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Physics 9BA
Solution Set #2
Cole
[Q15.13]
The intensity of the wave is the average power per area perpendicular to the direction the
wave is traveling.
For the surface water wave the perpenidular area is a segment of a right circular
cylinder
centered at the point the rock hit the water.
Area
º
= 2π
rh.
As wave expands, the radius
grows, this area grows, and the energy in the wave front must be spread more thinly causing the
amplitude to decrease as
I á 1/r
and
I
á
A
2
õ
A á
1/
r
.
[Q15.20]
As you stretch a rubber band the frequency goes up!
As a string (fixed both ends) is
made longer the frequency of the fundamental goes down.
f
v
L
1
2
=
.
For a rubberband, when it is
stretched, the tension goes up.
If we assume Hooke's law

F  = kx
, the frequency should still
decrease
f
kx
x
x
1
2
1
=
∝
/
μ
.
The solution is to realize that the rubber band is not perfectly elastic
but becomes stiffer as it is stretched:
F  = kx + cx
3
.
The
x
3
term is responsible for the frequency
going up.
[P15.26]
In 15 ms the wave travels
x = (
40 cm/s)(0.015 s) = 6 mm and 2mm each .005 s after.
a) The wave inverts at the reflection.
In (b) the wave does not invert.
[15.28]
In 1/4 second each wave will travel 0.5 cm.
After the first 1/4 s the two waves will be just touching.
After 1/2 s they will be overlapping by 1/2, after 3/4 s
they will be exactly on top of eachother which means the
amplitude will be 2 cm, after 1 s they will only half
overlap and after 1.25 s they will again be just touching
at they move away from eachother.
Phy 9B #2, 10/14/05
Page 1 of 4
UC Davis
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View Full Document[15.43]
Trick:
Put all the things that change on one side of the equation and all the constants on
the other.
a)
f
v
L
1
2
=
Here, the tension in the string is constant so
v
is constant.
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 Fall '07
 Cole
 Power

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