A maior rede de estudos do Brasil

Grátis
6 pág.
EXERCÍCIOS RESOLVIDOS - ONDULATÓRIA - FÍSICA

Pré-visualização | Página 1 de 1

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= =