13.1: a)
s.rad1038.12s,1055.4
3
2
3
1


πfωT
T
π
f

b)
s.rad1053.52s,1014.1
33
Hz)220(4
1


πfω
13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the
amplitude. b) The glider will return to its original position after another 0.80 s, so the
period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)),

s60.1
1
f
Hz.625.0
13.3: The period is
s1014.1

glider is at the equilibrium position (see Fig. (13.8));
m.N292.0kg)200.0(
s)60.2(2
22
2
2
2

















π
m
T
π
mωk
13.7: a)


π
T
π
mk
13.9: From Eq. (13.12) and Eq. (13.10),
Hz,66.2s,375.02
1
mN140
kg500.0

T
fπT
s.rad7.162  πfω
13.10: a)
)(so,)sin(
22
2
2
txxωβωtAωa
dt
xd
x

is a solution to Eq. (13.4) if
ωAaω
m
k
2b).
2

37
max
j
13.12: a) From Eq. (13.19),
m.98.0
00

mk
v
ω
v
A
b) Equation (13.18) is
indeterminant, but from Eq. (13.14),
,
2



and from Eq. (13.17), sin
.so,0
2
π


c)
)).)srad sin((12.2m)98.0(so,sin ))2((cos txωtπωt 
13.13: With the same value for
ω
, Eq. (13.19) gives

x
b) From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is
0.715 rad. The angular frequency is
rad/s,7.152 πf
so
.rad)715.0rad/s)((15.7cos)cm/s359(
rad)715.0rad/s)((15.7sin )scm9.22(
rad)715.0rad/s)((15.7coscm)46.1(
2



ta
tv
tx
x
x
13.15: The equation describing the motion is
;sinωtAx 
this is best found from either
inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the
arctangent). Even so,
x
is determined only up to the sign, but that does not affect the
result of this exercise. The distance from the equilibrium position is
      
m.353.054sinm600.02sin  πTtπA
13.16: Empty chair:
k
m

ππ
kT
m
km
πT
13.17:
kg400.0,2  mkmπT
s09.22
N/m60.3
m300.0
)m/s70.2kg)(400.0(
gives
:calculate tom/s70.2Use
2
2





kmπT
x
ma
kmakx
ka
x
x
x
13.18: We have
).2)s4.71cm/s)sin((60.3()(


ω
A
(c )
2212
max
scm9.16)cm764.0()s71.4( 

Aωa
13.19:
rad)42.2s)radcos((4.16cm)40.7()(a)  ttx
22
2222
max
max
2
2
1
2
2
1
2
2
1
2
sm216.0sm)0125.0)(50.10.26(
.sm0.303isSpeed
sm303.0sm)0125.0()0740.0(50.10.26
m0125.0givess00.1at evaluated)(e)
N92.1sod)

 A
ω
v
Ax
Squaring and adding,
,
2
2
2
0
2
0
A
ω
v
x

which is the same as Eq. (13.19). b) At time
,0t
Eq. (13.21) becomes
,
2
1
2
1
2
1
2
1
2

J.1096.2s)mkg)(1.48107.2(
2
1
)(
2
1
5252
maxmax

 VmK
13.23: a) Setting
2
2
1
2
2
1
kxmv 
in Eq. (13.21) and solving for x gives
.
2
A
x 
Eliminating x in favor of v with the same relation gives
.2
2
2
ωA
x
mkAv 

m/s.11.1m)015.0(m)040.0(
kg0.500
N450
22
v
c) The extremes of acceleration occur at the extremes of motion, when
,Ax 
and
2
max
m/s36
kg)(0.500
m)N/m)(0.040450(

m
kA
a
d) From Eq. (13.4),
.m/s5.13
2
kg)(0.500
m)0.015N/m)(450(


x
a
e) From Eq. (13.31),
J. 36.0m)N/m)(0.040450(
2
2

c) The fraction of one period is
)21( π
arcsin
),0.180.12(
and so the time is
)2( πT
arcsin
1
1037.1)0.180.12(


s. Note that this is also arcsin
ωAx )(
.
d) The conservation of energy equation can be written
2
2
1
2
2
1
2
2
1
kxmvkA 
. We are
given amplitude, frequency in Hz, and various values of
x
. We could calculate velocity
from this information if we use the relationship

