Tài liệu Physics exercises_solution: Chapter 13 ppt

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

3
440
s50.0


and the angular frequency is

T
π
ω
2
s.rad1053.5
3

13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its
period is thus 2.0 s and its frequency
.s5.0period1
1

(b) The displacement varies
from
m,20.0 tom20.0 
so the amplitude is 0.20 m. (c) 2.0 s (see part a)
13.5: This displacement is
4
1
of a period.

s.0500.0sos,200.01  tfT
13.6: The period will be twice the time given as being between the times at which the

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)

kg.084.0c) s.rad7.372b) s.167.0
2
1

ω
k
f
mπfωT
13.8: Solving Eq. (13.12) for k,
m.N1005.1
s150.0
2
kg)600.0(
2
3
22
















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


a constant, so Eq. (13.4) is not satisfied. c)
,
)( βωti
dt
dx
x
iωv




mkωtxxωAeiωa
βωti
dt
dv
x
x
22)(2
if(13.4)Eq.osolution tais)(so,)(
13.11: a)
s,m8.29Hz))(440m)(210(3.0b) )Hz))(440((2cosmm)0.3(
3


πtπx
),Hz))(440sin((2)sm1034.6()(c) .sm1029.2Hz)440()mm)(20.3(
372422
tπtjπ 
.sm1034.6

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

m383.0A
and Eq. (13.18) gives
 
.rad02.1rad/s)(12.2cosm)(0.383and  tx
,58.5rad02.1
kgN/m/2.00300m)(0.200
m/s)4.00(
arctan












and x = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad).
13.14: For SHM,
 
.m/s71.2m)101.1(Hz)5.2(2)2(
22
2
22


πxπfxωa

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

πT 2
N/m993
s)(1.30
kg)5.42(44
2
2
2
2

π
T
m
π
k
With person in chair:
kg120kg5.42kg162
kg162
4
N/m)993(s)54.2(
4
2
person
2
2
2
2



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()(

1
πttv
x


Comparing this to the general
form of the velocity for SHM:
2
s4.71
cm/s60.3
1
π
ω
ωA




(a)
s33.1s71.422T
1


πωπ
(b)
cm764.0
s71.4
scm60.3scm60.3
1



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)

sm308.0gives
m0740.0cm40.7c)
mN0.26)2(so2b)
s1.51so 2s)rad(4.16,When








mkxa
xAmkv
xttx
kAFkxF
mkAvkAkxmv
A
T
πmkkmπT
T
πTTt
13.20: See Exercise 13.15;
s.0.0871))(20.36))(1.5arccos((  πt
13.21: a) Dividing Eq. (13.17) by
ω
,
.sin ,cos
0
0

 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

1
2
0
2
0
2
2
0
2
0
2
kxv
ω
k
kxmvkA

where
2
kωm 
(Eq. (13.10)) has been used. Dividing by
2k
gives Eq. (13.19).
13.22: a)
s.m1.48m)10Hz))(0.60392(2()2(
3
max


πAπfv
b)

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 

b) This
happens four times each cycle, corresponding the four possible combinations of + and –
in the results of part (a). The time between the occurrences is one-fourth of a period or
 
8
3
84
3
4
1
24
2
22
, ,c) .4/
kAkA
ω
π
ω
π
KUEKEUT 
13.24: a) From Eq. (13.23),
m/s.1.20m)040.0(
kg0.500
mN450
max
 A
m
k
v
b) From Eq. (13.22),

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

1
E
13.25: a)
 


max
22
2
22
max
.m/s5.13m)100.18(Hz)85.0(2)2( vAfAωa


m/s961.02  πfAωA
.
,m/s57.2)2(b)
22
 xπfa
x
 
m/s.833.0m)100.9(m)100.18(Hz)85.0(2
)2(
2222
22



π
xAπfv

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

222
4 fπωmk 
and rewrite the
conservation equation as
2
2
1
4
2
1
2
2
1
22
2
xA
fπ
v

. Using energy principles is generally a good
approach when we are dealing with velocities and positions as opposed to accelerations
and time when using dynamics is often easier.
13.26: In the example,
mM
M
AA


12
and now we want

,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

1
2
or ,, EEEAA 
is lost to
heat.
13.27: a)
J 0284.0
2
2
1
2
2
1
 kxmv
.
b)
m.014.0
kg)(0.150N/m)300(
m/s)300.0(
m)012.0(
2
2
2
2
0
2
0

ω
v

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 



ωtAω
Next square both this new equation and the acceleration equation and add them:
m840.0
m7054.0
)s742.3(
sm3.138sm3.138
sm3.138sm56.70sm77.67
)(cos)(sin
)sm40.8()sm232.8()(cos)(sin
2
41
42
4
42
2
42424224
2224
2222224224






A
ω
A
A
ω

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

mkT
13.30: Using
0
0
L
F
k 
from the calibration data,
kg.00.6
Hz))60.2((2
m)10(1.25N)200(
)2(
)(
2
1
2
00




ππf
LF
m
13.31: a)
m.N10153
m)120.0(
)sm(9.80kg)650(
Δ
3

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

 UKU
.
c) At equilibrium the spring is stretched half as much as it was for part (a), and so
J 98.0soandJ,1.96J)92.3(J, 0.98J)92.3(
2
1
grav
4
1
spring
 KUU
.
13.33: The elongation is the weight divided by the spring constant,
cm97.3
4
2
2
2

