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P16.2 (a) See ANS FIG P16.2(a); (b) See ANS FIG P16.2(b); (c) The graph (b) has the same amplitude and wavelength as graph (a). It differs just by being shifted toward larger x by 2.40 m; (d) The wave has traveled d = vt = 2.40 m to the right. P16.4 (a) longitudinal P wave; (b) 666 s P16.6 (a) See ANS FIG P16.6(a); (b) See ANS FIG P16.6(b); (c) See ANS FIG P16.6(c); (d) See ANS FIG P16.6(d); (e) See ANS FIG P16.6(e) P16.8 0.800 m/s P16.10 2.40 m/s P16.12 6.67 cm P16.14 (a) See ANS FIG P16.14; (b) 0.125 s; (c) This agrees with the period found in the example in the text. P16.16 (a) 0.100 sin (1.002–20.0t); (b) 3.18 Hz P16.18 (a) 0.0215 m; (b) 1.95 rad; (c) 5.41 m/s; (d) y(x, t) = (0.0215)sin (8.38x+80.0+1.95) P16.20 (a) See ANS FIG P16.20(a); (b) 18.0 rad/m; (c) 0.0833 s; (d) 75.4 rad/s; (e) 4.20 m/s; (f)y 0.200 m sin 18.0x/m 75.4t/s ; (g) y(x, t)=0.200 sin (18.0x+75.4t0.151), where x and y are in meters and t is in seconds. P16.22 520 m/s P16.24 (a) units are newtons; (b) The first T is period of time; the second is force of tension. P16.26 (a) y=(2.00104) sin (16.0x 3140t)where y and x are in meters and t is in seconds; (b) 158 N P16.28 The calculated gravitational acceleration of the Moon is almost twice that of the accepted value. P16.30 (a) v (30.4) m where v is in meters per second and m is in kilograms; (b) m = 3.89 kg P16.32 1.07 kW P16.34 55.1 Hz P16.36 (a) As for a string wave, the rate of energy transfer is proportional to the square of the amplitude to the speed. The rate of energy transfer stays constant because each wavefront carries constant energy, and the frequency stays constant. As the speed drops, the amplitude must increase; (b) The amplitude increases by 5.00 times P16.38 P16.40 P16.42 See P16.38 for the full explanation. 2P0 (a) A = 40.0; (b) A = 7.00, B = 0, and C = 3.00; (c) In order for two vectors to be equal, they must have the same magnitude and the same direction in three-directional space. All of their components must be equal, so all coefficients of the unit vectors must be equal; (d) A = 0, B = 7.00, C = 3.00, D = 4.00, E = 2.00; (e) Identify corresponding parts. In order for two functions to be identically equal, corresponding parts must be identical. The argument of the sine function must have no units or be equal to units of radians. P16.44 1 (a) See P16.44(a) for full explanation; (b) f ( x vt ) ( x vt )2 and 2 1 g ( x vt ) ( x vt )2 ; 2 1 1 (c) f ( x vt ) sin( x vt ) and g ( x vt ) sin( x vt ) 2 2 P16.46 1 min P16.48 (a) 375 m/s2; (b) 0.0450 N; (c) 46.9 N. The maximum transverse force is very small compared to the tension, more than a thousand times smaller. 2Mg ; (c) k 2 Mg 2Mg L0 k k P16.50 (a) 2 Mg; (b) L0 P16.52 (a) The energy a wave crest carries is constant in the absence of absorption. Then the rate at which energy passes a stationary point, which is the power of the wave, is constant; (b) The power is proportional to the square of the amplitude and to the wave speed. The speed decreases as the wave moves into shallower water near shore, so the amplitude must increase; (c) 8.31 m; (d) As the water depth goes to zero, our model would predict zero speed and infinite amplitude. In fact, the amplitude must be finite as the wave comes ashore. As the speed decreases, the wavelength also decreases. When it becomes comparable to the water depth, or smaller, our formula gd for wave speed no longer applies. P16.54 2 (a) P 0.0500 v y ,max ; where P is in watts and vy,max is in meters per second; (b) The power is proportional to the square of the maximum particle speed; 4 2 (c) E 7.5 10 vy ,max where E is in joules and vy,max is in meters per second; (d) 1 2 2 mvy ,max ; (e) E 0.300vy ,max where E is in joules and vy,max is 2 in meters per second P16.56 8.43 × 10−3 s P16.58 2 P16.60 3.86 × 10−4 P16.62 This is greater than the average ocean depth, about 4280 m. P16.64 (a) t 2 P16.66 (a) ( x) L g L g M m M ; (b) 2 L ; (c) g mL Mg 2L ( L 0 ) x L3/ 2 03/ 2 0 ; (b) t 2 T L 0 L