Chapter 15

# Chapter 15 - 15.1 a The period is twice the time to go from...

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15.1: a) The period is twice the time to go from one extreme to the other, and s m 1.2 or s, m 20 . 1 s) (5.0 m) 00 . 6 ( = = λ = λ = T f v to two figures. b) The amplitude is half the total vertical distance, m. 310 . 0 c) The amplitude does not affect the wave speed; the new amplitude is m. 150 . 0 d) For the waves to exist, the water level cannot be level (horizontal), and the boat would tend to move along a wave toward the lower level, alternately in the direction of and opposed to the direction of the wave motion. 15.2: v f = λ Hz 10 5 . 1 m 001 . 0 s m 1500 6 × = = λ = v f 15.3: a) m. 0.439 Hz) (784 s) m 344 ( = = = λ f v b) Hz. 10 5.25 m) 10 (6.55 s) m 344 ( 6 5 × = × = λ = - v f 15.4: Denoting the speed of light by c , , f c = λ and a) m. 87 . 2 b) . m 556 Hz 10 5 . 104 s m 10 00 . 3 Hz 10 540 s m 10 00 3 6 8 3 8 = = × × × × . 15.5: a) Hz) 000 , 20 ( s) m 344 ( m, 17.2 Hz) 0 . 20 ( ) s m 344 ( min max = λ = = λ cm. 72 . 1 = b) mm. 74.0 Hz) 000 , 20 ( s) m 1480 ( m, 74.0 Hz) 0 . 20 ( ) s m 1480 ( min max = = λ = = λ 15.6: Comparison with Eq. (15.4) gives a) mm, 50 . 6 cm, 0 . 28 b) Hz 8 . 27 c) s 0360 . 0 1 1 = = = T f and from Eq. (15.1), d) s m 7.78 Hz) m)(27.8 280 . 0 ( = = v , direction. e) x +

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15.7: a) Hz, 25.0 m) 320 . 0 ( s) m 00 . 8 ( = = λ = v f m. rad 19.6 m) 320 . 0 ( ) 2 ( 2 s, 10 4.00 Hz) 0 . 25 ( 1 1 2 = = λ = × = = = - π π k f T b) . m 320 . 0 Hz) (25.0 2 cos m) 0700 . 0 ( ) , ( + = x t π t x y c) [ ] cm. 95 . 4 m)) (0.320 m) (0.360 Hz) s)(25.0 ((0.150 2 cos m) 0700 . 0 ( - = + π d) The argument in the square brackets in the expression used in part (c) is ), 875 . 4 ( 2 π and the displacement will next be zero when the argument is ; 10 π the time is then s 0.1550 m)) (0.320 m) (0.360 Hz)(5 0 . 25 1 ( ) 5 ( = - = λ - x T and the elapsed time is s. 02 . 0 2 e) s, 0050 . 0 = T 15.8: a) b)
15.9: a) ) cos( ) sin( 2 2 2 ωt kx Ak x y ωt kx Ak x y + - = + - = ), cos( ) ( sin 2 2 2 ωt kx t y ωt kx t y + - = + - = and so ) , ( and , 2 2 2 2 2 2 t x y t y ω k x y = is a solution of Eq. (15.12) with . k ω v = b) ) sin( ) cos( 2 2 2 ωt kx Ak x y ωt kx Ak x y + - = + + = ), ( sin ) cos( 2 2 2 ωt kx t y ωt kx t y + - = + + = and so ) , ( and , 2 2 2 2 2 2 t x y t y ω k x y = is a solution of Eq. (15.12) with . k ω v = c) Both waves are moving in the x - -direction, as explained in the discussion preceding Eq. (15.8). d) Taking derivatives yields . ) ( sin ) , ( and ) ( cos ) , ( 2 ωt kx A ω t x a ωt kx ωA t x v y y + - = + - =

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15.10: a) The relevant expressions are ) cos( ) , ( t ω kx A t x y - = ) ( sin ωt kx ωA t y v y - = = ). ( cos 2 2 2 kx ωt A ω t v t y a y y - - = = = b) (Take A , k and ω to be positive. At , 0 = t x the wave is represented by (19.7(a)); point (i) in the problem corresponds to the origin, and points (ii)-(vii) correspond to the points in the figure labeled 1-7.) (i)
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## This note was uploaded on 09/14/2009 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.

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Chapter 15 - 15.1 a The period is twice the time to go from...

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