Mechanical Waves
15-23
15.70.
IDENTIFY:
The displacement of the string at any point is
SW
( , )
(
sin
)sin
.
yxt
A
k
x
t
ω
=
For the fundamental mode
2,
L
λ
=
so at the midpoint of the string
sin
sin(2
)(
2)
1,
kx
L
πλ
=
=
and
SW
sin
.
yA
t
=
The transverse velocity is
/
y
vy
t
=∂ ∂
and the transverse acceleration is
/
yy
av
t
=
∂∂
.
SET UP:
Taking derivatives gives
SW
cos
y
y
vA
t
t
∂
==
∂
, with maximum value
,max
SW
y
=
, and
2
SW
sin
y
y
v
aA
t
t
∂
−
∂
, with maximum value
2
, max
SW
y
=
.
EXECUTE:
32
3
,max
,max
(8.40 10 m s ) (3.80 m s)
2.21 10 rad s
×
=
×
, and then
33
SW
,max
(3.80 m s) (2.21 10 rad s)
1.72 10 m
y
Av
−
×
=
×
.
(b)
3
(2 )(
2 )
(0.386 m)(2.21 10 rad s)
272 m s
vf
L
L
λω
π
=
=
×
=
.
EVALUATE:
The maximum transverse velocity and acceleration will have different (smaller) values at other
points on the string.
15.71.
IDENTIFY:
To show this relationship is valid, take the second time derivative.
SET UP:
sin
cos
tt
t
∂
=
∂
.
cos
sin
t
ωω
∂
=−
∂
.
EXECUTE:
(a)
22
SW
SW
(,)
[(
sin
)sin
]
[(
sin
)cos
]
Ak
xt
x
t
t
∂
∂
[]
2
sw
2
(
sin
)sin
( , )
x
t
y
x
t
t
∂
∂
. This equation shows that
2
y
ay
. This is characteristic of
simple harmonic motion; each particle of the string moves in simple harmonic motion.
(b)
Yes, the traveling wave is also a solution of this equation. When a string carries a traveling wave each point on
the string moves in simple harmonic motion.
EVALUATE:
A standing wave is the superposition of two traveling waves, so it is not surprising that for both
types of waves the particles on the string move in SHM.
15.72.
IDENTIFY
and
SET UP:
Carry out the analysis specified in the problem.
EXECUTE:
(a)
The wave moving to the left is inverted and reflected; the reflection means that the wave moving
to the left is the same function of
,
x
−
and the inversion means that the function is
(
).
f
x
−
−
(b)
The wave that is the sum is
( )
(
)
f
xfx
−
−
(an inherently odd function), and for any
, (0)
( 0)
0.
ff
f
−−=
(c)
The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in
Figure 15.21 in the textbook is
(
).
f
x
+−
(d)
()
( )
( ) ( )
(()
)
x
x
dy
d
df x
df
x
df x
df
x d
x
df
df
fx f x
dx
dx
dx
dx
dx
d
x
dx
dx
dx
=−
−−
−
=+
−
=
+
=
+
=
−
−
.
At
0
x
=
, the terms are the same and the derivative is zero.
EVALUATE:
Our results verify the behavior shown in Figures 15.20 and 15.21 in the textbook.
15.73.
IDENTIFY:
Carry out the derivation as done in the text for Eq.(15.28). The transverse velocity is
/
y
t
and
the transverse acceleration is
/
.
t
=∂
∂
(a)
SET UP:
For reflection from a free end of a string the reflected wave is
not
inverted, so
12
( , )
( , ),
yx t
y x t
where
1
(,
)
c
o
s
(
)
yxt A
k
x t
(traveling to the left)
2
)
c
o
s
(
)
k
(traveling to the right)
Thus ( , )
[cos(
)
cos(
)].
A
k
x
t
k
x
t
+
−
EXECUTE:
Apply the trig identity cos(
)
cos cos
sin sin
ab
a b
±
=
∓
with
ak
x
=
and
:
bt
=
cos(
)
cos
cos
sin
sin
kx
t
kx
t
kx
t
+=
−
and
cos(
)
cos
cos
sin
sin
.
kx
t
kx
t
kx
t
−=
+
Then ( , )
(2 cos
)cos
A
k
x
t
=
(the other two terms cancel)
(b)
For
0,
x
=
cos
1
kx
=
and
( , )
2 cos
.
A
t
=