is the wave speed hen, Mg = t and *16.14 v = g = Mt = 1.60 ( kg) 3.00 kg ( s) = 1.64 /s µ = µv = ρav = ρπ r v = (890 kg/ 3 )(π)( ) (00 /s) = 631 N Since µ is constant, µ = v = 1 v 1 and = v 1 = 30.0 /s (6.00 N) = 13.5 N v /s Goal Solution G: Since v F, the new tension ust be about twice as uch as the original to achieve a 50% increase in the wave speed. O: he equation for the speed of a transverse wave on a string under tension can be used if we assue that the linear density of the string is constant. hen the ratio of the two wave speeds can be used to find the new tension. A: he wave speeds can be written as: v 1 = F 1 µ and v = F µ Dividing, v v 1 = F F 1 so F = v F1 = 30.0 /s (6.00 N) = 13.5 N v /s : he new tension is slightly ore than twice the original, so the result agrees with our initial prediction and is therefore reasonable.

5 Chapter 16 Solutions he period of the pendulu is = π g et F represent the tension in the string (to avoid confusion with the period) when the pendulu is vertical and stationary. he speed of waves in the string is then: v = F µ = Mg (/) = Mg Since it ight be difficult to easure precisely, we eliinate = g π so v = Mg g π = g π If the tension in the wire is, the tensile stress is M Stress = /A so = A(stress) he speed of transverse waves in the wire is v = µ = A(Stress) = / Stress /A = Stress /(Volue) = Stress ρ where ρ is the density. he axiu velocity occurs when the stress is a axiu: v ax = Pa 7860 kg/3 = 586 /s Fro the free-body diagra, g = sin θ g = sin θ he angle θ is found fro cos θ = 3/8 / = 3 4 θ = 41.4 θ θ g (a) v = µ = g µ sin 41.4 = 9.80 /s ( kg/) sin 41.4 or v = 30.4 /s kg v = 60.0 = 30.4 and = 3.89 kg

6 6 Chapter 16 Solutions First, observe fro the geoetry shown in the figure that d + = D, or d = D 4 hus, cos θ = = 1 3 = = 0.50, and θ = 70.5 D /4 /4 θ A B θ d / d Now, consider a free body diagra of point A: F x = 0 becoes = 1 cos θ, and M 1 M F y = 0 becoes Mg = 1 sin θ θ A Dividing the second of these equations by the first gives: Mg = tan θ or = 19.6 N tan 70.5 = 6.94 N Mg he linear density of the string is: µ = = kg 3.00 = kg/ so the speed of transverse waves in the string between points A and B is: v = µ = 6.94 N kg/ = 45.6 /s he tie for the pulse to travel 1.50 fro A to B is: t = /s = s = 3.9 s 16.0 Refer to the diagras given in the solution for Proble 19 above. Fro the free-body diagra of point A: F y = 0 1 sin θ = Mg and F x = 0 1 cos θ = Cobining these equations to eliinate 1 gives the tension in the string connecting points A Mg and B as: = tan θ he speed of transverse waves in this segent of string is then v = µ = Mg/tan θ = / Mg tan θ and the tie for a pulse to travel fro A to B is t = / v = tan θ 4Mg

7 Chapter 16 Solutions 7 o evaluate tan θ, refer to the geoetry shown in the first diagra: tan θ = (/4) d d = 1 4d Also, d = D /, which gives 4d = D hus, tan θ = 1 D he travel tie for the pulse going fro A to B is then t = tan θ where tan θ = 4Mg 1 D 16.1 he total tie is the su of the two ties. In each wire t = v = µ where µ = ρa = πρd 4 hus, t = πρd 4 1/ For copper, t = (0.0) (π)(890)( ) 1/ = s (4)(150) For steel, t = (30.0) (π)(7860)( ) 1/ = 0.19 s (4)(150) he total tie is = 0.39 s *16. (a) If the end is fixed, there is inversion of the pulse upon reflection. hus, when they eet, they cancel and the aplitude is zero. If the end is free, there is no inversion on reflection. When they eet, the aplitude is A = (0.150 ) = (a) See figure at right. y (c) = π ω = π 50.3 = 0.15 s t (s)

