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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= =