15
Oscillatory Motion
CHAPTER OUTLINE
15.1
15.2
15.3
15.4
15.5
15.6
15.7
Q15.4
Motion of an Object
Attached to a Spring
Mathematical Representation
of Simple Harmonic Motion
Energy of the Simple
Harmonic Oscillator
Comparing Simple Harmonic
Motion with Uniform Circular
Motion
The Pendulum
Damped Oscillations
Forced Oscillations
ANSWERS TO QUESTIONS
Q15.1
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.
Q15.2
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 .
Q15.3
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.
(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.
Q15.5
No. It is necessary to know both the position and velocity at time zero.
Q15.6
The motion will still be simple harmonic motion, but the period of oscillation will be a bit larger. The
F kI
effective mass of the system in ω = G
H m JK
12
will need to include a certain fraction of the mass of the
eff
spring.
439
440
Q15.7
Oscillatory Motion
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.
Q15.8
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
Q15.9
Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2
are valid.
Equation
x t = A cos ωt + φ
Information given by equation
position as a function of time
v t = −ωA sin ωt + φ
velocity as a function of time
af
b g
af
b g
va x f = ±ω e A − x j
aat f = −ω A cosbωt + φ g
aat f = −ω xat f
2
2 12
velocity as a function of position
2
acceleration as a function of time
2
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.
Q15.10
Lf
Li
2 Li
and T f =
=
= 2Ti . The period gets larger by
g
g
g
mass has no effect on the period of a simple pendulum.
We have Ti =
2 times. Changing the
Q15.11
(a)
Q15.12
No, the equilibrium position of the pendulum will be shifted (angularly) towards the back of the car.
The period of oscillation will increase slightly, since the restoring force (in the reference frame of the
accelerating car) is reduced.
Q15.13
The motion will be periodic—that is, it will repeat. The period is nearly constant as the angular
amplitude increases through small values; then the period becomes noticeably larger as θ increases
farther.
Q15.14
Shorten the pendulum to decrease the period between ticks.
Q15.15
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.
Period decreases.
(b)
Period increases.
(c)
No change.
Chapter 15
441
Q15.16
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.
Q15.17
The phase constant must be π rad .
Q15.18
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
is greater
the board is greater than the stiffness constant describing the end. And then f =
2π m
for you bouncing on the center of the board.
FG IJ
H K
Q15.19
The release of air from one side of the parachute can make the parachute turn in the opposite
direction, causing it to release air from the opposite side. This behavior will result in a periodic driving
force that can set the parachute into side-to-side oscillation. If the amplitude becomes large enough,
the parachute will not supply the needed air resistance to slow the fall of the unfortunate skydiver.
Q15.20
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.
Q15.21
We assume the diameter of the bob is not very small compared to the length of the cord supporting
it. As the water leaks out, the center of mass of the bob moves down, increasing the effective length
of the pendulum and slightly lowering its frequency. As the last drops of water dribble out, the
center of mass of the bob hops back up to the center of the sphere, and the pendulum frequency
quickly increases to its original value.
SOLUTIONS TO PROBLEMS
Section 15.1
P15.1
Motion of an Object Attached to a Spring
(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
a
f
2 4.00 m
2x
=
= 0.909 s
g
9.80 m s 2
f
This equals one-half the period, so T = 2 0.909 s = 1.82 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.
442
Oscillatory Motion
Section 15.2
P15.2
P15.3
Mathematical Representation of Simple Harmonic Motion
π
6
IJ
K
x = 5.00 cm cos 2t +
(b)
v=
dx
π
= − 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
IJ
K
j FGH
e
f a
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=
f
b
2π
ω
=
4.33 cm
2π
= 3.14 s
2
g
x = 4.00 m cos 3.00πt + π Compare this with x = A cos ωt + φ to find
(a)
ω = 2π f = 3.00π
or
*P15.4
f FGH
a
(a)
T=
f = 1.50 Hz
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)
The spring constant of this spring is
a
f a
f a
f
k=
F 0.45 kg 9.8 m s 2
=
= 12.6 N m
x
0.35 m
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)
a f
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
443.5 rad = 70 2π + 3.62 rad
a
f
Distance moved = 70 72 cm + 18 + 15.9 cm = 50.7 m
(c)
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.
Chapter 15
P15.5
(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
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
P15.6
t=
T
1
=
s
2
3
t=
3
T = 0.500 s
4
= 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
(b)
443
2
s and A = 2.00 cm, the particle will travel 8.00 cm in this time.
