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Unformatted text preview: 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 ( = = = = T f v to two figures. b) The amplitude is half the total vertical distance, m. 310 . c) The amplitude does not affect the wave speed; the new amplitude is m. 150 . 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 = Hz 10 5 . 1 m 001 . s m 1500 6 = = = v f 15.3: a) m. 0.439 Hz) (784 s) m 344 ( = = = f v b) Hz. 10 5.25 m) 10 (6.55 s) m 344 ( 6 5 = = = v f 15.4: Denoting the speed of light by c , , f c = 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) . 20 ( ) s m 344 ( min max = = = cm. 72 . 1 = b) mm. 74.0 Hz) 000 , 20 ( s) m 1480 ( m, 74.0 Hz) . 20 ( ) s m 1480 ( min max = = = = 15.6: Comparison with Eq. (15.4) gives a) mm, 50 . 6 cm, . 28 b) Hz 8 . 27 c) s 0360 . 1 1 = = = T f and from Eq. (15.1), d) s m 7.78 Hz) m)(27.8 280 . ( = = v , direction. e) x + 15.7: a) Hz, 25.0 m) 320 . ( s) m 00 . 8 ( = = = v f m. rad 19.6 m) 320 . ( ) 2 ( 2 s, 10 4.00 Hz) . 25 ( 1 1 2 = = = = = = k f T b) . m 320 . Hz) (25.0 2 cos m) 0700 . ( ) , ( + = x t t x y c) [ ] cm. 95 . 4 m)) (0.320 m) (0.360 Hz) s)(25.0 ((0.150 2 cos m) 0700 . ( = + d) The argument in the square brackets in the expression used in part (c) is ), 875 . 4 ( 2 and the displacement will next be zero when the argument is ; 10 the time is then s 0.1550 m)) (0.320 m) (0.360 Hz)(5 . 25 1 ( ) 5 ( = =  x T and the elapsed time is s. 02 . 2 e) s, 0050 . = T 15.8: a) b) 15.9: a) ) cos( ) sin( 2 2 2 t kx Ak x y t kx Ak x y + = + = ), cos( ) ( sin 2 2 2 t kx A t y t kx A t y + = + = and so ) , ( and , 2 2 2 2 2 2 t x y t y k x y = is a solution of Eq. (15.12) with . k v = b) ) sin( ) cos( 2 2 2 t kx Ak x y t kx Ak x y + = + + = ), ( sin ) cos( 2 2 2 t kx A t y t kx A t y + = + + = and so ) , ( and , 2 2 2 2 2 2 t x y t y k x y = is a solution of Eq. (15.12) with . k v = c) Both waves are moving in the xdirection, as explained in the discussion preceding Eq. (15.8). d) Taking derivatives yields . ) ( sin ) , ( and ) ( cos ) , ( 2 t kx A t x a t kx A t x v y y + = + = 15.10: a) The relevant expressions are ) cos( ) , ( t kx A t x y = ) ( sin t kx A t y v y = = ). ( cos 2 2 2 kx t A t v t y a y y = = = b) (Take A , k and to be positive. At , = 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 17.) (i) in the figure labeled 17....
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 Fall '07
 Nearing
 Physics

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