Phys 2101
Homework 13 Solution Spring `11
These solutions use the parameter values from the problems printed in the book, not those that
appear in your personal homework assignment. The logic used to get to the answer is the same,
however.
1. (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:
11
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
2. (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
where
= 0.40 m.
Thus,
k
=
2
=
16 rad/m .
(c) The angular frequency is found from
=
k
v
= (16 rad/m)(15 m/s) = 2.4×10
2
rad/s.
(d) We choose the minus sign (between
kx
and
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
) .
3.
(a) The amplitude is
y
m
= 6.0 cm.
(b) We find
from 2
/
= 0.020
:
= 1.0×10
2
cm.
(c) Solving 2
f
=
= 4.0
, we obtain
f
= 2.0 Hz.
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View Full Document(d) The wave speed is
v
=
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
+
t
instead of
kx
–
t
(as in Eq. 162).
(f) The maximum transverse speed (found from the time derivative of
y
) is
1
max
2
4.0 s
6.0cm
75cm s.
m
u
fy
(g)
y
(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020
(3.5) + 4.0
(0.26)] =
–
2.0 cm.
4. (a) The wave speed is given by
v
=
/
T
=
/
k
, where
is the wavelength,
T
is the period,
is
the angular frequency (2
/
T
), and
k
is the angular wave number (2
/
). The displacement has the
form
y
=
y
m
sin(
kx
+
t
), so
k
= 2.0 m
–
1
and
= 30 rad/s. Thus
v
= (30 rad/s)/(2.0 m
–
1
) = 15 m/s.
(b) Since the wave speed is given by
v
=
, where
is the tension in the string and
is the
linear mass density of the string, the tension is
2
24
1.6 10 kg m 15m s
0.036N.
v
5.
The pulses have the same speed
v
. Suppose one pulse starts from the left end of the wire at time
t
= 0. Its coordinate at time
t
is
x
1
=
vt
. The other pulse starts from the right end, at
x
=
L
, where
L
is
the length of the wire, at time
t
= 30 ms. If this time is denoted by
t
0
, then the coordinate of this
wave at time
t
is
x
2
=
L
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 Spring '07
 GROUPTEST
 Work, 2 m, 2.0 m, 3 kg, 0.000845 kg, 8.40 m

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