,So.
2
1
1
2
1
2
mM
M
AA


or
Mm 3
. For the energy,
2
2
2
1
2
kAE 
, but since
1
4
3
1
4
1
21
2

x
c)
 sm615.0mAkωA
13.28: At the time in question we have
22
sm40.8)(cos
sm20.2)sin(
m600.0)(cos






ωtAωa
ωtωAv
ωtAx
Using the displacement and acceleration equations:
222
sm40.8m)600.0()(cos  ωωtAω

12
s742.3and0.14

 ωω
To find A, multiply the velocity equation by
:ω
212
sm232.8)sm(2.20)s742.3()(sin 


ωtωt
Aω
ωt
AωωtAω


The object will therefore travel
m0.240m600.0m840.0 
to the right before stopping
at its maximum amplitude.
13.29:
mkAv 
max
sm509.0Then
m.0405.0)(so
:find to Use
s158)2(so2
:find to Use
max
maxmax
max
22




mkAv
mkaAmkAa
Aa
TπmkkmπT

2

l
mg
k
b)
s.695.0
sm9.80
m120.0
222
2
 π
g
l
π
k
m
πT

13.32: a) At the top of the motion, the spring is unstretched and so has no potential
energy, the cat is not moving and so has no kinetic energy, and the gravitational potential
energy relative to the bottom is
J 3.92m)050.0()m/skg)(9.8000.4(22
2
mgA
.
This is the total energy, and is the same total for each part.
b)
J 92.3so,0,0
springgrav

3

and so
the moment of inertia would decrease by a factor of
)2431()31)(271(
2

, and for the
same spring constant, the frequency and angular frequency would increase by a factor of
6.15243 
. b) The torsion constant would need to be decreased by a factor of 243, or
changed by a factor of 0.00412 (approximately).
13.35: a) With the approximations given,
,mkg1072.2
282


mRI
28
mkg102.7or 

to two figures.
b)
radmN103.4)mkg1072.2(Hz)22()2(
62822


πIπfκ
.
13.36: Solving Eq. (13.24) for

I
T
π
13.37:
 
.mkg0152.0
s)(265125)(2
m/radN450.0
)2(
2
2
2





π
πf
I
13.38: The equation
)t(cos φωθ 
describes angular SHM. In this problem,
.0φ
a)
).cos(and)sin(
2
2
2
tωωtωω

2
and,
2
3
)sin(since
2
3
2
2
2





tω
ω
dt
θd
t
ω
ω
dt
dθ
This corresponds to a displacement of
60
.
13.39: Using the same procedure used to obtain Eq. (13.29), the potential may be
expressed as
].)1(2)1[(





















2
00
2
00
0
2
76
612
2
1312

0
/72 RUk 
has been used. Note that terms in
2
u
from Eq. (13.28) must be
kept ; the fact that the first-order terms vanish is another indication that
0
R
is an extreme
(in this case a minimum) of U.
13.40:
   
Hz.1033.1
kg)1066.1(008.1
N/m)580(2
2
1
22
1
14
27






m
k

m
k





and
0
,
4
0
2
1
w
kg
ω 
which is solved for
kwL 4
. But L is the length of the stretched
spring; the unstretched length is
   
m.00.2N/m50.1N00.133
0
 kwkwLL
13.44:
13.45: The period of the pendulum is
 
s.36.1100s136 T
Then,

13.48: a) Solving Eq. (13.39) for I,
 
 
 
.mkg0987.0m250.0sm80.9kg1.80
2
s940.0
2
22
22















π
mgd
π
T
I










π
T
π
13.49: Using the given expression for I in Eq. (13.39), with d=R (and of course m=M),
s.58.0352  gRπT
13.50: From Eq. (13.39),
 