8 8 Chapter 16 Solutions 16.4 Using data fro the observations, we have λ = 1.0 and f = s. herefore, v = λf = (1.0 ) s = /s 16.5 f = 40.0 vibrations 30.0 s = 4 3 Hz 45 c v = 10.0 s = 4.5 c/s λ = v f 4.5 c/s = = 31.9 c = Hz *16.6 At tie t, the phase of y = (15.0 c) cos(0.157x 50.3t) at coordinate x is φ = (0.157 rad/c)x (50.3 rad/s)t. Since 60.0 = π rad, the requireent for point B is that 3 φ B = φ A ± π 3 rad, or (since x A = 0), (0.157 rad/c)x B (50.3 rad/s)t = 0 (50.3 rad/s)t ± π 3 rad ±π rad his reduces to x B = = ±6.67 c 3(0.157 rad/c) 16.7 v = fλ = (4.00 Hz)(60.0 c) = 40 c/s =.40 /s 16.8 (a) y () 0. t = x () 0.1 k = π λ = π = 18.0 rad/ = 1 f = 1 1.0/s = s ω = π f = π 1.0/s = 75.4 rad/s v = f λ = (1.0/s)(0.350 ) = 4.0 /s

9 Chapter 16 Solutions 9 (c) y = A sin (kx + ωt + φ) specializes to y = 0.00 sin (18.0 x/ t/s + φ) at x = 0, t = 0 we require = 0.00 sin (+φ) φ = 8.63 = rad so y(x, t) = (0.00 ) sin (18.0 x/ t/s rad) 16.9 y = 0.50 sin(0.300x 40.0t) Copare this with the general expression y = A sin (kx ωt) (a) A = 0.50 ω = 40.0 rad/s (c) k = rad/ (d) λ = π k = π rad/ = 0.9 (e) v = fλ = ω π λ = 40.0 rad/s (0.9 ) = 133 /s π (f) he wave oves to the right, in +x direction y = (0.10 ) sin π 8 x + 4π t (a) v = dy dt = (0.10)(4π) cos π 8 x + 4π t v(0.00 s, 1.60 ) = 1.51 /s a = dv dt = ( 0.10 )(4π) sin π 8 x + 4π t a = (0.00 s, 1.60 ) = 0 k = π 8 = π λ ω = 4π = π λ = 16.0 = s v = λ = s = 3.0 /s

10 10 Chapter 16 Solutions (a) A = y ax = 8.00 c = : k = π λ = π ( ) = ω = π f = π (3.00) = 6.00π rad/s herefore, y = A sin (kx + ω t) or y = (0.0800) sin (7.85x + 6π t) [where y(0, t) = 0] In general, y = sin (7.85x + 6π t + φ) Assuing y(x, 0) = 0 at x = 0.100, then we require that 0 = sin ( φ) or φ = herefore, y = sin (7.85x + 6π t 0.785) 16.3 (a) et us write the wave function as y(x, t) = A sin (kx + ω t + φ) y(0, 0) = A sin φ = dy = Aω cos φ =.00 /s dt 0, 0 Also, ω = π = π s = 80.0 π/s A = x i + (v i /ω) = (0.000 ) +.00 /s 80.0π/s A = A sin φ A cos φ = /80.0π =.51 = tan φ Your calculator's answer tan 1 (.51) = 1.19 rad has a negative sine and positive cosine, just the reverse of what is required. You ust look beyond your calculator to find φ = π 1.19 rad = 1.95 rad

11 Chapter 16 Solutions 11 (c) (d) v y,ax = Aω = (80.0π /s) = 5.41 /s λ = v x = (30.0 /s)(0.050 s) = k = π λ = π = 8.38/ ω = 80.0π /s y(x, t) = (0.015 ) sin (8.38x rad/ π t rad/s rad) (a) f = v λ = (1.00 /s).00 = Hz ω = π f = π (0.500/s) = 3.14 rad/s (c) k = π λ = π.00 = 3.14 rad/ y = A sin (kx ω t + φ) becoes y = (0.100 ) sin (3.14 x/ 3.14 t/s + 0) (d) For x = 0 the wave function requires y = (0.100 ) sin ( 3.14 t/s) (e) y = (0.100 ) sin (4.71 rad 3.14 t/s) (f) v y = y t = ( 3.14/s) cos (3.14 x/ 3.14 t/s) he cosine varies between +1 and 1, so v y /s y = (0.150 ) sin(3.10x 9.30t) SI units v = ω k = = 3.00 /s s = vt = 30.0 in positive x-direction