π
gT
mω
mg
k
w
l

.
13.34: See Exercise 9.40. a) The mass would decrease by a factor of
271)31(

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


in terms of the period,
m/rad.N1091.1
)m)10kg)(2.201000.2)(21((
s00.1
2
2
5
223
2
2


















π

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

dt
θd
dt
d
θ

b) When the angular displacement is
)cos(, tω
, and this occurs at
,0t
so
1.cos(0)since,and0,sin(0) since 0
2
2
2


ω
dt
θd
dt
dθ
When the angular displacement is
).cos(or ),cos(,2
2
1
2
tωtω 

.21)cos(since,

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[(

6
0
12
00

 RxRxUU
Note that at
.,
00
UURr 
Using the appropriate forms of the binomial theorem for
||
0
Rx
<< 1,
 
  
 
 
  
 










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

121
RxRx
RxRx
UU








2
2
0
0
36
1
x
R
U

.
2
1
0
2
Ukx 
where
2

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

f
13.41:
,2 gLT


so for a different acceleration due to gravity
,g

 
s.60.2sm71.3sm80.9s60.1
22




ggTT
13.42: a) To the given precision, the small-angle approximation is valid. The highest
speed is at the bottom of the arc, which occurs after a quarter period,
s.25.0
24

g
LT

b) The same as calculated in (a), 0.25 s. The period is independent of amplitude.
13.43: Besides approximating the pendulum motion as SHM, assume that the angle
is sufficiently small that the length of the spring does not change while swinging in the
arc. Denote the angular frequency of the vertical motion as
L
gkg

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,

 
 
.sm67.10
s1.36
m5.44
2
2
2
2
2

π
T
Lπ
g
13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is
 
.13.39Eq.in with ,22so,2
222
RdgRπTMRMRMRI 
Solving for R,
m.496.08
2
2
 πgTR
13.47: For the situation described,
LdmLI  and
2
in Eq. (13.39); canceling the factor
of m and one factor of L in the square root gives Eq. (13.34).

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

b) The small-angle approximation will not give three-figure accuracy for
rad.0.400Θ 
From energy considerations,
 
.Ω
2
1
cos1
2
max
Imgd 
Expressing
max

in terms of the period of small-angle oscillations, this becomes
    
.srad66.2rad0.40cos1
s940.0
2
2cos1
2
2
22
max
















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
















ππ
T
mgdI
13.51: a) From Eq. (13.43),
 
 
 
 
Hz.393.0
2
so,srad47.2
kg300.04
skg90.0
kg300.0
mN50.2

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

1



















A
A
t
m
b
As a check, note that the oscillation frequency is the same as the undamped frequency to
valid.is(13.42)Eq.so,%108.4
3

13.53: a) With

.(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

2
'cos
4
2
2
2
)2(




















tω
m
b

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











m
k
m
b
A
ω
m
b
Aa
x
(Note that this is
.))(
00
mkxbv 
This will be negative if
.2ifpositiveand2ifzero,2 kmbkmbkmb 
The graph in the three cases will
be curved down, not curved, or curved up, respectively.
13.54: At resonance, Eq. (13.46) reduces to
.2b) .a) .

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)(

maxmaxd
kFkFAkb 

as shown in Fig.(13.27).
13.56: The resonant frequency is
Hz,22.2srad139)kg108)mN102.1(
6
mk
and this package does not meet the criterion.
13.57: a)
.sm1072.6
minrev
srad
30
)minrev3500(
2
m100.0
2
3
2
2










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

J.6.75)sm3.18)(kg45)(.(
2
2
1
2
2
1
 mvK
d) At the midpoint of the stroke, cos(
ω
t)=0
and so
s,rad))(minrev(35002 thus,2
3
350
minrev
srad
30
ππ
ω. ωπtπωt 
so
W.101.76s)(J6.75or ,Then s.
4
2(350)
3
)350(2
3
 PtKPt
e) If the frequency doubles, the acceleration and hence the needed force will quadruple
(12.1

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








g
l
πTT
g
l
πT
k
mM
πT
13.59: a) For SHM, the period, frequency and angular frequency are independent of
amplitude, and are not changed. b) From Eq. (13.31), the energy is decreased by a factor
of
4
1
. c) From Eq. (13.23), the maximum speed is decreased by a factor of
2
1
d) Initially,
the speed at
4
1
A
was
;
1

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

g
L
k
m
πT 
which is the period of oscillation of a simple pendulum of lentgh L.
13.61: a) Rewriting Eq. (13.22) in terms of the period and solving,
s.68.1
2
22



v
xAπ
T
b) Using the result of part (a),
m.0904.0
2
2
2










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

at the time when
the cowboy leaves, the position of the saddle as a function of time is given by Eq.
(13.13), with
;cos
2
Aω
g


this is checked by setting
0t
and finding that
.
22
ω
a
ω
g
x 
The cowboy’s position is
.)2(
2
00c
tgtvxx 
Finding the time at which
the cowboy and the saddle are again in contact involves a transcendental equation which
must be solved numerically; specifically,
rad),11.1s)rad42.9((cosm)25.0()sm90.4(s)m11.2(m)110.0(
22
 ttt

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

