Oscillatory Motion
CHAPTER OUTLINE
12.1
12.2
12.3
12.4
12.5
12.6
12.7
12.8
Motion of a Particle
Attached to a Spring
Mathematical
Representation of Simple
Harmonic Motion
Energy Considerations
in Simple Harmonic
Motion
The Simple Pendulum
The Physical Pendulum
Damped Oscillations
Forced Oscillations
Context Connection—
Resonance in Structures
ANSWERS TO QUESTIONS
Q12.1
Neither are examples of simple harmonic motion, although they are
both periodic motion. In neither case is the acceleration proportional
to the position. Neither motion is so smooth as SHM. The ball’s
acceleration is very large when it is in contact with the floor, and the
student’s when the dismissal bell rings.
Q12.2
The two will be equal if and only if the position of the particle at time
zero is its equilibrium position, which we choose as the origin of
coordinates.
Q12.3
You can take φ = π , or equally well, φ = −π . At t = 0 , the particle is at its turning point on the
negative side of equilibrium, at x = − A .
Q12.4
No. It is necessary to know both the position and velocity at time zero.
Q12.5
(a)
In simple harmonic motion, one-half of the time, the velocity is in the same direction as the
displacement away from equilibrium.
(b)
Velocity and acceleration are in the same direction half the time.
(c)
Acceleration is always opposite to the position vector, and never in the same direction.
Q12.6
We assume that the coils of the spring do not hit one another. The frequency will be higher than f by
the factor 2 . When the spring with two blocks is set into oscillation in space, the coil in the center
of the spring does not move. We can imagine clamping the center coil in place without affecting the
motion. We can effectively duplicate the motion of each individual block in space by hanging a
single block on a half-spring here on Earth. The half-spring with its center coil clamped—or its other
half cut off—has twice the spring constant as the original uncut spring, because an applied force of
the same size would produce only one-half the extension distance. Thus the oscillation frequency in
space is
FG 1 IJ FG 2 k IJ
H 2π K H m K
12
= 2 f . The absence of a force required to support the vibrating system in
orbital free fall has no effect on the frequency of its vibration.
331
332
Q12.7
Oscillatory Motion
No; Kinetic, Yes; Potential, No. For constant amplitude, the total energy
1 2
kA stays constant. The
2
1
mv 2 would increase for larger mass if the speed were constant, but here the greater
2
mass causes a decrease in frequency and in the average and maximum speed, so that the kinetic and
potential energies at every point are unchanged.
kinetic energy
Q12.8
Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2
are valid.
Equation
Information given by equation
x t = A cos ω t + φ
position as a function of time
af
b g
vat f = −ωA sinbω t + φ g
va x f = ±ω e A − x j
aat f = −ω A cosbω t + φ g
aa x f = −ω xat f
2
2
2
2 12
velocity as a function of time
velocity as a function of position
acceleration as a function of time
acceleration as a function of position
The angular frequency ω appears in every equation. It is a good idea to figure out the value of angular
frequency early in the solution to a problem about vibration, and to store it in calculator memory.
Lf
Li
2Li
and T f =
=
= 2Ti . The period gets larger by
g
g
g
mass has no effect on the period of a simple pendulum.
Q12.9
We have Ti =
Q12.10
Shorten the pendulum to decrease the period between ticks.
Q12.11
No. If the resistive force is greater than the restoring force of the spring (in particular, if b 2 > 4mk ),
the system will be overdamped and will not oscillate.
Q12.12
Higher frequency. When it supports your weight, the center of the diving board flexes down less
than the end does when it supports your weight. Thus the stiffness constant describing the center of
1
k
the board is greater than the stiffness constant describing the end. And then f =
is greater
2π m
for you bouncing on the center of the board.
2 times. Changing the
FG IJ
H K
Q12.13
Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in
time. Without damping, the amplitude would increase without limit at resonance.
Q12.14
An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in the
wind, the fibers in its stem provide a restoring torque. If the frequency of the breeze matches the
natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven
into a large-amplitude resonance vibration. Note that it is not the size of the driving force that sets
the leaf into resonance, but the frequency of the driving force. If the frequency changes, another leaf
will be set into resonant oscillation.
Chapter 12
333
SOLUTIONS TO PROBLEMS
Section 12.1 Motion of a Particle Attached to a Spring
P12.1
(a)
Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then
repeat the motion over and over again. Thus, the motion is periodic .
(b)
To determine the period, we use: x =
1 2
gt .
2
The time for the ball to hit the ground is t =
a
f
2 4.00 m
2x
=
= 0.904 s
g
9.80 m s 2
a
f
This equals one-half the period, so T = 2 0.904 s = 1.81 s .
(c)
No . The net force acting on the ball is a constant given by F = − mg (except when it is in
contact with the ground), which is not in the form of Hooke’s law.
Section 12.2 Mathematical Representation of Simple Harmonic Motion
P12.2
P12.3
f FGH
a
IJ
6K
π
(a)
x = 5.00 cm cos 2t +
(b)
v=
π
dx
= − 10.0 cm s sin 2t +
dt
6
(c)
a=
π
dv
= − 20.0 cm s 2 cos 2t +
dt
6
(d)
A = 5.00 cm
a
g FGH
b
j FGH
e
f b
IJ
K
IJ
K
f FGH π6 IJK =
a
At t = 0 ,
x = 5.00 cm cos
At t = 0 ,
v = −5.00 cm s
At t = 0 ,
a = −17.3 cm s 2
and
T=
b
g
g
2π
ω
=
2π
= 3.14 s
2
x = 4.00 m cos 3.00π t + π compare this with x = A cos ω t + φ to find
(a)
ω = 2π f = 3.00π
or
f = 1.50 Hz
T=
1
= 0.667 s
f
(b)
A = 4.00 m
(c)
φ = π rad
(d)
x t = 0.250 s = 4.00 m cos 1.75π = 2.83 m
a
f a
f a
f
4.33 cm
334
P12.4
Oscillatory Motion
(a)
(b)
20.0 cm
v max = ω A = 2π fA = 94.2 cm s
This occurs as the particle passes through equilibrium.
(c)
b g
a max = ω 2 A = 2π f
2
A = 17.8 m s 2
This occurs at maximum excursion from equilibrium.
P12.5
(a)
At t = 0 , x = 0 and v is positive (to the right). Therefore, this situation corresponds to
x = A sin ωt
and
v = vi cos ωt
Since f = 1.50 Hz ,
ω = 2π f = 3.00π
a
(b)
a
f
v max = vi = Aω = 2.00 3.00π = 6.00π cm s = 18.8 cm s
The particle has this speed at t = 0 and next at
(c)
a
a max = Aω 2 = 2.00 3.00π
f
2
Since T =
FG
H
(a)
1
T
=
s
2
3
t=
3
T = 0.500 s
4
2
s and A = 2.00 cm, the particle will travel 8.00 cm in this time.