12 1 Chapter 16 Solutions y = (0.000 ) sin (.11x 3.6t) SI units A =.00 c k =.11 rad/ λ = π k =.98 ω = 3.6 rad/s f = ω = Hz π v = f λ = ω π π k = ω k = = 1.7 /s (a) ω = πf = π(500) = 3140 rad/s, k = ω/v = (3140)/(196) = 16.0 rad/ y = ( ) sin (16.0x 3140t) v = 196 /s = kg/ = 158 N (a) at x =.00, y = (0.100 ) sin(1.00 rad 0.0t) y = (0.100 ) sin(0.500x 0.0t) = A sin(kx ωt) so ω = 0.0 rad/s and f = ω = 3.18 Hz π f = v λ = = 60.0 Hz ω = π f = 10π rad/s = 1 µω A v = (10 π) 3.60 (0.100) (30.0) = 1.07 kw Suppose that no energy is absorbed or carried down into the water. hen a fixed aount of power is spread thinner farther away fro the source, spread over the circuference π r of an expanding circle. he power-per-width across the wave front π r is proportional to aplitude squared so aplitude is proportional to π r

13 Chapter 16 Solutions = constant; v = µ ; = 1 µω A (a) If is doubled, v reains constant and is constant. If A is doubled and ω is halved, ω A reains constant. (c) If λ and A are doubled, the product ω A A /λ reains constant, so reains constant. (d) If and λ are halved, then ω 1/λ is quadrupled, so is quadrupled. (Changing doesn't affect ) A = µ = kg/ = 300 W = 100 N herefore, v = µ = 50.0 /s = 1 µω A v ω = µa v = (300) ( )( ) (50.0) ω = 346 rad/s f = ω = 55.1 Hz π *16.4 µ = 30.0 g/ = kg/ λ = 1.50 f = 50.0 Hz ω = π f = 314 s 1 A = A = (a) y = A sin π λ x ω t y = ( ) sin (4.19x 314t) = 1 µω A v = 1 ( )(314) ( ) W = 65 W

14 14 Chapter 16 Solutions (a) v = fλ = ω π = ω π k k = 50.0 /s = 6.5 /s λ = π k = π = 7.85 (c) f = 50.0 π = 7.96 Hz (d) = 1 µω A v = 1 ( )(50.0) (0.150) (6.5) W = 1.1 W Originally, o = 1 µω A v o = 1 µω A µ o = 1 ω A µ he doubled string will have doubled ass-per-length. Presuing that we hold tension constant, it can carry power larger by ties. o = 1 ω A µ *16.45 (a) A = ( )4.00 yields A = 40.0 (c) In order for two vectors to be equal, they ust have the sae agnitude and the sae direction in three-diensional space. All of their coponents ust be equal. hus, 7.00i + 0j k = Ai + Bj + Ck requires A = 7.00, B = 0, and C = In order for two functions to be identically equal, they ust be equal for every value of every variable. hey ust have the sae graphs. In A + B cos(cx + Dt + E) = cos(3.00x t +.00), the equality of average values requires that A = 0. he equality of axiu values requires B = he equality for the wavelength or periodicity as a function of x requires C = 3.00 rad/. he equality of period requires D = 4.00 rad/s, and the equality of zero-crossings requires E =.00 rad.

15 Chapter 16 Solutions Equation 16.6, with v = /µ is y x = 1 y v t If y = e b(x vt), then y t = bveb(x vt) and y x = beb(x vt) y t = b v eb(x vt) and y x = b eb(x vt) herefore, y t = v y x, and eb(x vt) is a solution Fro Equation 16.5, µ y t = y x o verify that y = ln[b(x vt)] is a solution, find the first and second derivatives of y with respect to x and t and substitute into Equation 16.5: y t = [b(x vt)] 1 ( bv); y t = v (x vt) y x = [b(x vt)] 1 ; y t = 1 (x vt) Substituting into Equation 16.5 we have µ v (x vt) = 1 (x vt) But v = µ, therefore the given function is a solution (a) Fro y = x + v t, evaluate Does y t y x = x y t = v t = 1 v y t? y x = y t = v By substitution: equation. = 1 v v and this is true, so the wave function does satisfy the wave

