Nov. 8, 2010
PHYS 2101
HW#11
WileyPlus Problem Solutions
1
HW #11 Solutions – or at least most of them…
(USING BOOK VALUES 9
th
Edition in problems; problem # indicated by [])
1) [16CQ4]
2)
[163] . (a) The angular wave number is
1
2
2
3.49m .
1.80m
k
−
π
π
=
=
=
λ
(b) The speed of the wave is
(
)
(
)
1.80m
110rad s
31.5m s.
2
2
v
f
ω
λ
=
λ
=
=
=
π
π
3) [165] (a) The motion from maximum displacement to zero is onefourth of a cycle. Onefourth of a
period is 0.170 s, so the period is
T
= 4(0.170 s) = 0.680 s.
(b) The frequency is the reciprocal of the period:
1
1
1.47Hz.
0.680s
f
T
=
=
=
(c) A sinusoidal wave travels one wavelength in one period:
1.40m
2.06m s.
0.680s
v
T
=
=
=
λ
4) [169] (a) The amplitude
y
m
is half of the 6.00 mm vertical range shown in the figure, that is,
3.0 mm.
m
y
=
(b) The speed of the wave is
v = d/t =
15 m/s, where
d
= 0.060 m and
t
= 0.0040 s.
The angular wave
number is
k =
2
π
/
λ
where
λ
= 0.40 m.
Thus,
k
=
2
π
λ
=
16 rad/m .
(c) The angular frequency is found from
ω
=
k
v
= (16 rad/m)(15 m/s) = 2.4
×
10
2
rad/s.
(d) We choose the minus sign (between
kx
and
ω
t
) in the argument of the sine function because the wave
is shown traveling to the right (
in the +
x
direction,
see Section 165).
Therefore, with SI units understood,
we obtain
y
=
y
m
sin(
kx
−
kvt
)
≈
0.0030 sin(16
x
−
2.4
×
10
2
t
) .
5) [1610] (a) The amplitude is
y
m
= 6.0 cm.
(b) We find
λ
from 2
π
/
λ
= 0.020
π
:
λ
= 1.0
×
10
2
cm.
(c) Solving 2
π
f
=
ω
= 4.0
π
, we obtain
f
= 2.0 Hz.
(d) The wave speed is
v
=
λ
f
= (100 cm) (2.0 Hz) = 2.0
×
10
2
cm/s.
(e) The wave propagates in the –
x
direction, since the argument of the trig function is
kx
+
ω
t
instead of
kx
–
ω
t
(as in Eq. 162).
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