3
Hence, in 1.00 s =
*P12.6
t=
= 18.0π 2 cm s 2 = 178 cm s 2
This positive value of acceleration first occurs at
(d)
f
x = 2.00 cm sin 3.00π t
Also, A = 2.00 cm, so that
IJ
K
3
T , the particle will travel
2
8.00 cm + 4.00 cm = 12.0 cm .
For constant acceleration. position is given as a function of time by
x = xi + v xi t +
b
1
ax t 2
2
x
4.5
ga f 12 e−0.32 m s ja4.5 sf
2
= 0.27 m + 0.14 m s 4.5 s +
t
2
= −2.34 m
FIG. P12.6(a)
ja f
e
(b)
v x = v xi + a x t = 0.14 m s − 0.32 m s 2 4.5 s = −1.30 m s
(c)
For simple harmonic motion we have instead x = A cos ω t + φ and v = − Aω sin ω t + φ
2
2
2
a
f
b
g
b
g
where a = −ω x , so that −0.32 m s = −ω 0.27 m , ω = 1.09 rad s. At t = 0 , 0.27 m = A cos φ
0.14 m s
and 0.14 m s = −A 1.09 s sin φ . Dividing gives
= − 1.09 s tan φ , tan φ = −0.476,
0.27 m
φ = −25.5° . Still at t = 0 , 0.27 m = A cos −25.5° , A = 0.299 m . Now at t = 4.5 s,
continued on next page
b
g
a
f
b
g
Chapter 12
a
f b
= a0.299 mf cos 255° =
ga f
a
f a
335
f
x = 0.299 m cos 1.09 rad s 4.5 s − 25.5° = 0.299 m cos 4.90 rad − 25.5°
a
(d)
fb
−0.076 3 m
g
x
v = − 0.299 m 1.09 s sin 255° = +0.315 m s
4.5
t, s
FIG. P12.6(d)
P12.7
FG v IJ sin ω t
HωK
af
i
The proposed solution
x t = xi cos ω t +
implies velocity
v=
dx
= − xiω sin ω t + vi cos ω t
dt
and acceleration
a=
v
dv
= − xiω 2 cos ω t − viω sin ω t = −ω 2 xi cos ω t + i sin ω t = −ω 2 x
dt
ω
FG
H
FG IJ
H K
IJ
K
(a)
The acceleration being a negative constant times position means we do have SHM, and its
angular frequency is ω. At t = 0 the equations reduce to x = xi and v = vi so they satisfy all
the requirements.
(b)
v 2 − ax = − xiω sin ω t + vi cos ω t
b
g − e− x ω
2
i
2
jFGH
cos ω t − vi sin ω t xi cos ω t +
FG v IJ sin ω tIJ
HωK K
i
v 2 − ax = xi2ω 2 sin 2 ω t − 2 xi viω sin ω t cos ω t + vi2 cos 2 ω t
+ xi2ω 2 cos 2 ω t + xi viω cos ω t sin ω t + xi viω sin ω t cos ω t + vi2 sin 2 ω t = xi2ω 2 + vi2
So this expression is constant in time. On one hand, it must keep its original value vi2 − ai xi .
On the other hand, if we evaluate it at a turning point where v = 0 and x = A , it is
A 2ω 2 + 0 2 = A 2ω 2 . Thus it is proved.
P12.8
P12.9
(a)
T=
12.0 s
= 2.40 s
5
(b)
f=
1
1
=
= 0.417 Hz
T 2. 40
(c)
ω = 2π f = 2π 0.417 = 2.62 rad s
f=
a
ω
2π
=
1
2π
Solving for k,
k
m
f
or
T=
k=
1
m
= 2π
f
k
4π 2 m
T
2
=
b
4π 2 7.00 kg
a2.60 sf
2
g=
40.9 N m .
336
*P12.10
Oscillatory Motion
The mass of the cube is
ja
e
f
m = ρV = 2.7 × 10 3 kg m3 0.015 m
3
= 9.11 × 10 −3 kg
The spring constant of the strip of steel is
P12.11
(a)
ω=
k
=
m
k=
14.3 N
F
=
= 52.0 N m
x 0.027 5 m
f=
1
ω
=
2π 2π
8.00 N m
= 4.00 s −1
0.500 kg
t=
FG 1 IJ sin FG x IJ and when
H 4.00 K H 10.0 K
−1
P12.12
Using t =
s 9.11 × 10 −3 kg
= 12.0 Hz
a
v max = 40.0 cm s
2
amax = 160 cm s 2 .
x = 6.00 cm , t = 0.161 s.
a f
a = −160 sin 4.00a0.161f =
v = 40.0 cos 4.00 0.161 = 32.0 cm s
−96.0 cm s 2 .
FG 1 IJ sin FG x IJ
H 4.00 K H 10.0 K
−1
when x = 0 , t = 0 and when
x = 8.00 cm , t = 0.232 s.
Therefore,
∆t = 0.232 s .
m = 1.00 kg , k = 25.0 N m, and A = 3.00 cm . At t = 0 , x = −3.00 cm
(a)
ω=
k
=
m
so that,
(b)
25.0
= 5.00 rad s
1.00
T=
2π
ω
=
2π
= 1.26 s
5.00
b
g
v max = Aω = 3.00 × 10 −2 m 5.00 rad s = 0.150 m s
b
amax = Aω 2 = 3.00 × 10 −2 m 5.00 rad s
(c)
g
2
= 0.750 m s 2
Because x = −3.00 cm and v = 0 at t = 0 , the required solution is x = − A cos ω t
or
a
f
x = −3.00 cos 5.00t cm
a
a
f
f
dx
= 15.0 sin 5.00t cm s
dt
dv
a=
= 75.0 cos 5.00t cm s 2
dt
v=
f
x = 10.0 sin 4.00t cm .