16 16 Chapter 16 Solutions Note 1 (x + vt) + 1 (x vt) = 1 x + xvt + 1 v t + 1 x xvt + 1 v t = x + v t as required So f(x + vt) = 1 (x + vt) and g(x vt) = 1 (x vt) (c) y = sin x cos vt akes y x = cos x cos vt y x = sin x cos vt y t = v sin x sin vt y t = v sin x cos vt hen y x = 1 y v t becoes sin x cos vt = 1 v v sin x cos vt which is true as required. Note sin (x + vt) = sin x cos vt + cos x sin vt sin (x vt) = sin x cos vt cos x sin vt So sin x cos vt = f(x + vt) + g(x vt) with f(x + vt) = 1 sin (x + vt) and g(x vt) = 1 sin (x vt) *16.49 Assue a typical distance between adjacent people ~1. hen the wave speed is v = x t ~ 1 ~ 10 /s 0.1 s Model the stadiu as a circle with a radius of order 100. hen, the tie for one circuit around the stadiu is = πr v ~ π(10 ) 10 /s = 63 s ~ 1 in

17 Chapter 16 Solutions Copare the given wave function y = 4.00 sin (.00x 3.00t) c to the general for y = A sin (kx ωt) to find (a) aplitude A = 4.00 c = k = π λ =.00 c 1 and λ = π c = (c) ω = πf = 3.00 s 1 and f = Hz (d) = 1 f =.09 s (e) he inus sign indicates that the wave is traveling in the positive x-direction (a) et u = 10π t 3π x + π 4 du dt = 10π 3π dx dt = 0 dx dt = 10 3 = 3.33 /s he velocity is in the positive x-direction. y(0.100, 0) = (0.350 ) sin 0.300π + π 4 = = 5.48 c (c) k = π λ = 3π λ = ω = π f = 10π f = 5.00 Hz (d) v y = y t = (0.350)(10π) cos 10π t 3π x + π 4 v y, ax = (10π)(0.350) = 11.0 /s

18 18 Chapter 16 Solutions *16.5 he equation v = λf is a special case of speed = (cycle length)(repetition rate) hus, v = frae 4.0 fraes s = /s Assuing the incline to be frictionless and taking the positive x-direction to be up the incline: F x = Mg sin θ = 0 or the tension in the string is = Mg sin θ he speed of transverse waves in the string is then v = µ = Mg sin θ = / Mg sin θ and the tie for a pulse to travel the length of the string is t = v = Mg sin θ = Mg sin θ (a) v = µ = 80.0 N ( = 179 /s kg/.00 ) Fro Equation 16.1, = 1 µvω A and ω = π v λ = 1 µva πv λ π µa v 3 = λ = π ( kg/.00 )( ) (179 /s) 3 (0.160 ) = W = 17.7 kw Energy is conserved as the block oves down distance x: (K + U g + U s ) top + E = (K + U g + U s ) botto 0 + Mgx = kx x = Mg k (a) = kx = Mg = (.00 kg)(9.80 /s ) = 39. N

19 Chapter 16 Solutions 19 = 0 + x = 0 + Mg k = N 100 N/ = 0.89 (c) v = v = µ = 39. N kg v = 83.6 /s Mgx = 1 kx (a) = kx = Mg = 0 + x = 0 + Mg k (c) v = µ = = Mg 0 + Mg k v = µ and in this case = g ; therefore, = µv g Fro Equation 16.13, v = ω/k so that = µ g ω k (a) µ = d d 0.50 kg/ = 18π s /s 0.750π 1 = 14.7 kg = ρa dx dx = ρa v = /µ = /ρa = /[ρ(ax + b)] = /[ρ(10 3 x + 10 )c ] With all SI units, v = /[ρ(10 3 x + 10 )10 4 ] /s v x = 0 = 4.0/[(700)( )(10 4 )] = 94.3 /s v x = 10.0 = 4.0/[(700)( )(10 4 )] = 66.7 /s

20 0 Chapter 16 Solutions v = µ where = µxg, the weight of a length x, of rope. herefore, v = gx But v = dx dt dt = dx gx so that and t = 0 dx gx = g At distance x fro the botto, the tension is = (xg/) + Mg, so the wave speed is: hen v = /µ = / = xg + (Mg/) = dx dt ( a ) t = t dt = [xg + (Mg/)] 1/ dx 0 0 t = 1 g [xg + (Mg/)] 1/ 1 x = x = 0 t = g [(g + Mg/)1/ (Mg/) 1/ ] t = g + M M When M = 0, t = g 0 = g as in Proble 59.

