15.1: a) The period is twice the time to go from one extreme to the other, and
sm1.2or s,m20.1s)(5.0m)00.6(  Tfv
to two figures. b) The amplitude
is half the total vertical distance,
m.310.0
c) The amplitude does not affect the wave
speed; the new amplitude is
m.150.0
d) For the waves to exist, the water level cannot be
level (horizontal), and the boat would tend to move along a wave toward the lower level,
alternately in the direction of and opposed to the direction of the wave motion.
15.2:
vf 

Hz105.1
m001.0
sm1500
6



v
f
15.3: a)
m.0.439Hz)(784s)m344(  fv
b)
Hz.105.25m)10(6.55s)m344(
65


vf

minmax

15.6: Comparison with Eq. (15.4) gives a)
mm,50.6 cm,0.28b)
Hz8.27c)
s0360.0
11

T
f
and from Eq. (15.1), d)
sm7.78Hz)m)(27.8280.0( v
,
direction.e) x
15.7: a)
Hz,25.0m)320.0(s)m00.8(  vf
m.rad19.6m)320.0()2(2s,104.00Hz)0.25(11
2


ππkfT
b)
.
m320.0
Hz)(25.02cosm)0700.0(),(






y







),cos()(sin
2
2
2
ωtkxAω
t
y
ωtkxAω
t
y






and so
),(and,
2
2
2
2





),(sin)cos(
2
2
2
ωtkxAω
t
y
ωtkxAω
t
y






and so
),(and,
2
2
2
2
2
2
txy
t

y





).(cos
2
2
2
kxωtAω
t
v
t
y
a
y
y








b) (Take A, k and
ω
to be positive. At
,0t

22
AωπAωa
y

and the particle is speeding up.
(iv)
,2)43cos( ωAπωAv
y

and the particle is moving down.
,2)43sin(
22
AωπAωa
y

and the particle is slowing down (
y
v
is becoming less
negative). (v)
ωAπωAv
y
 )cos(
and the particle is moving down.
,0)sin(
2
 πAωa
y
and the particle is instantaneously not accelerating.
(vi)

2)47sin(
22
AωπAωa
y

and the particle is slowing down (
y
v
and
y
a
have
opposite signs).
15.11: Reading from the graph, a)
s.040.0b) ,mm0.4  TA
c) A displacement of
0.090 m corresponds to a time interval of 0.025 s; that is, the part of the wave represented
by the point where the red curve crosses the origin corresponds to the point where the
blue curve crosses the t-axis
)0( y
at
s,025.0t
and in this time the wave has traveled
0.090 m, and so the wave speed is
sm6.3
and the wavelength is
m14.0)s040.0)(sm6.3( vT
. d)
sm6.0s0.015m090.0 
and the wavelength is

πA
2
cos2cos

 
,
2
cos
vtxA 



where
vf
T


has been used.
b)
 
.
λ
2
sin
λ
2
vtx
π
A
πv

b) i) t = 0.400 s:
_______________________________________________________________
x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00
_______________________________________________________________
y(cm) 0.285 0.136 0.093 0.267 0.285 0.136 0.093 0.267 0.285
_______________________________________________________________


:s800.0ii t
________________________________________________________________
x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00
________________________________________________________________
y(cm) 0.176 0.296 0.243 0.047 0.176 0.296 0.243 0.047 0.176
________________________________________________________________
15.14: Solving Eq. (15.13) for the force
,F
 
.2.43))m750.0()Hz0.40((
m2.50
kg120.0
2
2
2








the time for the
pulse to reach the other end is
s.390.0
)sm80.9)(kg50.7(
)m0.14)(kg800.0(
)(
2

Mg
mL
LmMg
L
v
L
t
15.18: a) The tension at the bottom of the rope is due to the weight of the load, and
the speed is the same
sm5.88
as found in Example 15.4 b) The tension at the middle
of the rope is (21.0
N8.205)sm809()kg
2
.
(keeping an extra figure) and the speed of
the rope is
s.m7.90
c) The tension at the top of the rope
is
sm6.215)sm80.9)(kg0.22(
2

22
ave
2
1
AωμFP 

232
3
)m106.1())Hz0.120(2()N0.25(
m0.80
kg1000.3
2
1














= 0.223 W,
or 0.22 W to two figures. b) Halving the amplitude quarters the average power, to 0.056
W.