3
3
Hence, in 1.00 s = T , the particle will travel
8.00 cm + 4.00 cm = 12.0 cm .
2
Since T =
FG
H
IJ
K
af
FG v IJ sin ωt
HωK
i
The proposed solution
x t = xi cos ωt +
implies velocity
v=
dx
= − x iω sin ωt + vi cos ωt
dt
and acceleration
a=
v
dv
= − x iω 2 cos ωt − viω sin ωt = −ω 2 x i cos ωt + i sin ωt = −ω 2 x
ω
dt
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 = − x iω sin ωt + vi cos ωt
b
g − e− x ω
2
i
2
jFGH
cos ωt − vi sin ωt xi cos ωt +
FG v IJ sin ωtIJ
HωK K
i
v 2 − ax = xi2ω 2 sin 2 ωt − 2 xi viω sin ωt cos ωt + vi2 cos 2 ωt
+ x i2ω 2 cos 2 ωt + x i viω cos ωt sin ωt + xi viω sin ωt cos ωt + vi2 sin 2 ωt = x i2ω 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.
P15.7
(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
a
f
444
*P15.8
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
k=
f=
P15.9
f=
ω
1
=
π
2
2π
k
m
14.3 N
F
=
= 52.0 N m
x 0.027 5 m
ω
2π
k=
x = A cos ωt
52 kg
k
1
=
m 2π
1
2π
T=
or
Solving for k,
*P15.10
=
s 2 9.11 × 10 −3 kg
= 12.0 Hz
1
m
= 2π
f
k
4π 2 m
A = 0.05 m
T
2
=
b
4π 2 7.00 kg
a2.60 sf
2
g=
40.9 N m .
a = − Aω 2 cos ωt
v = − Aω sin ωt
If f = 3 600 rev min = 60 Hz , then ω = 120π s −1
a
f
v max = 0.05 120π m s = 18.8 m s
P15.11
(a)
ω=
k
=
m
8.00 N m
= 4.00 s −1
0.500 kg
From this we find that
(b)
t=
FG 1 IJ sin FG x IJ and when
H 4.00 K H 10.0 K
−1
Using t =
f
2
m s 2 = 7.11 km s 2
a
a f
a = −160 sina 4.00t f cm s
v = 40.0 cos 4.00t cm s
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 =
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 .
f
x = 10.0 sin 4.00t cm .
so position is given by
v = 40.0 cos 4.00 0.161 = 32.0 cm s
We find
(c)
a
amax = 0.05 120π
−96.0 cm s 2 .
Chapter 15
P15.12
445
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
25.0
= 5.00 rad s
1.00
2π
2π
T=
=
= 1.26 s
ω 5.00
so that,
(b)
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 ωt
a
f
x = −3.00 cos 5.00t cm
or
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=
P15.13
The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.
ω=
b
2π
= 6. 28 s
T
f
ga
and v max = ωA = 6.28 s 0.100 m = 0.628 m s .
P15.14
(a)
v max = ωA
A=
(b)
Section 15.3
P15.15
(a)
v max
ω
=
v
ω
x = − A sin ωt = −
FG v IJ sin ωt
HωK
Energy of the Simple Harmonic Oscillator
Energy is conserved for the block-spring system between the maximum-displacement and
the half-maximum points:
aK + U f = aK + U f
1
b6.50 N mga0.100 mf
2
i
32.5 mJ =
k
=
m
0+
f
b
2
1
m 0.300 m s
2
b
=
1
m 0.300 m s
2
g
+ 8.12 mJ
2
6.50 N m
= 3.46 rad s
0.542 kg
(b)
ω=
(c)
amax = Aω 2 = 0.100 m 3.46 rad s
b
g
2
g
2
+
1 2 1
1
kA = mv 2 + kx 2
2
2
2
b
ge
1
6.50 N m 5.00 × 10 −2 m
2
m=
a
2 24.4 mJ
9.0 × 10
∴T =
= 1.20 m s 2
j
2π
ω
=
−2
f
2
m s2
2
= 0.542 kg
2π rad
= 1.81 s
3.46 rad s
446
P15.16
P15.17
Oscillatory Motion
b
P15.19
k = mω 2 = 0.200 kg 25.1 rad s
(b)
E=
kA 2
⇒A=
2
2E
=
k
2
126
Choose the car with its shock-absorbing bumper as the system; by conservation of energy,
k
= 3.16 × 10 −2 m
m
e
v=x
j
e
j
(a)
(b)
v max = Aω
(c)
a max = Aω 2 = 3.50 × 10 −2 m 22.4 s −1
(a)
E=
(b)
v = ω A2 − x2 =
e
b
j
2
ge
250
= 22.4 s −1
0.500
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
j
2
= 28.0 mJ
k
A2 − x2
m
35.0
e4.00 × 10 j − e1.00 × 10 j = 1.02 m s
50.0 × 10
1
1
1
1
mv = kA − kx = a35.0 fLe 4.00 × 10 j − e3.00 × 10 j O =
MN
PQ
2
2
2
2
v=
(c)
= 0.153 J
k
=
m
ω=
where
5.00 × 10 6
= 2.23 m s
10 3
2
−2
kA 2 250 N m 3.50 × 10 m
E=
=
2
2
.