21 Chapter 16 Solutions 1 (c) As 0 we expand to obtain + M = M (1 + /M) 1/ = M /M 1 8 /M +... t = g M + 1 / M 1 8 /M 3/ +... M t 1 g M = Mg (a) he speed in the lower half of a rope of length is the sae function of distance (fro the botto end) as the speed along the entire length of a rope of length (/). hus, the tie required = ' g with ' = and the tie required = g = g It takes the pulse ore that 70% of the total tie to cover 50% of the distance. By the sae reasoning applied in part (a), the distance clibed in τ is given by d = gτ 4 For τ = t = g, we find the distance clibed = 4 In half the total trip tie, the pulse has clibed 1 4 of the total length. Goal Solution G: he wave pulse travels faster as it goes up the rope because the tension higher in the rope is greater (to support the weight of the rope below it). herefore it should take ore than half the total tie t for the wave to travel halfway up the rope. ikewise, the pulse should travel less than halfway up the rope in tie t/. O: By using the tie relationship given in the proble and aking suitable substitutions, we can find the required tie and distance.

22 Chapter 16 Solutions A: (a) Fro the equation given, the tie for a pulse to travel any distance, d, up fro the botto of a rope is t d = d g. So the tie for a pulse to travel a distance / fro the botto is t / = g = g ikewise, the distance a pulse travels fro the botto of a rope in a tie t d is d = gt d 4. So the distance traveled by a pulse after a tie t d = /g is d = g(/g) 4 = 4 : As expected, it takes the pulse ore than 70% of the total tie to cover 50% of the distance. In half the total trip tie, the pulse has clibed only 1/4 of the total length (a) v = ω k = 15.0 = 5.00 /s in positive x-direction 3.00 v = 15.0 = 5.00 /s in negative x-direction 3.00 (c) v = 15.0 = 7.50 /s in negative x-direction.00 (d) v = 1.0 1/ = 4.0 /s in positive x-direction Young s odulus for the wire ay be written as Y = /A, where is the tension aintained in / the wire and is the elongation produced by this tension. Also, the ass density of the wire ay be expressed as ρ = µ A. he speed of transverse waves in the wire is then v = µ = /A µ/a = Y( /) ρ and the strain in the wire is = ρv Y If the wire is aluinu and v = 100 /s, the strain is = ( kg/ 3 )(100 /s) N/ =

23 Chapter 16 Solutions (a) For an increent of spring length dx and ass d, F = a becoes k dx = a d, or k (d/dx) = a But d dx = µ so a = k µ Also, a = dv dt = v t when v i = 0. But = vt, so a = v. Equating the two expressions for a, we have k µ = v or v = k µ Using the expression fro part (a) v = k µ = k = (100 N/)(.00 ) = 31.6 /s kg (a) v = (/µ) 1/ = ( 0 /µ o ) 1/ = v 0 where v 0 ( 0 /µ 0 ) 1/ v' = (/µ') 1/ = ( 0 /3µ 0 ) 1/ = v 0 /3 t left = / v = v 0 = t 0 = 0.354t 0 where t 0 v 0 t right = / v ' = v 0 /3 = t 0 /3 = 0.61t 0 t left + t right = 0.966t (a) (x) = 1 µω A v = 1 µω A 0 e bx ω k = µω k A 0 e bx (0) = µω k A 0 (c) (x) (0) = e bx v = 4450 k 9.50 h = 468 k/h = 130 /s

24 4 Chapter 16 Solutions d = v g = (130 /s) (9.80 /s ) = 1730

25 Chapter 16 Solutions 5

Physics 41 Homework #2 Chapter 16. fa. Here v is the speed of the wave. 16. The speed of a wave on a massless string would be infinite!

Physics 41 Homework #2 Chapter 16. fa. Here v is the speed of the wave. 16. The speed of a wave on a massless string would be infinite! Physics 41 Hoewor # Chapter 1 Serway 7 th Conceptual: Q: 3,, 8, 11, 1, Probles P: 1, 3, 5, 9, 1, 5, 31, 35, 38, 4, 5, 57 Conceptual 3. (i) d=e, f, c, b, a (ii) Since, the saller the (the coefficient of