so position is given by
v = 40.0 cos 4.00t cm s
We find
(c)
2
a f
a = −160 sina 4.00tf cm s
From this we find that
(b)
52 kg
1
k
=
m 2π
Chapter 12
337
Section 12.3 Energy Considerations in Simple Harmonic Motion
P12.13
Choose the car with its shock-absorbing bumper as the system; by conservation of energy,
1
1
mv 2 = kx 2 :
2
2
P12.14
P12.15
j
2π
2π
=
= 25.1 rad s
0.250
T
b
g = 126 N m
2a 2.00 f
= 0.178 m
2
(a)
k = mω 2 = 0.200 kg 25.1 rad s
(b)
E=
(a)
Energy is conserved for the block-spring system between the maximum-displacement and
the half-maximum points:
1
1
1
K +U i = K +U f
0 + kA 2 = mv 2 + kx 2
2
2
2
2
1
1
1
2
2
6.50 N m 0.100 m = m 0.300 m s + 6.50 N m 5.00 × 10 −2 m
2
2
2
2 24.4 mJ
1
2
32.5 mJ = m 0.300 m s + 8.12 mJ
m=
= 0.542 kg
2
9.0 × 10 −2 m 2 s 2
kA 2
⇒A=
2
2E
=
k
f a
b
P12.17
e
m = 200 g , T = 0.250 s, E = 2.00 J ; ω =
a
P12.16
5.00 × 10 6
= 2.23 m s
10 3
k
= 3.16 × 10 −2 m
m
v=x
126
f
f
ga
b
b
g
6.50 N m
k
=
= 3.46 rad s
m
0.542 kg
ω=
(c)
amax = Aω 2 = 0.100 m 3.46 rad s
(a)
−2
kA 2 250 N m 3.50 × 10 m
E=
=
2
2
(b)
v max = Aω
(c)
amax = Aω 2 = 3.50 × 10 −2 m 22.4 s −1
(a)
E=
(b)
v = ω A2 − x2 =
b
g
e
2
j
ω=
where
e
∴T =
ge
2π
ω
=
f
j
2π rad
= 1.81 s
3.46 rad s
= 1.20 m s 2
2
= 0.153 J
k
250
=
= 22.4 s −1
m
0.500
j
2
v max = 0.784 m s
= 17.5 m s 2
1 2 1
kA = 35.0 N m 4.00 × 10 −2 m
2
2
b
ge
a
g
(b)
v=
b
j
2
= 28.0 mJ
k
A2 − x2
m
35.0
50.0 × 10 −3
e4.00 × 10 j − e1.00 × 10 j = 1.02 m s
1
1
− kx = a35.0 fLe 4.00 × 10 j − e3.00 × 10 j O =
MN
2
2
QP
−2 2
(c)
1
1
mv 2 = kA 2
2
2
(d)
1 2
1
kx = E − mv 2 = 15.8 mJ
2
2
2
−2 2
−2 2
−2 2
12.2 mJ
338
P12.18
*P12.19
Oscillatory Motion
F
20.0 N
=
= 100 N m
x 0.200 m
(a)
k=
(b)
ω=
(c)
v max = ωA = 50.0 0.200 = 1.41 m s at x = 0
(d)
a max = ω 2 A = 50.0 0.200 = 10.0 m s 2 at x = ± A
(e)
E=
(f)
v = ω A 2 − x 2 = 50.0
(g)
a = ω 2 x = 50.0
(a)
E=
k
= 50.0 rad s
m
so
a
f
a
f
a fa
1 2 1
kA = 100 0.200
2
2
FG 0.200 IJ =
H 3 K
f
2
f=
ω
= 1.13 Hz
2π
= 2.00 J
a
8
0.200
9
f
2
= 1.33 m s
3.33 m s 2
a f
1
1 2
kA , so if A ′ = 2 A , E ′ = k A ′
2
2
2
=
a f
1
k 2A
2
2
= 4E
Therefore E increases by factor of 4 .
*P12.20
(b)
v max =
k
A , so if A is doubled, v max is doubled .
m
(c)
a max =
k
A , so if A is doubled, a max also doubles .
m
(d)
T = 2π
(a)
y f = yi + v yi t +
m
is independent of A, so the period is unchanged .
k
1
ayt 2
2
1
−11 m = 0 + 0 + −9.8 m s 2 t 2
2
e
t=
(b)
j
22 m ⋅ s 2
= 1.50 s
9.8 m
Take the initial point where she steps off the bridge and the final point at the bottom of her
motion.
eK + U
g
+ Us
j = eK + U
i
g
+ Us
j
f
1 2
kx
2
1
65 kg 9.8 m s 2 36 m = k 25 m
2
k = 73.4 N m
0 + mgy + 0 = 0 + 0 +
a
continued on next page
f
2
Chapter 12
(c)
The spring extension at equilibrium is x =
339
F 65 kg 9.8 m s 2
=
= 8.68 m , so this point is
k
73.4 N m
11 + 8.68 m = 19.7 m below the bridge and the amplitude of her oscillation is
36 − 19.7 = 16.3 m .
73.4 N m
k
=
= 1.06 rad s
m
65 kg
(d)
ω=
(e)
Take the phase as zero at maximum downward extension. We find what the phase was 25 m
higher, where x = −8.68 m:
In x = A cos ω t ,
FG
H
−8.68 m = 16.3 m cos 1.06
t
s
16.3 m = 16.3 m cos 0
IJ
K
t
1.06 = −122° = −2.13 rad
s
t = −2.01 s
Then +2.01 s is the time over which the spring stretches.
(f)
P12.21
total time = 1.50 s + 2.01 s = 3.50 s
Model the oscillator as a block-spring system.
v2 + ω 2x2 = ω 2 A2
From energy considerations,
v max = ω A and v =
ωA
2
From this we find x 2 =
3 2
A
4
so
FG ω A IJ
H2K
and
x=
2
+ ω 2x2 = ω 2 A2
3
A = ±2.60 cm where A = 3.00 cm
2
Section 12.4 The Simple Pendulum
Section 12.5 The Physical Pendulum
P12.22
The period in Tokyo is
TT = 2π
LT
gT
and the period in Cambridge is
TC = 2π
LC
gC
We know
TT = TC = 2.00 s
For which, we see
LT LC
=
gT gC
or
g C LC 0.994 2
=
=
= 1.001 5
g T LT 0.992 7
340
P12.23
Oscillatory Motion
Using the simple harmonic motion model:
A = rθ = 1 m 15°
π
180°
= 0.262 m
g
9.8 m s 2
=
= 3.13 rad s
L
1m
ω=
(a)
v max = Aω = 0.262 m 3.13 s = 0.820 m s
(b)
a max = Aω 2 = 0.262 m 3.13 s
b
a tan = rα
α=
g
2
r
= 2.57 m s 2
a tan 2.57 m s 2
=
= 2.57 rad s 2
1m
r
r
F = ma = 0.25 kg 2.57 m s 2 = 0.641 N
(c)
FIG. P12.23
More precisely,
(a)
1
2
mv max
2
mgh =
a
h = L 1 − cos θ
and
a
f
f
∴ v max = 2 gL 1 − cos θ = 0.817 m s
Iα = mgL sin θ
(b)
α max =
P12.25
2
mL
=
g
sin θ i = 2.54 rad s 2
L
a fa
f
Fmax = mg sin θ i = 0.250 9.80 sin 15.0° = 0.634 N
(c)
P12.24
mgL sin θ
ω=
2π
:
T
T=
ω=
g
:
L
L=
2π
ω
g
ω
2
=
=
2π
= 1.42 s
4.43
9.80
a4.43f
2
= 0.499 m
v v v
mg mg mg
Referring to the sketch we have
F = − mg sin θ
and
tan θ =
x
R
For small displacements,
tan θ ≈ sin θ
and
F=−
mg
x = − kx
R
Since the restoring force is proportional to the displacement from
equilibrium, the motion is simple harmonic motion.
Comparing toF = − mω 2 x shows ω =
k
=
m
g
.