P15.20
g = 126 N m
2a 2.00f
= 0.178 m
(a)
1
1
mv 2 = kx 2 :
2
2
P15.18
2π
2π
=
= 25.1 rad s
T
0. 250
m = 200 g , T = 0.250 s, E = 2.00 J ; ω =
−2 2
−3
2
2
−2 2
−2 2
2
(d)
1 2
1
kx = E − mv 2 = 15.8 mJ
2
2
(a)
k=
(b)
ω=
(c)
v max = ωA = 50.0 0. 200 = 1.41 m s at x = 0
(d)
amax = ω 2
(e)
E=
(f)
v = ω A 2 − x 2 = 50.0
(g)
a = ω 2 x = 50.0
−2 2
F
20.0 N
=
= 100 N m
x 0.200 m
k
= 50.0 rad s
m
1 2
kA
2
f=
so
a f
A = 50.0a0.200 f = 10.0 m s
1
= a100 fa0.200 f = 2.00 J
2
2
at x = ± A
2
FG 0.200 IJ =
H 3 K
a
8
0.200
9
f
2
3.33 m s 2
= 1.33 m s
ω
2π
= 1.13 Hz
12.2 mJ
Chapter 15
P15.21
(a)
E=
a f
1 2
1
kA , so if A ′ = 2 A , E ′ = k A ′
2
2
2
=
a f
1
k 2A
2
2
447
= 4E
Therefore E increases by factor of 4 .
*P15.22
(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
0 + mgy + 0 = 0 + 0 + kx 2
2
1
65 kg 9.8 m s 2 36 m = k 25 m
2
k = 73. 4 N m
a
(c)
The spring extension at equilibrium is x =
f
2
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 .
k
=
m
73.4 N m
= 1.06 rad s
65 kg
(d)
ω=
(e)
Take the phase as zero at maximum downward extension. We find what the phase was 25 m
higher when x = −8.68 m:
In x = A cos ωt ,
FG
H
t
−8.68 m = 16.3 m cos 1.06
s
t = −2.01 s
IJ
K
16.3 m = 16.3 m cos 0
t
1.06 = −122° = −2.13 rad
s
Then +2.01 s is the time over which the spring stretches.
(f)
total time = 1.50 s + 2.01 s = 3.50 s
448
P15.23
Oscillatory Motion
Model the oscillator as a block-spring system.
v2 + ω 2x2 = ω 2 A2
From energy considerations,
v max = ωA and v =
ωA
2
From this we find x 2 =
P15.24
3 2
A
4
so
FG ωA IJ
H2K
and
x=
2
+ ω 2x2 = ω 2 A2
3
A = ±2.60 cm where A = 3.00 cm
2
The potential energy is
a f
1 2 1 2
kx = kA cos 2 ωt .
2
2
Us =
The rate of change of potential energy is
a f
a f
dU s 1 2
1
= kA 2 cos ωt −ω sin ωt = − kA 2ω sin 2ωt .
2
2
dt
(a)
This rate of change is maximal and negative at
2ωt =
Then, t =
π
π
π
, 2ωt = 2π + , or in general, 2ωt = 2nπ + for integer n.
2
2
2
a
f
π 4n + 1
π
4n + 1 =
4ω
4 3.60 s −1
a
f
e
j
For n = 0 , this gives t = 0.218 s while n = 1 gives t = 1.09 s .
All other values of n yield times outside the specified range.
