P17_008.PDF
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Chapter 17 - 17.8
8. (a) The figure in the book makes it clear that the period is
T
= 10 s and the amplitude is
y
m
=4
.
0cm.
The phase constant
φ
is more subtly determined by that figure: what is shown is 4 sin
ωt
,yet what
follows from Eq. 17-2 (without the phase constant) should be 4 sin(
−
ωt
)at
x
= 0. Thus,we need
ωt
+
π
) = 4 sin(
ωt
)). Therefore,we use Eq. 17-2 (modified
by the inclusion of
φ
)with
k
=2
π/λ
=
π/
10 (in inverse centimeters) and
ω
=2
π/T
=
π/
5(in
inverse seconds). In the graph below we plot the equation for
t
= 0 over the range 0
−
≤
x
≤
20 cm,
making sure our calculator is in radians mode.
4
y_centimeters
2
0
2 4 6 8 10 12 14 16 18 20
x_centimeters
–2
–4
(b) Since the frequency is
f
=1
/T
=0
.
10 s,the speed of the wave is
v
=
fλ
=2
.
0cm/s.
(c) Using the observations made in part (a),Eq. 17-2 becomes
y
=4
.
0 sin
πx
10
−
πt
5
+
π
=
−
4
.
0 sin
πx
10
−
πt
5
where
y
and
x
are in centimeters and
t
is in seconds.
(d) Taking the derivative of
y
with respect to
t
,we find
u
=
∂y
∂t
=4
.
0
π
t
cos
πx
10
−
πt
5
which (evaluated at (
x, t
)=(0
,
5
.
0),making sure our calculator is in radians mode) yields
u
=
−
the phase constant
φ
=
π
since 4 sin(
2
.
5cm/s.
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