R
r
mg
FIG. P12.25
Chapter 12
*P12.26
T=
(a)
total measured time
50
T 2, s 2
The measured periods are:
a f
af
Length, L m
Period, T s
T = 2π
(b)
L
g
4
1.000 0.750 0.500
1.996 1.732 1.422
3
2
g=
so
2
4π L
T2
1
0
The calculated values for g are:
af
Period, T s
e
g m s
2
j
9.91
From T 2 =
Thus, g =
P12.27
0.25
1.996 1.732 1.422
9.87
Thus, g ave = 9.85 m s 2
(c)
0.5
0.75
1.0
L, m
FIG. P12.26
9.76
this agrees with the accepted value of g = 9.80 m s 2 within 0.5%.
F 4π I L , the slope of T
GH g JK
2
2
versus L graph =
4π 2
= 4.01 s 2 m .
g
4π 2
= 9.85 m s 2 . This is the same as the value in (b).
slope
f = 0.450 Hz , d = 0.350 m, and m = 2.20 kg
T=
1
;
f
T = 2π
I =T
2
I
;
mgd
mgd
4π
2
4π 2 I
mgd
T2 =
F 1I
=G J
H fK
2
mgd
4π
2
=
a fa f =
e0.450 s j
2.20 9.80 0.350
4π
2
−1 2
0.944 kg ⋅ m 2
r
mg
FIG. P12.27
P12.28
(a)
The parallel-axis theorem:
I = I CM + Md 2 =
=M
FG 13 m IJ
H 12 K
a
f
1
1
ML2 + Md 2 =
M 1.00 m
12
12
2
a
f
+ M 1.00 m
2
2
e
j
M 13 m 2
I
13 m
T = 2π
= 2π
= 2π
= 2.09 s
Mgd
12 Mg 1.00 m
12 9.80 m s 2
(b)
a
f
e
j
FIG. P12.28
For the simple pendulum
T = 2π
341
1.00 m
= 2.01 s
9.80 m s 2
difference =
2.09 s − 2.01 s
= 4.08%
2.01 s
342
*P12.29
Oscillatory Motion
(a)
The parallel axis theorem says directly I = I CM + md 2
so
(b)
eI
I
= 2π
T = 2π
mgd
CM
+ md 2
j
mgd
When d is very large T → 2π
d
gets large.
g
When d is very small T → 2π
I CM
gets large.
mgd
So there must be a minimum, found by
j bmgdg
F 1I
FG 1 IJ eI
= 2π e I
+ md j G − J bmgd g
mg + 2π bmgd g
H 2K
H 2K
−π e I
+ md jmg
2π md mgd
=
+
=0
eI + md j bmgdg eI + md j bmgdg
dT
d
2π I CM + md 2
=0=
dd
dd
e
−1 2
2 12
CM
−3 2
2 12
32
CM
This requires
− I CM − md 2 + 2md 2 = 0
or
I CM = md 2 .
Section 12.6 Damped Oscillations
P12.30
1
1
mv 2 + kx 2
2
2
The total energy is
E=
Taking the time-derivative,
dE
d2x
= mv 2 + kxv
dt
dt
Use Equation 12.28:
md 2 x
= − kx − bv
dt 2
a
f
dE
= v − kx − bv + kvx
dt
Thus,
−1 2
2
CM
CM
12
dE
= − bv 2 < 0
dt
2 12
32
CM
+ md 2
j
−1 2
2md
Chapter 12
P12.31
b
θ i = 15.0°
g
θ t = 1 000 = 5.50°
x1 000
x = Ae − bt 2 m
xi
Ae − bt 2 m 5.50
− b 1 000 g
=
=e b
15.0
A
=
2m
FG 5.50 IJ = −1.00 = −bb1 000g
H 15.0 K
2m
ln
∴
P12.32
b
= 1.00 × 10 −3 s −1
2m
b
Show that
x = Ae − bt 2 m cos ω t + φ
is a solution of
− kx − b
where
ω=
b
x = Ae − bt 2 m cos ω t + φ
FG
H
g
dx
d2x
=m 2
dt
dt
(1)
FG IJ
H K
(2)
k
b
−
m
2m
2
.
g
(3)
IJ b g
b g
K
FG − b IJ cosbω t + φ g − Ae ω sinbω t + φ gOP
H 2m K
Q
L
FG − b IJ ω sinbω t + φ g + Ae ω
− M Ae
H 2m K
N
dx
b
cos ω t + φ − Ae − bt 2 mω sin ω t + φ
= Ae − bt 2 m −
2m
dt
LM
N
d2x
b
=−
Ae − bt 2 m
2
2
m
dt
(4)
− bt 2 m
− bt 2 m
− bt 2 m
2
b
cos ω t + φ
gOPQ
(5)
Substitute (3), (4) into the left side of (1) and (5) into the right side of (1);
2
b g 2bm Ae cosbω t + φ g + bωAe
FG − b IJ cosbω t + φ g − Ae ω sinbω t + φ gOP
H 2m K
Q
ω sinbω t + φ g − mω Ae
cosbω t + φ g
− kAe − bt 2 m cos ω t + φ +
=−
+
LM
N
− bt 2 m
b
Ae − bt 2 m
2
b − bt 2 m
Ae
2
− bt 2 m
b
sin ω t + φ
− bt 2 m
2
b
− bt 2 m
g
b
g
Compare the coefficients of Ae − bt 2 m cos ω t + φ and Ae − bt 2 m sin ω t + φ :
cosine-term: − k +
sine-term:
FG
H
F
GH
IJ
K
I
JK
b2
b
b
b2
k
b2
b2
=−
−
− mω 2 =
−m
−
=
−
k
+
2m
2
2m
4m
2m
m 4m 2
bω = +
af af
b
b
ω + ω = bω
2
2
b
g
Since the coefficients are equal, x = Ae − bt 2 m cos ω t + φ is a solution of the equation.
g
343
344
Oscillatory Motion
Section 12.7 Forced Oscillations
P12.33
b g
F = 3.00 sin 2π t N
2π
= 2π rad s
T
(a)
ω=
(b)
In this case,
and
k = 20.0 N m
so
T = 1.00 s
20.0
= 3.16 rad s
2.00
k
=
m
ω0 =
The equation for the amplitude of a driven oscillator,
P12.34
(a)
2
− ω 02
j
−1
a f
3
4π 2 − 3.16
2
Thus
A = 0.050 9 m = 5.09 cm .
=
2 −1
For resonance, her frequency must match
ω0
1
=
2π 2π
k
1
=
m 2π
4.30 × 10 3 N m
= 2.95 Hz .
