10. Yes. Any function of (x - vt) will represent wave motion because it will satisfy the wave equation, Eq

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1 CHAPER 5: Wae Motion Responses to Questions 5. he speed of sound in air obeys the equation B. If the bulk modulus is approximately constant and the density of air decreases with temperature, then the speed of sound in air should increase with increasing temperature. 6. First, estimate the number of wae crests that pass a gien point per second. his is the frequency of the wae. hen, estimate the distance between two successie crests, which is the waelength. he product of the frequency and the waelength is the speed of the wae. 7. he speed of sound is defined as B, where B is the bulk modulus and ρ is the density of the material. he bulk modulus of most solids is at least 0 6 times as great as the bulk modulus of air. his difference oercomes the larger density of most solids, and accounts for the greater speed of sound in most solids than in air. 0. Yes. Any function of (x - t) will represent wae motion because it will satisfy the wae equation, Eq AM radio waes hae a much longer waelength than FM radio waes. How much waes bend, or diffract, around obstacles depends on the waelength of the wae in comparison to the size of the obstacle. A hill is much larger than the waelength of FM waes, and so there will be a shadow region behind the hill. Howeer, the hill is not large compared to the waelength of AM signals, so the AM radio waes will bend around the hill. Solutions to Problems 3. he elastic and bulk moduli are taken from able -. he densities are taken from able 3-. (a) For water: N m B kg m 400 m s (b) For granite: N m E kg m 400 m s (c) For steel: N m E kg m 500 m s

2 6. o find the time for a pulse to trael from one end of the cord to the other, the elocity of the pulse on the cord must be known. For a cord under tension, we hae Eq. 5-, F. x F x 8.0 m t t F 40 N 0.65kg 8.0 m 0.9s 8. he speed of the water wae is gien by B, where B is the bulk modulus of water, from able -, and is the density of sea water, from able 3-. he wae traels twice the depth of the ocean during the elapsed time. 9 l t t B.8s.0 0 N m l 3 3 t.050 kg m m. (a) he shape will not change. he wae will moe.0 meters to the right in.00 seconds. See the graph. he parts of the string that are moing up or down are indicated. centimeters Earlier 4 meters Later down up down up down

3 (b) At the instant shown, the string at point A will be moing down. As the wae moes to the right, the string at point A will moe down by cm in the time it takes the alley between m and m to moe to the right by about 0.5 m. y cm 4cm s t 0.5m.0 m s his answer will ary depending on the alues read from the graph.. We assume that the wae will be transerse. he speed is gien by Eq. 5-. he tension in the wire is equal to the weight of the hanging mass. he linear mass density is the olume mass density times the cross-sectional area of the wire. he olume mass density is found in able 3-. F m g m g V Al l l 5.0 kg9.80 m s kg m m ball ball 89 m s 5. From Eq. 5-7, if the speed, medium density, and frequency of the two waes are the same, then the intensity is proportional to the square of the amplitude. I I E E A A 3 A A 3.73 he more energetic wae has the larger amplitude. 6. (a) Assume that the earthquake waes spread out spherically from the source. Under those conditions, Eq. (5-8ab) applies, stating that intensity is inersely proportional to the square of the distance from the source of the wae.

4 I I 5km 45km 0. 45km 5km (b) he intensity is proportional to the square of the amplitude, and so the amplitude is inersely proportional to the distance from the source of the wae. A A 5km 45km km 5km. (a) he only difference is the direction of motion., 0.05sin5 00 D x t x t (b) he speed is found from the wae number and the angular frequency, Eq rad s k 5rad m 48m s 4. he traeling wae is gien by D x t 0.sin (a) he waelength is found from the coefficient of x. 5.6 m. m.m 5.6 m (b) he frequency is found from the coefficient of t. 34s 34s f f 5.4Hz 5.4 Hz (c) he elocity is the ratio of the coefficients of t and x. 34s f 6.07m s 6.m s 5.6 m Because both coefficients are positie, the elocity is in the negatie x direction.

5 (d) he amplitude is the coefficient of the sine function, and so is 0. m. (e) he particles on the cord moe in simple harmonic motion with the same frequency as the wae. From Chapter 4, D fd. max max 34s fd 0.m 7.5m s he minimum speed is when a particle is at a turning point of its motion, at which time the speed is 0. 0 min 6. he displacement of a point on the cord is gien by the wae, Dx t x t D elocity of a point on the cord is gien by. t D 0.60 m,0.0s 0. m sin 3.0m 0.60 m 5.0s 0.0s 0.m, 0.sin he D 0. m5.0s cos 3.0x 5.0t t D 0.60 m, 0.0s 0. m5.0s cos 3.0m 0.60 m 5.0s 0.0s 0.65m s t 8. (a) he waelength is the speed diided by the frequency. 345m s f 54 Hz m t (b) In general, the phase change in degrees due to a time difference is gien by. 360 t 90 ft t 360 f Hz s

6 x (c) In general, the phase change in degrees due to a position difference is gien by. 360 x x m m 9. he amplitude is 0.00 cm, the waelength is m, and the frequency is 54 Hz. he displacement is at its most negatie alue at x = 0, t = 0, and so the wae can be represented by a cosine that is phase shifted by half of a cycle., cos D x t A kx t f 54 Hz A k f 345m s 0.00cm ; 9.54 m ; 54 Hz 390 rad s D x, t 0.00 cm cos 9.54 m x 390 rad s t, x in m, t in s Other equialent expressions include the following., 0.00cmcos9.54 m 390 rad s D x t x t, 0.00cmsin 9.54 m 390 rad s 3 D x t x t 3. o be a solution of the wae equation, the function must satisfy Eq. 5-6, D D. x t (a) D Alnx t A A A x x t x t x t t D ; D ; D A ; D x t x t his gies D D, x t and so yes, the function is a solution. (b) D x t 4 D D D D 4 ; 4 ; x x t t 3 3 x t x t ; x t x t