 
 
.kg.m129.0
2
100s120
m200.0sm9.80kg80.1
2
2
2
2
2










π
ω
fω
b)
  
.skg73.1kg300.0mN50.222  kmb
13.52: From Eq. (13.42)
 
,for Solving.exp
2
12
btAA
m
b

s.kg0220.0
m100.0
m300.0
ln
)s00.5(
)kg050.0(2
ln
2
2

.(0)0, Ax 

b)
,sin cos
2
)2(











tωωtω
m
b
Ae
dt
dx
v
tmb
x
and at
down.slopes0near versusofgraph the;2,0  ttxmAbvt
c)
,sin

ω
tω
m
b
Ae
dt
dv
a
tmb
x
x

and at
,0t
.
24
2
2
2
2
2










1
3
dmax
1
AbFA
A


Note that
the resonance frequency is independent of the value of b (see Fig. (13.27)).
13.55: a) The damping constant has the same units as force divided by speed, or
 
   
 .skgsmsmkg
2
b)The units of
km
are the same as
 
,skgkg]]][skg[[
212

the same as those for
 
.52.0so ,2.0(i).c) .
maxmaxd
2
d
kFkFAkbωmkωb 
,5.2)4.0(so,0.4 ii)(











π
Aωa
b)
.sm3.18
minrev
srad
30
)m05(.)minrev3500( c)N.1002.3
3








π
ωA
ma

N).10
3

The maximum speed increases by a factor of 2 since
,α ωv
so the speed
will be 36.7 m/s. Because the kinetic energy depends on the square of the velocity, the
kinetic energy will increase by a factor of four (302 J). But, because the time to reach the
midpoint is halved, due to the doubled velocity, the power increases by a factor of eight
(141 kW).
13.58: Denote the mass of the passengers by m and the (unknown) mass of the car by M.
The spring cosntant is then
lmgk 
. The period of oscillation of the empty car is
kMπT 2
E

and the period of the loaded car is
 
 
s.003.12
so,22
2
2
LE
2
2
EL



4
15
ωA
after the amplitude is reduced, the speed is
   
1
4
3
2
1
2
1
42 ωAAAω 
, so the speed is decreased by a factor of
5
1
(this result is
valid at
4
1
Ax 
as well). e) The potential energy depends on position and is
unchanged. From the result of part (d), the kinetic energy is decreased by a factor of
5
1
.
13.60: This distance
;is kmgLL 
the period of the oscillatory motion is
,22

vT
Ax
c) If the block is just on the verge of slipping, the friction force is its maximum,
.
ss
mgμnμf 
Setting this equal to
   
.143.02gives2
22
 gTπAμTπmAma
s
13.62: a) The normal force on the cowboy must always be upward if he is not holding on.
He leaves the saddle when the normal force goes to zero (that is, when he is no longer in
contact with the saddle, and the contact force vanishes). At this point the cowboy is in
free fall, and so his acceleration is
g
; this must have been the acceleration just before
he left contact with the saddle, and so this is also the saddle’s acceleration.
b)
m.110.0))Hz50.1(2)sm80.9()2(
222
 πfπax
c) The cowboy’s speed will
be the saddle’s speed,
s.m11.2)2(
22
 xAπfv
d) Taking
0t

which has as its least non-zero solution
s.538.0t
e) The speed of the saddle is
,sm72.1)(sin s)m36.2( 

tω
and the cowboy’s speed is (2.11
)sm80.9()sm
2

s,m16.3s)538.0( 
giving a relative speed of
sm87.4
(extra figures were kept in
the intermediate calculations).
13.63: The maximum acceleration of both blocks, assuming that the top block does not
slip, is
),(
max
MmkAa 
and so the maximum force on the top block is
 
.)(isamplitudemaximum thesoand,
smaxs
kgMmμAmgμkA
Mm
m

