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
to get
(see plots)
y=
x − vt =x − 4.5t
6
( x − 4.5t ) 2 + 3


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
2R sin 30.0
=
° 2 6.37 × 106 m sin 30.0
=
° 6.37 × 106 m
at a speed of
7 800 m/s
Therefore, it takes
6.37 × 106 m
= 817 s
7 800 m s
The wave that travels along the Earth’s surface must travel
π

6
θ R  rad=
=
s R=
 6.67 × 10 m
3
a distance of
at a speed of
4 500 m/s
Therefore, it takes
6.67 × 106
= 1 482 s
4 500
The time difference is
5. (a)
Let u = 10π t − 3π x +
π
4
665 s = 11.1 min
du
dx
10π − 3π
0 at a point of constant phase
=
=
dt
dt
dx 10
= =
dt
3
3.33 m s
The velocity is in the positive x -direction .
(b)
(c)
(d)
π
−0.054 8 m =
−5.48 cm
y ( 0.100, 0) =
( 0.350 m ) sin  −0.300π +  =
4
k
=
2π
= 3π : λ = 0.667 m
λ
=
ω 2=
π f 10π : f = 5.00 H z
∂y
π

vy , max
vy = =( 0.350)( 10π ) cos  10π t − 3π x + =

∂t
4
10π )( 0.350)
(=
11.0 m s