12.5 kg
dx
dv
= − Aω sin ω t , and a =
= − Aω 2 cos ω t , the maximum
dt
dt
acceleration is Aω 2 . When this becomes equal to the acceleration due to gravity, the normal
force exerted on her by the mattress will drop to zero at one point in the cycle:
From x = A cos ωt , v =
2
Aω = g
P12.35
0
A=
f0 =
(b)
FG F IJ eω
H mK
with b = 0, gives
F0 cos ω t − kx = m
b
x = A cos ω t + φ
or
d2x
dt 2
A=
g
ω2
ω0 =
=
g
k
m
e9.80 m s jb12.5 kg g =
A=
2
gm
=
k
4.30 × 10 3 N m
2.85 cm
k
m
(1)
g
b
(2)
dx
= − Aω sin ω t + φ
dt
g
d2x
= − Aω 2 cos ω t + φ
dt 2
b
(3)
g
(4)
b
g e
j b
Substitute (2) and (4) into (1):
F0 cos ω t − kA cos ω t + φ = m − Aω 2 cos ω t + φ
Solve for the amplitude:
ekA − mAω j cosbω t + φ g = F cos ω t
2
0
These will be equal, provided only that φ must be zero and kA − mAω 2 = F0
Thus, A =
F0 m
b k mg − ω
2
g
Chapter 12
P12.36
P12.37
345
Fext m
A=
eω
2
− ω 02
j + b bω m g
2
2
Fext m
Fext m
With b = 0,
A=
Thus,
ω 2 = ω 02 ±
This yields
ω = 8.23 rad s or ω = 4.03 rad s
Then,
f=
e
j
2
ω 2 − ω 02
ω
2π
=
e
±ω
2
− ω 02
j
=±
Fext m
ω 2 − ω 02
Fext m k Fext 6.30 N m
1.70 N
= ±
=
±
0.150 kg
A
m mA
0.150 kg 0.440 m
b
gives either f = 1.31 Hz
or
f
ga
f = 0.641 Hz
The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm:
f=
1
2π
g
1
=
L 2π
9.80 m s 2
= 1.74 Hz .
0.082 1 m
Section 12.8 Context Connection—Resonance in Structures
P12.38
For the resonance vibration with the occupants in the car, we have for the spring constant of the
suspension
f=
1
2π
k
m
e j d1 130 kg + 4b72.4 kg gi = 1.82 × 10
F 4b72.4 kg ge9.8 m s j
x= =
= 1.56 × 10 m
k = 4π 2 f 2 m = 4π 2 1.8 s −1
2
5
kg s 2
2
Now as the occupants exit
P12.39
k
1.82 × 10 5 kg s 2
−2
Suppose a 100-kg biker compresses the suspension 2.00 cm.
Then,
k=
980 N
F
=
= 4.90 × 10 4 N m
x 2.00 × 10 −2 m
If total mass of motorcycle and biker is 500 kg, the frequency of free vibration is
f=
1
2π
1
k
=
m 2π
4.90 × 10 4 N m
= 1.58 Hz
500 kg
If he encounters washboard bumps at the same frequency, resonance will make the motorcycle
bounce a lot. Assuming a speed of 20.0 m/s, we find these ridges are separated by
20.0 m s
1.58 s −1
= 12.7 m ~ 10 1 m .
In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can also
vibrate at higher frequencies by rocking back and forth between front and rear wheels, by having
just the front wheel bounce inside its fork, or by doing other things. Other spacing of bumps will
excite all of these other resonances.
346
Oscillatory Motion
Additional Problems
P12.40
(a)
Total energy =
b
f
ga
1 2 1
kA = 100 N m 0.200 m
2
2
2
= 2.00 J
At equilibrium, the total energy is:
b
g
b
g
b
g
1
1
m1 + m 2 v 2 = 16.0 kg v 2 = 8.00 kg v 2 .
2
2
Therefore,
b8.00 kg gv
2
= 2.00 J , and v = 0.500 m s .
This is the speed of m1 and m 2 at the equilibrium point. Beyond this point, the mass m 2
moves with the constant speed of 0.500 m/s while mass m1 starts to slow down due to the
restoring force of the spring.
The energy of the m1 -spring system at equilibrium is:
(b)
b
gb
1
1
m1 v 2 = 9.00 kg 0.500 m s
2
2
This is also equal to
g
2
= 1.125 J .
a f
1
2
k A′ , where A ′ is the amplitude of the m1 -spring system.
2
Therefore,
a fa f
1
100 A ′
2
2
= 1.125 or A ′ = 0.150 m.
The period of the m1 -spring system is T = 2π
m1
= 1.885 s
k
1
T = 0.471 s after it passes the equilibrium point for the spring to become fully
4
stretched the first time. The distance separating m1 and m 2 at this time is:
and it takes
D=v
P12.41
F d xI
GH dt JK
2
= Aω 2
2
max
fmax = µ sn = µ s mg = mAω 2
A=
µsg
= 6.62 cm
ω2
FG T IJ − A ′ = 0.500 m s a0.471 sf − 0.150 m = 0.085 6 =
H 4K
8.56 cm .
µs
B
P
nr
n
rr
fff
B
r
mg
mg
FIG. P12.41
Chapter 12
*P12.42
(a)
347
The spring constant of this spring is
k=
F 0.45 kg 9.8 m s 2
=
= 12.6 N m
0.35 m
x
we take the x-axis pointing downward, so φ = 0
x = A cos ωt = 18.0 cm cos
(d)
12.6 kg
0.45 kg ⋅ s 2
84. 4 s = 18.0 cm cos 446.6 rad = 15.8 cm
a f
Now 446.6 rad = 71 × 2π + 0.497 rad . In each cycle the object moves 4 18 = 72 cm , so it has
a
f a
f
moved 71 72 cm + 18 − 15.8 cm = 51.1 m .
(b)
By the same steps, k =
x = A cos
(e)
0.44 kg 9.8 m s 2
= 12.1 N m
0.355 m
k
12.1
t = 18.0 cm cos
84.4 = 18.0 cm cos 443.5 rad = −15.9 cm
m
0.44
a f
443.5 rad = 70 2π + 3.62 rad
a
f
Distance moved = 70 72 cm + 18 + 15.9 cm = 50.7 m
(c)
P12.43
The answers to (d) and (e) are not very different given the difference in the data about the
two vibrating systems. But when we ask about details of the future, the imprecision in our
knowledge about the present makes it impossible to make precise predictions. The two
oscillations start out in phase but get completely out of phase.
Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of one
proton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as D
and to the diatomic molecule of hydrogen-1 as H.
ωD
=
ωH
MD = 2MH
P12.44
ω=
k 2π
=
T
m
(a)
k = ω 2m =
(b)
m′ =
a f
k T′
4π 2
4π 2 m
T2
2
= m
FG T ′ IJ
HTK
2
k
MD
k
MH
=
MH
=
MD
1
2
fD =
fH
2
= 0.919 × 10 14 Hz
348
*P12.45
Oscillatory Motion
(a)
The period of the swinging rod is
T = 2π
I
= 2π
mgd
b1 3gml
2
mgl 2
θ max
2l
3g
= 2π
stride length
T
2l
The time for one half a cycle is = π
. The
2
3g
distance traveled in this time is the stride length
2 l sin θ max , so the speed is
2 l3 g sin θ max
d 2 l sin θ max
=
=
t
π 2l 3 g
(b)
6 gl sin θ max
=
π
π
We use the more precise expression
b
g
6 gl cos θ max 2 sin θ max
π
(c)
FIG. P12.45(a)
=
ja
e
f
6 9.8 m s 2 0.85 m cos 14° sin 28°
π
With
v old =
v new =
b
g
6 gl old cos θ max 2 sin θ max
b
π
g
6 gl new cos θ max 2 sin θ max
π
dividing gives
v new
=
v old
l new
l old
=2
l new
= 22
0.85 m
l new = 3.40 m
= 1.04 m s
349
Chapter 12
*P12.46
(a)
Consider the first process of
spring compression. It
continues as long as glider 1 is
moving faster than glider 2.