2
1
mW0.1
mW11.0
12

Ι
Ι
rr
so it is possible to
move
m5.0m5.2m5.7
21
 rr
closer to the source.
15.23: a)
2
22
2
11
rΙrΙ 
2222
2112
mW050.0)m3.1m3.4)(mW026.0()(  rrΙΙ
b)
W04.6)mW026.0()m3.4(44
222
 πΙπrP
Energy =
J102.2)s3600)(W04.6(

.ωtkx 
(f) The linear mass density is
mkg10504.2)m35.1()kg1038.3(
33 
μ
, so the tension is
N3.28)sm3.106)(mkg10504.2(
232


μvF
(keeping an extra figure in
v
for
accuracy).
23
2
1
22
2
1
av
)srad742()N3.28)(mkg1050.2()g(

 AωμFP
W.39.0)m1030.2(
23


15.25:

, where


vωkπTπωA s,rad1072.1s)1065.3(22mm,46.2
33
m.rad5.15)sm111)(srad1072.1(
3

15.31: a) The nodes correspond to the places where
integeran ,or 0sin is, that );1.15(Eq.in allfor 0
nodenode
nnπkxkxty 
.
nxπk m)333.1( m,rad75.0With
node

and for
m,...6.67m,5.33m,4.00m,2.67m,1.333,0...,,2,1,0
node
 xn
b) The antinodes
correspond to the points where cos
,0kx
which are halfway between any two adjacent
nodes, at 0.667 m, 2.00 m, 3.33 m, 4.67 m, 6.00 m, ...
15.32: a)
 
,sinsin
sw
2



and
.
k
ω
v 
b) A standing wave is built up by the superposition of traveling waves, to which the
relationship
kωv 
applies.
15.33: a) The amplitude of the standing wave is
cm,85.0
sw
A
the wavelength is
twice the distance between adjacent antinodes, and so Eq. (15.28) is
).cm0.302sin())s0.0752sin(()cm85.0(),( πxtπtxy 

c)
cm.688.0))cm0.30(cm)(10.52sin()cm0.850( π

m/s.00.4)s0750.0()cm30.0(λλb)  Tfv
15.34:
)]cos()cos([
21
ωtkxωtkxAyy 

]sinsincoscossinsincoscos[ ωtkxωtkxωtkxωtkxA 



x
y
,
2
2
2
2
1
2
x
y
x
y






2
2



t
y
.
2
2

2
2
1
2
x
y
x
y












2
1
v



2
1
2
t










2
2
2
2
1
2
t
y
t
y








2
1
v

Hz000,10
24 theso,5.24
harmonic may be heard, but not the
.25
th
15.37: a) In the fundamental mode,
s.m0.96)m60.1)(Hz0.60(soandm60.12  fvL

b)
.461)m800.0()kg0400.0()sm0.96(
222
 LmvμvF
15.38: The ends of the stick are free, so they must be displacement antinodes.
1
st
harmonic:

m0.42
2
1
11
 LL
2
nd
harmonic:

m0.21
22
 LL
3

So,
cm.8184)cmrad0340.0(2  π

cm277)3(3 L
d)
(c)part fromcm,185

Hz96.72sosrad0.50  πωfω
period
s126.01  fT

scm1470 fv
e)
ωtkxωAdtdyv
y
cossin
sw

scm280cm)s)(5.60rad0.50(
SWmax,
 ωAv
y
f)
Hz65.2so,3Hz96.7
113
 fff
is the fundamental
srad1332Hz;2.218
8818
 πfωff