(b)
Section 15.4
P15.25
dU s
dt
=
max
1 2
1
kA ω = 3.24 N m 5.00 × 10 −2 m
2
2
b
ge
j e3.60 s j =
2
−1
14.6 mW
Comparing Simple Harmonic Motion with Uniform Circular Motion
(a)
The motion is simple harmonic because the tire is rotating with constant velocity and you
are looking at the motion of the bump projected in a plane perpendicular to the tire.
(b)
Since the car is moving with speed v = 3.00 m s , and its radius is 0.300 m, we have:
ω=
3.00 m s
= 10.0 rad s .
0.300 m
Therefore, the period of the motion is:
T=
2π
ω
=
2π
b10.0 rad sg =
0.628 s .
Chapter 15
P15.26
The angle of the crank pin is θ = ωt .
Its x-coordinate is
ω
Piston
x = A cos θ = A cos ωt
A
where A is the distance from the
center of the wheel to the crank pin.
This is of the form x = A cos ωt + φ ,
so the yoke and piston rod move
with simple harmonic motion.
b
Section 15.5
P15.27
(a)
P15.28
P15.29
g
FIG. P15.26
The Pendulum
T = 2π
L=
(b)
x = –A
gT 2
4π 2
L
g
e9.80 m s ja12.0 sf
2
=
Tmoon = 2π
4π 2
L
= 2π
g moon
2
= 35.7 m
35.7 m
1.67 m s 2
= 29.1 s
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
The swinging box is a physical pendulum with period T = 2π
I
.
mgd
The moment of inertia is given approximately by
I=
1
mL2 (treating the box as a rod suspended from one end).
3
Then, with L ≈ 1.0 m and d ≈
T ≈ 2π
1
3
L
,
2
mL2
mg
ch
L
2
= 2π
a
f
2 1.0 m
2L
= 2π
= 1.6 s or T ~ 10 0 s .
2
3g
3 9.8 m s
e
j
x ( t)
449
450
P15.30
P15.31
Oscillatory Motion
ω=
2π
:
T
T=
ω=
g
:
L
L=
2π
=
ω
2π
= 1.42 s
4. 43
g
9.80
=
= 0.499 m
2
ω2
4.43
a f
Using the simple harmonic motion model:
A = rθ = 1 m 15°
π
= 0. 262 m
180°
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
(c)
g
2
= 2.57 m s 2
a tan 2.57 m s 2
=
= 2.57 rad s 2
r
1m
α=
FIG. P15.31
F = ma = 0.25 kg 2.57 m s 2 = 0.641 N
More precisely,
(a)
1
mv 2
and
2
∴ v max = 2 gL 1 − cos θ = 0.817 m s
mgh =
a
(b)
f
f
Iα = mgL sin θ
α max =
P15.32
a
h = L 1 − cos θ
mgL sin θ
2
mL
=
g
sin θ i = 2.54 rad s 2
L
a fa
f
(c)
Fmax = mg sin θ i = 0.250 9.80 sin 15.0° = 0.634 N
(a)
The string tension must support the weight of the bob, accelerate it upward, and also provide
the restoring force, just as if the elevator were at rest in a gravity field 9.80 + 5.00 m s 2
a
T = 2π
L
5.00 m
= 2π
g
14.8 m s 2
T = 3.65 s
(b)
T = 2π
(c)
g eff =
5.00 m
e9.80 m s
2
− 5.00 m s 2
j
= 6.41 s
e9.80 m s j + e5.00 m s j
T = 2π =
2 2
5.00 m
11.0 m s 2
2 2
= 4.24 s
= 11.0 m s 2
f
Chapter 15
P15.33
Referring to the sketch we have
x
R
For small displacements,
tan θ ≈ sin θ
mg
F=−
and
x = − kx
R
Since the restoring force is proportional to the displacement from
equilibrium, the motion is simple harmonic motion.
F = − mg sin θ
tan θ =
and
Comparing toF = − mω 2 x shows ω =
P15.34
T=
(a)
g
.