The spring instantaneously
has maximum compression
when both gliders are moving
with the same speed v a .
1
2
1
2
FIG. P12.46(a)
Momentum conservation:
m 1 v 1i + m 2 v 2 i = m 1 v 1 f + m 2 v f
b0.24 kg gb0.74 m sg + b0.36 kg gb0.12 m sg = b0.24 kg gv + b0.36 kg gv
a
0.220 8 kg ⋅ m s
= va
0.60 kg
(b)
r
v a = 0.368 m s $i
Energy conservation:
bK
1
+ K 2 + Us
g = bK
i
1
+ K 2 + Us
g
f
b
g
gb
1
1
1
1
m1 v12i + m 2 v 22i + 0 = m1 + m 2 v a2 + kx 2
2
2
2
2
1
1
2
0.24 kg 0.74 m s + 0.36 kg 0.12 m s
2
2
1
0.068 3 J = 0.040 6 J + 45 N m x 2
2
b
gb
g
b
b
F 2b0.027 7 Jg I
x=G
H 45 N m JK
(c)
a
g
g
2
=
b
gb
1
0.60 kg 0.368 m s
2
g
2
+
b
g
1
45 N m x 2
2
12
= 0.035 1 m
Conservation of momentum guarantees that the center of mass moves with constant
velocity. Imagine viewing the gliders from a reference frame moving with the center of
mass. We see the two gliders approach each other with momenta in opposite directions of
equal magnitude. Upon colliding they compress the spring and then together bounce,
extending and compressing it cyclically.
b
gb
(d)
1
1
2
m tot vCM
= 0.60 kg 0.368 m s
2
2
(e)
1 2 1
kA = 45 N m 0.035 1 m
2
2
b
gb
g
2
g
2
= 0.040 6 J
= 0.027 7 J
350
P12.47
Oscillatory Motion
r
Hy
We draw a free-body diagram of the pendulum.
r
The force H exerted by the hinge causes no torque
about the axis of rotation.
τ = Iα
r
Hx
d 2θ
= −α
dt 2
and
r
kx
The negative sign appears because clockwise θ is
counted as positive in the diagram.
τ = MgL sin θ + kxh cos θ = − I
r
mg
2
d θ
dt 2
FIG. P12.47
For small amplitude vibrations, use the
approximations: sin θ ≈ θ , cos θ ≈ 1, and x ≈ s = hθ .
Therefore,
d 2θ
dt 2
=−
F MgL + kh Iθ = −ω θ
GH I JK
2
2
MgL + kh 2
ω=
ML2
1
2π
f=
P12.48
(a)
In
b
g
x = A cos ω t + φ ,
v = −ω A sin φ = − v max
This requires φ = 90° , so
x = A cos ω t + 90°
And this is equivalent to
x = − A sin ω t
Numerically we have
ω=
and v max = ωA
20 m s = 10 s −1 A
In
b
continued on next page
g
50 N m
k
=
= 10 s −1
m
0.5 kg
e
a
1
1
1
mv 2 + kx 2 = kA 2 ,
2
2
2
j
f e
FG
H
1 2
1
kx = 3 mv 2
2
2
A=2m
j
g
L
IJ
K
11 2 1 2 1 2
kx + kx = kA
32
2
2
x=±
ω=
ML2
x = −2 m sin 10 s −1 t
implies
(c)
MgL + kh 2
g
we have at t = 0
So
(b)
b
v = −ω A sin ω t + φ
= 2π f
L=
g
ω2
3
A = ±0.866 A = ±1.73 m
4
=
9.8 m s 2
e10 s j
−1 2
= 0.098 0 m
4 2
x = A2
3
Chapter 12
(d)
a
f e
351
j
x = −2 m sin 10 s −1 t
In
the particle is at x = 0 at t = 0 , at 10t = π s , and so on.
The particle is at
x =1 m
when
−
with solutions
e10 s jt = − π6
1
= sin 10 s −1 t
2
e
j
−1
e10 s jt = π + π6 , and so on.
−1
FIG. P12.48(d)
The minimum time for the motion is ∆t in
10 ∆t =
FG π IJ s
H 6K
∆t =
P12.49
(a)
FG π IJ s =
H 60 K
0.052 4 s
At equilibrium, we have
F LI
∑ τ = 0 − mg GH 2 JK + kx0 L
where x 0 is the equilibrium compression.
FIG. P12.49
After displacement by a small angle,
F LI
F LI
∑ τ = −mg GH 2 JK + kxL = −mg GH 2 JK + kbx0 − Lθ gL = − kθL2
But,
1
3
∑ τ = Iα = mL2
d 2θ
.
dt 2
So
d 2θ
3k
= − θ.
m
dt 2
The angular acceleration is opposite in direction and proportional to the displacement, so
3k
we have simple harmonic motion with ω 2 =
.
m
(b)
f=
ω
1
=
2π 2π
3k
1
=
m 2π
b
g=
3 100 N m
5.00 kg
1.23 Hz
352
P12.50
Oscillatory Motion
As it passes through equilibrium, the 4-kg object has speed
v max = ωA =
100 N m
k
2 m = 10.0 m s.
A=
4 kg
m
In the completely inelastic collision momentum of the two-object system is conserved. So the new
10-kg object starts its oscillation with speed given by
b
g b g b
g
4 kg 10 m s + 6 kg 0 = 10 kg v max
v max = 4.00 m s
(a)
The new amplitude is given by
1
1
2
mv max
= kA 2
2
2
b
10 kg 4 m s
g = b100 N mgA
2
2
A = 1.26 m
(b)
(c)
Thus the amplitude has decreased by
2.00 m − 1.26 m = 0.735 m
The old period was
T = 2π
4 kg
m
= 2π
= 1.26 s
100 N m
k
The new period is
T = 2π
10 2
s = 1.99 s
100
The period has increased by
1.99 m − 1.26 m = 0.730 s
The old energy was
1
1
2
mv max
= 4 kg 10 m s
2
2
The new mechanical energy is
1
10 kg 4 m s
2
b gb
b
gb
g
2
g
2
= 200 J
= 80 J
The energy has decreased by 120 J .
P12.51
(d)
The missing mechanical energy has turned into internal energy in the completely inelastic
collision.
(a)
T=
(b)
E=
(c)
At maximum angular displacement
2π
ω
= 2π
L
= 3.00 s
g
a fa f
1
1
mv 2 = 6.74 2.06
2
2
a
h = L − L cos θ = L 1 − cos θ
2
f
= 14.3 J
mgh =
1
mv 2
2
cos θ = 1 −
h
L
h=
v2
= 0.217 m
2g
θ = 25.5°
Chapter 12
P12.52
353
One can write the following equations of motion:
T − kx = 0
(describes the spring)
mg − T ′ = ma = m
a
f
R T′ −T = I
d 2x
dt 2
(for the hanging object)
d 2θ I d 2 x
=
dt 2 R dt 2
(for the pulley)
with I =
FIG. P12.52
1
MR 2
2
Combining these equations gives the equation of motion
FG m + 1 MIJ d x + kx = mg .
