a period, except that the period is shifted by an amount of
kx
, and therefore the integral is
not changed because of it.
Therefore,
dk
dt
=
1
2
μv
ω
2
y
2
m
cos
2
(
kx

ω
t
)
=
1
4
μv
ω
2
y
2
m
We won’t show this here, but
dP
dt
=
dK
dt
so that
dE
dt
=
1
2
μv
ω
2
y
2
m
Exercise
: Find the above average by averaging over a wavelength (instead of averaging over
a period).
Change of linear mass density
Usually the frequency
f
is determined by the external force.
As the speed depends only
on the “hardware,” not on the frequency, then
λ
=
v
f
implies that increasing
f
results in a
smaller
λ
. Consider a string with two parts: one with linear mass density
μ
1
and the other
with
μ
2
.
How is the wave changing? Assume that the wave propagates initially in the medium
μ
1
.
It is the local
physics that determines the speed. Therefore,
v
1
=
τ
μ
1
,
v
2
=
τ
μ
2
. What
about the interface? Continuity of the string requires that the string moves the same way
on either side of the interface, i.e., the amplitude is unchanged (
f
is also the same from the
requirement of continuity at the interface; also, it is required by conservation of energy).
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 Spring '09
 LIORBURKO
 Physics, Fundamental physics concepts, linear mass density, ψi, ψR

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