7 D D his gies, x t and so yes, the function is a solution. 34. Find the arious deriaties for the linear combination.,,, D x t C D C D C f x t C f x t D f f D f f C C C C x x x x x x ; D f f D f f C C C C t t t t t t ; o satisfy the wae equation, we must hae satisfy the wae equation. D D. x t Use the fact that both f and f D f f f f f f D C C C C C C x x x t t t t t hus we see that D D, x t and so D satisfies the wae equation (a) For the wae in the lighter cord, D x, t 0.050m sin 7.5m x.0s t. k 7.5m 0.84 m (b) he tension is found from the elocity, using Eq. 5-. F F.0s 7.5m 0.0 kg m 0.6 N k (c) he tension and the frequency do not change from one section to the other. F m F k k k 7.5m

8 4. (a) he resultant wae is the algebraic sum of the two component waes. sin sin sin sin sin cos D D D A kx t A kx t A kx t A kx t A kx t kx t kx t kx t Asin kx t cos Acos sin kx t (b) he amplitude is the absolute alue of the coefficient of the sine function, Acos. he wae is purely sinusoidal because the dependence on x and t is sin kx t. (c) If 0,, 4, L, n, then the amplitude is n Acos Acos Acosn A A, which is constructie interference. If,3,5, L, n, then the amplitude is interference. n Acos Acos Acosn 0, which is destructie (d) If, then the resultant wae is as follows. D Acos sin kxt Acos sin kxt Asin kxt his wae has an amplitude of A, is traeling in the positie x direction, and is shifted to the left by an eighth of a cycle. his is halfway between the two original waes. he displacement is A at the origin at t = he oscillation corresponds to the fundamental. he frequency of that oscillation is f Hz. he bridge, with both ends fixed, is similar to a ibrating string, and so.5s 3

9 n f nf n Hz, n,,3 K. he periods are the reciprocals of the frequency, and so 3 n.5s, n,,3 K. n 48. Adjacent nodes are separated by a half-waelength, as examination of Figure 5-6 will show. 96 m s x node f f 445Hz 0.m 5. he speed of the wae is gien by Eq. 5-, F. he waelength of the fundamental is F l. hus the frequency of the fundamental is f. l Each harmonic is present in a ibrating string, and so n F,,,3, K l f nf n n. 54. he standing wae is gien by D x t.4cm sin 0.60 cos 4. (a) he distance between nodes is half of a waelength. d 5.36cm k 0.60cm 5. cm (b) he component waes trael in opposite directions. Each has the same frequency and speed, and each has half the amplitude of the standing wae. 4s A.4 cm. cm ; f Hz 6.7 Hz ; node f d f 5.36 cm Hz 70.0cm s 70 cm s sig. fig. D (c) he speed of a particle is gien by. t

10 D t t D t.4 cmsin 0.60x cos 4t 4 rad s.4 cmsin 0.60xsin 4t 3.0cm,.5s 4 rad s.4 cmsin 0.60cm 3.0cmsin4 rad s.5s 9cm s 55. (a) he gien wae is D x t 4.sin o produce a standing wae, we simply need to add a wae of the same characteristics but traeling in the opposite direction. his is the appropriate wae. D 4.sin 0.84x47t. (b) he standing wae is the sum of the two component waes. We use the trigonometric identity that sin sin sin cos. 4. sin sin x t x t cos 0.84x47t.0.84x47t. D D D x t x t 4. sin x t x t 8.4 sin cos sin cos 47 We note that the origin is NO a node. 60. (a) he maximum swing is twice the amplitude of the standing wae. hree loops is.5 waelengths, and the frequency is gien. f A 8.00cm 4.00cm ; 0 Hz 750 rad s ; 3 k ;.64 m.09 m ; k 5.75m.09 m D A sin kx cos t 4.00 cm sin 5.75 m x cos 750 rad s t (b) Each component wae has the same waelength, the same frequency, and half the amplitude of the standing wae.

11 .00 cm sin 5.75 m 750 rad s D x t.00 cm sin 5.75 m 750 rad s D x t 63. he standing wae is the sum of the two indiidual standing waes. We use the trigonometric identities for the cosine of a difference and a sum. ; cos cos cos sin sin cos cos cos sin sin D D D Acos kx t Acos kx t A cos kx t cos kx t A cos kxcos t sin kxsin t cos kxcos t sin kxsin t Acos kxcos t hus the standing wae is D Acoskxcos t. he nodes occur where the position term forces D Acoskxcos t 0 cos kx 0 kx n, n 0,,, L. hus, for all time. hus since k.0m, we hae x n m, n 0,,, L. 65. he speed in the second medium can be found from the law of refraction, Eq sin sin sin km s 5. km s sin sin sin he angle of refraction can be found from the law of refraction, Eq he relatie elocities can be found from the relationship gien in the problem. sin sin sin 33 sin sin sin sig. fig. 68. (a) Eq. 5-9 gies the relationship between the angles and the speed of sound in the two media. For total internal reflection (for no sound to enter the water), 90 or sin. he water water air is the incident media. hus the incident angle is gien by the following.

12 sin sin ; sin sin sin sin air air air air i air i water im water water water water r (b) diagram. From the angle of incidence, the distance is found. See the air sin sin 3.8 air M water 343m s 440 m s x tan x air M.8 m tan m.8m.8m x air M

Physics 207 Lecture 28

Physics 207 Lecture 28 Goals: Lecture 28 Chapter 20 Employ the wae model Visualize wae motion Analyze functions of two ariables Know the properties of sinusoidal waes, including waelength, wae number, phase, and frequency. Work