H 2 K dt
2
2
af
mg
mg
(where
arises because of the extension of the spring due to
k
k
the weight of the hanging object), with frequency
The solution is x t = A sin ωt +
f=
P12.53
ω
1
=
2π 2π
1
k
=
1
m + 2 M 2π
(a)
For M = 0
f = 3.56 Hz
(b)
For M = 0.250 kg
f = 2.79 Hz
(c)
For M = 0.750 kg
f = 2.10 Hz
(a)
∑ F = −2T sin θ $j
r
where θ = tan −1
100 N m
.
0.200 kg + 12 M
FG y IJ
H LK
Therefore, for a small displacement
sin θ ≈ tan θ =
(b)
y
L
FIG. P12.53
r −2Ty
∑ F = L $j
and
The total force exerted on the ball is opposite in direction and proportional to its
displacement from equilibrium, so the ball moves with simple harmonic motion. For a
spring system,
r
r
∑ F = −kx
becomes here
Therefore, the effective spring constant is
2T
and
L
r
∑F = −
ω=
2T r
y.
L
k
=
m
2T
.
mL
354
P12.54
Oscillatory Motion
(a)
r
v
For each segment of the spring
dK =
vx =
Also,
x
v
l
a f
1
dm v x2 .
2
and
m
dx .
l
dm =
FIG. P12.54
Therefore, the total kinetic energy of the block-spring system is
K=
(b)
ω=
k
m eff
*P12.55
(a)
T=
z FGH
l
0
FG
H
I
JK
FG
H
IJ
K
1
x2v2 m
m 2
dx =
M+
v .
l
2
3
l2
IJ
K
1
1
m 2
m eff v 2 =
M+
v
2
2
3
and
Therefore,
1
1
Mv 2 +
2
2
2π
ω
M + m3
= 2π
.
k
∆K + ∆ U = 0
M
R
Thus, K top + U top = K bot + U bot
where K top = U bot = 0
Therefore, mgh =
θ θ
1 2
Iω , but
2
a
h = R − R cos θ = R 1 − cos θ
ω=
and
I=
r
v
f
FIG. P12.55
v
R
MR 2 mr 2
+
+ mR 2
2
2
Substituting we find
f 12 FGH MR2 + mr2 + mR IJK Rv
L M mr + m OPv
mgRa1 − cos θ f = M +
N 4 4R 2 Q
a1 − cos θ f
v = 4 gR
e + + 2j
Rg a1 − cos θ f
v=2
b M m g + er R j + 2
2
a
mgR 1 − cos θ =
2
2
2
and
so
continued on next page
2
M
m
r2
R2
2
2
m
2
2
2
2
Chapter 12
(b)
I
mT gd CM
T = 2π
mT = m + M
T = 2π
P12.56
(a)
1
2
dCM =
af
mR + M 0
m+M
MR 2 + 12 mr 2 + mR 2
mgR
Assuming a Hooke’s Law type spring,
F = Mg = kx
and empirically
Mg = 1.74x − 0.113
so
(b)
k = 1.74 N m ± 6% .
M , kg
0.020 0
x, m
0.17
Mg , N
0.196
0.040 0
0.293
0.392
0.050 0
0.353
0.49
0.060 0
0.413
0.588
0.070 0
0.080 0
0.471
0.493
0.686
0.784
We may write the equation as theoretically
T2 =
4π 2
4π 2
M+
ms
k
3k
FIG. P12.56
and empirically
T 2 = 21.7 M + 0.058 9
so
k=
4π 2
= 1.82 N m ± 3%
21.7
Time, s T , s
7.03 0.703
M , kg
0.020 0
T 2 , s2
0.494
9.62
0.962
0.040 0
0.925
10.67
1.067
0.050 0
1.138
11.67
1.167
0.060 0
1.362
12.52
1.252
0.070 0
1.568
13.41
1.341
0.080 0
1.798
continued on next page
355
356
Oscillatory Motion
The k values 1.74 N m ± 6%
and
1.82 N m ± 3% differ by 4%
so
(c)
they agree.
Utilizing the axis-crossing point,
ms = 3
FG 0.058 9 IJ kg =
H 21.7 K
8 grams ± 12%
in agreement with 7.4 grams.
P12.57
(a)
We require Ae − bt 2 m =
A
2
e + bt 2 m = 2
bt
= ln 2
2m
or
or
0.100 kg s
t = 0.693
2 0.375 kg
b
∴ t = 5.20 s
g
The spring constant is irrelevant.
(b)
We can evaluate the energy at successive turning points, where
b
g
cos ωt + φ = ±1 and the energy is
e + bt m = 2
or
(c)
From E =
dE
dt
E
=
e
d 1
dt 2
1
2
FG
H
1 2 1 2 −2 bt 2 m
1
1 1 2
kx = kA e
. We require kA 2 e − bt m =
kA
2
2
2
2 2
∴t =
a
IJ
K
f
m ln 2 0.375 kg 0.693
=
= 2.60 s .
b
0.100 kg s
1 2
kA , the fractional rate of change of energy over time is
2
kA 2
kA 2
j = ka 2 Af
1
2
1
2
kA 2
dA
dt
=2
dA
dt
A
two times faster than the fractional rate of change in amplitude.
P12.58
(a)
The block moves with the board in what we take as the positive x direction, stretching the
spring until the spring force − kx is equal in magnitude to the maximum force of static
µ mg
friction µ sn = µ s mg . This occurs at x = s
.
k
(b)
Since v is small, the block is nearly at the rest at this break point. It starts almost immediately
to move back to the left, the forces on it being − kx and + µ k mg . While it is sliding the net
force exerted on it can be written as
− kx + µ k mg = − kx +
FG
H
where x rel is the excursion of the block away from the point
continued on next page
IJ
K
kµ k mg
µ mg
= −k x − k
= − kx rel
k
k
µ k mg
k
.
357
Chapter 12
Conclusion: the block goes into simple harmonic motion centered about the equilibrium
µ mg
position where the spring is stretched by k
.
k
(d)
The amplitude of its motion is its original displacement, A =
rest at spring extension
µ k mg
b2 µ
k
g
− µ s mg
k
−
µ k mg
k
. It first comes to
. Almost immediately at this point it
k
k
latches onto the slowly-moving board to move with the board. The board exerts a force of
static friction on the block, and the cycle continues.
