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25.
(a) The displacement of the string is assumed to have the form
y
(
x, t
)=
y
m
sin(
kx
−
ωt
). The velocity of
a point on the string is
u
(
x, t
)=
∂y/∂t
=
−
ωy
m
cos(
kx
−
ωt
) and its maximum value is
u
m
=
ωy
m
.
For this wave the frequency is
f
= 120 Hz and the angular frequency is
ω
=2
πf
=2
π
(120 Hz) =
754 rad
/
s. Since the bar moves through a distance of 1
.
00 cm, the amplitude is half of that, or
y
m
=5
.
00
×
10
−
3
m. The maximum speed is
u
m
= (754 rad
/
s)(5
.
00
×
10
−
3
m) = 3
.
77 m
/
s.
(b) Consider the string at coordinate
x
and at time
t
and suppose it makes the angle
θ
with the
x
axis.
The tension is along the string and makes the same angle with the
x
axis. Its transverse component
is
τ
trans
=
τ
sin
θ
.Now
θ
is given by tan
θ
=
∂y/∂x
=
ky
m
cos(
kx
−
ωt
) and its maximum value is
given by tan
θ
m
=
ky
m
. We must calculate the angular wave number
k
.I
ti
sg
i
v
e
nb
y
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 Fall '08
 SPRUNGER
 Physics

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