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Physics 41 Homework #2 Chapter 16 Sereway 8th Edition OQ: 3, 6 CQ: 6, 7, 8 P: 1, 4, 5, 6, 11, 16, 18, 20, 21, 29, 32, 42, 48, 53, 54 OQ: 3, 6 CQ: 6, 7, 8 P: 1, 4, 5, 6, 11, 16, 18, 20, 21, 29, 32, 42, 48, 53, 54 1. At t = 0, a transverse pulse in a wire is described by the function y = 6 x2 + 3 where x and y are in meters. Write the function y(x, t) that describes this pulse if it is traveling in the positive x direction with a speed of 4.50 m/s. P16.1 Replace x by 4.5x vt x t− = − to get ( )2 6 4.5 3 y x t = − + (see plots) 4.Two points A and B on the surface of the Earth are at the same longitude and 60.0° apart in latitude. Suppose that an earthquake at point A creates a P wave that reaches point B by traveling straight through the body of the Earth at a constant speed of 7.80 km/s. The earthquake also radiates a Rayleigh wave, which travels along the surface of the Earth in an analogous way to a surface wave on water, at 4.50 km/s. (a) Which of these two seismic waves arrives at B first? (b) What is the time difference between the arrivals of the two waves at B? Take the radius of the Earth to be 6 370 km. P16.4 (a) The longitudinal wave travels a shorter distance and is moving faster, so it will arrive at point B first. The transverse takes more time and carries more energy! (b) The wave that travels through the Earth must travel a distance of ( )6 62 sin 30.0 2 6.37 10 m sin 30.0 6.37 10 mR ° = × ° = × at a speed of 7 800 m/s Therefore, it takes 66.37 10 m 817 s 7 800 m s × = The wave that travels along the Earth’s surface must travel a distance of 6 rad 6.67 10 m 3 s R R πθ = = = × at a speed of 4 500 m/s Therefore, it takes 66.67 10 1 482 s 4 500 × = The time difference is 665 s 11.1 min= 5. (a) Let 10 3 4 u t x ππ π= − + 10 3 0du dx dt dt π π= − = at a point of constant phase 10 3.33 m s 3 dx dt = = The velocity is in the positive -directionx . (b) ( ) ( )0.100, 0 0.350 m sin 0.300 0.054 8 m 5.48 cm 4 y ππ = − + = − = − (c) 2 3k π π λ = = : 0.667 mλ = 2 10fω π π= = : 5.00 Hzf = (d) ( ) ( )0.350 10 cos 10 3 4y y v t x t π π π π ∂ = = − + ∂ ( ) ( ), max 10 0.350 11.0 m syv π= =

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