(c)
−A=
µ s mg
The graph of
the motion
looks like this:
FIG. P12.58(c)
(e)
b
g
2 A 2 µ s − µ k mg
=
.
v
kv
The time for which the block is springing back is one half a cycle of simple harmonic motion,
The time during each cycle when the block is moving with the board is
F
GH
I
JK
1
m
m
2π
=π
. We ignore the times at the end points of the motion when the speed of
2
k
k
2A
, these
the block changes from v to 0 and from 0 to v. Since v is small compared to
π
m
k
times are negligible. Then the period is
T=
(f)
T=
a
fb
ge
2 0.4 − 0.25 0.3 kg 9.8 m s 2
b0.024 m sgb12 N mg
Then
f=
b
g
2 µ s − µ k mg
j +π
b
g
2 µ s − µ k mg
kv
+π
0.3 kg
= 3.06 s + 0.497 s = 3.56 s
12 N m
1
= 0.281 Hz .
T
+π
m
increases as m increases, so the frequency decreases .
k
(g)
T=
(h)
As k increases, T decreases and f increases .
(i)
As v increases, T decreases and f increases .
(j)
As µ s − µ k increases, T increases and f decreases .
b
kv
g
m
.
k
358
P12.59
Oscillatory Motion
(a)
When the mass is displaced a distance x from
equilibrium, spring 1 is stretched a distance x1 and
spring 2 is stretched a distance x 2 .
By Newton’s third law, we expect
k1 x1 = k 2 x 2 .
When this is combined with the requirement that
x = x1 + x 2 ,
FIG. P12.59
LM k OP x
Nk + k Q
L k k OP x = ma
F =M
Nk + k Q
x1 =
we find
The force on either spring is given by
1
2
1
2
1 2
1
2
where a is the acceleration of the mass m.
(b)
This is in the form
F = k eff x = ma
and
T = 2π
b
so that
(a)
(b)
g
In this case each spring is distorted by the distance x which the mass is displaced. Therefore,
the restoring force is
g
F = − k1 + k 2 x
P12.60
b
m k1 + k 2
m
= 2π
k eff
k1 k 2
T = 2π
b
m
k1 + k 2
and
g
k eff = k 1 + k 2
.
Newton’s law of universal gravitation is
F=−
Thus,
F=−
Which is of Hooke’s law form with
k=
The sack of mail moves without friction according to
−
GMm
r2
=−
FG
H
FG 4 π ρ GmIJ r
H3
K
4
π ρ Gm
3
FG 4 IJ π ρ Gmr = ma
H 3K
F 4I
a = −G J π ρ Gr = −ω r
H 3K
2
continued on next page
IJ
K
Gm 4 3
πr ρ
r2 3
Chapter 12
359
Since acceleration is a negative constant times excursion from equilibrium, it executes SHM
with
4π ρ G
3
ω=
The time for a one-way trip through the Earth is
T
=
2
We have also
g=
so
(c)
T=
and period
4ρ G
g
=
3
π Re
b g
and
2π
ω
3π
ρG
=
3π
4ρ G
GM e
R e2
=
G 4π R e3 ρ
3 R e2
=
4
π ρ GR e
3
Re
T
6.37 × 10 6 m
=π
=π
= 2.53 × 10 3 s = 42.2 min .
2
g
9.8 m s 2
For an object moving without friction in a circular orbit just above the Earth’s surface, we
2
GMm
v 2 m 2π R e
have ∑ F = ma
=m
=
Re Re
T
Re2
FG
H
mg =
4π 2 mR e
T
2
T = 2π
Re
g
so
IJ
K
Re
T
=π
2
g
The treetop satellite takes the same time to reach the antipode as the sack of mail.
They arrive at the same time.
ANSWERS TO EVEN PROBLEMS
P12.2
(a) 4.33 cm; (b) −5.00 cm s ;
(c) −17.3 cm s 2 ; (d) 5.00 cm; 3.14 s as the
particle passes through equilibrium
P12.4
(a) 20.0 cm;
(b) 94.2 cm/s as the particle passes through
equilibrium;
(c) 17.8 m s 2 at maximum excursion from
equilibrium
P12.6
(a) –2.34 m; (b) –1.30 m/s; (c) –0.076 3 m;
(d) 0.315 m/s
P12.8
(a) 2.40 s; (b) 0.417 Hz; (c) 2.62 rad/s
P12.10
12.0 Hz
P12.12
(a) 1.26 s; (b) 0.150 m s; 0.750 m s 2 ;
(c) x = −3.00 cos 5.00t cm ,
a
f
a f
a = 75.0 cosa5.00t f cm s
v = 15.0 sin 5.00t cm s ,
2
P12.14
(a) 126 N m; (b) 0.178 m
P12.16
(a) 0.153 J; (b) 0.784 m s; (c) 17.5 m s 2
P12.18
(a) 100 N m; (b) 1.13 Hz;
(c) 1.41 m s as the block passes through
equilibrium;
(d) 10.0 m s 2 at maximum excursion from
equilibrium;
(e) 2.00 J; (f) 1.33 m s ; (g) 3.33 m s 2
360
P12.20
P12.22
Oscillatory Motion
(a) 1.50 s; (b) 73.4 N m;
(c) 19.7 m below the bridge; (d) 1.06 rad s;
(e) 2.01 s; (f) 3.50 s
g Cambridge
g Tokyo
= 1.001 5
P12.46
(a) 0.368 $i m s ; (b) 3.51 cm;
(c) see the solution; (d) 0.040 6 J ;
(e) 0.027 7 J
P12.48
(a) x = −2 m sin 10 s −1 t ;
a
f e
j
(b) at x = ±1.73 m ; (c) 0.098 0 m ;
(d) 0.052 4 s
P12.24
1.42 s; 0.499 m
P12.26
(a) see the solution;
(b) 9.85 m s 2 agreeing with the accepted
value within 0.5%;
(c) 9.85 m s 2
P12.50
(a) decreased by 0.735 m;
(b) increased by 0.730 s;
(c) decreased by 120 J;
(d) see the solution
P12.28
(a) 2.09 s; (b) 4.08%
P12.52
(a) 3.56 Hz ; (b) 2.79 Hz; (c) 2.10 Hz
P12.30
see the solution
P12.54
(a)
P12.32
see the solution
P12.34
(a) 2.95 Hz; (b) 2.85 cm
P12.56
P12.36
either 1.31 Hz or 0.641 Hz
(a) k = 1.74 N m ± 6% ;
(b) k = 1.82 N m ± 3% showing agreement;
(c) 8 grams ± 12% showing agreement
P12.38
1.56 cm
P12.58
P12.40
(a) 0.500 m s ; (b) 8.56 cm
P12.42
(a) 15.8 cm; (b) –15.9 cm;
(c) see the solution; (d) 51.1 m; (e) 50.7 m
(a) see the solution; (b) see the solution;
(c) see the solution; (d) see the solution;
(e) see the solution; (f) 0.281 Hz;
(g) decreases; (h) increases; (i) increases;
(j) decreases
P12.60
(a) see the solution; (b) 42.2 min ;
(c) they arrive at the same time
P12.44
(a)
FG IJ
H K
T′
4π 2 m
; (b) m
2
T
T
2
FG
H
IJ
K
M + m3
m 2
1
M+
v ; (b) 2π
2
3
k