R
k
=
m
total measured time
50
a f
Period, T asf
Length, L m
4
1.000 0.750 0.500
3
1.996 1.732 1.422
2
L
4π 2 L
T = 2π
so
g=
g
T2
The calculated values for g are:
(b)
af
0
1.996 1.732 1.422
g m s2
9.91
j
9.87
From T 2 =
Thus, g =
0.25
0.5
0.75
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 =T2
I
;
mgd
mgd
4π
2
=
T2 =
FG 1 IJ
H fK
2
1.0
L, m
FIG. P15.34
Thus, g ave = 9.85 m s 2
(c)
1
Period, T s
e
FIG. P15.33
T2, s2
The measured periods are:
P15.35
451
4π 2 I
mgd
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
FIG. P15.35
452
P15.36
Oscillatory Motion
(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
13 m
I
= 2π
= 2π
= 2.09 s
T = 2π
12 Mg 1.00 m
Mgd
12 9.80 m s 2
(b)
(a)
e
j
FIG. P15.36
1.00 m
9.80 m s
2
difference =
= 2.01 s
2.09 s − 2.01 s
= 4.08%
2.01 s
The parallel axis theorem says directly I = I CM + md 2
so
(b)
f
For the simple pendulum
T = 2π
P15.37
a
T = 2π
eI
I
= 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
mg + 2π bmgd g
= 2π e I
+ md j G − J bmgd g
H 2K
H 2K
−π e I
+ md jmg
2π md mgd
=
+
=0
+
+
I
md
mgd
I
md
mgd
b
g
b
g
e
j
e
j
dT
d
=0=
2π I CM + md 2
dd
dd
e
12
−1 2
2 12
CM
−1 2
2
CM
2 12
CM
−3 2
3 2
CM
2 12
3 2
This requires
− I CM − md 2 + 2md 2 = 0
or
P15.38
ICM = md 2 .
We suppose the stick moves in a horizontal plane. Then,
b
f
ga
1
1
2.00 kg 1.00 m
mL2 =
12
12
I
T = 2π
I=
κ
κ=
4π 2 I
T
2
=
e
4π 2 0.167 kg ⋅ m 2
a180 sf
2
j=
2
= 0.167 kg ⋅ m 2
203 µN ⋅ m
CM
+ md 2
j
−1 2
2md
Chapter 15
P15.39
e
je
(a)
I = 5.00 × 10 −7 kg ⋅ m 2
(b)
I
d 2θ
= −κθ ;
dt 2
e
Section 15.6
θ
π I
JK
jFGH 0.2250
FIG. P15.39
2
= 3.16 × 10 −4
N⋅m
rad
Damped Oscillations
1
1
mv 2 + kx 2
2
2
2
dE
d x
= mv 2 + kxv
dt
dt
2
md x
= − kx − bv
dt 2
dE
= v − kx − bv + kvx
dt
dE
= − bv 2 < 0
dt
E=
The total energy is
Taking the time-derivative,
Use Equation 15.31:
a
Thus,
P15.41
2
κ
2π
=ω =
I
T
κ = Iω 2 = 5.00 × 10 −7
P15.40
j
T = 0.250 s, I = mr 2 = 20.0 × 10 −3 kg 5.00 × 10 −3 m
f
b
θ i = 15.0°
g
θ t = 1 000 = 5.50°
x1 000 Ae − bt 2 m 5.50
− b 1 000 g
=
=
=e b
15.0
xi
A
x = Ae − bt 2 m
2m
FG 5.50 IJ = −1.00 = −bb1 000g
H 15.0 K
2m
ln
∴
P15.42
b
= 1.00 × 10 −3 s −1
2m
b
x = Ae − bt 2 m cos ωt + φ
Show that
is a solution of
− kx − b
where
ω=
b
x = Ae − bt 2 m cos ωt + φ
FG
H
g
g
2
dx
d x
=m 2
dt
dt
FG IJ
H K
k
b
−
m
2m
(1)
2
.
(2)
IJ b g
b g
K
d x
b L
FG − b IJ cosbωt + φ g − Ae ω sinbωt + φ gOP
=−
Ae
H 2m K
2m MN
dt
Q
L
FG − b IJω sinbωt + φ g + Ae ω cosbωt + φ gOP
− M Ae
H 2m K
N
Q
dx
b
= Ae − bt 2 m −
cos ωt + φ − Ae − bt 2 mω sin ωt + φ
dt
2m
2
− bt 2 m
2
− bt 2 m
continued on next page
(3)
(4)
− bt 2 m
− bt 2 m
2
(5)
453
454
Oscillatory Motion
Substitute (3), (4) into the left side of (1) and (5) into the right side of (1);
2
b g 2bm Ae cosbωt + φ g + bωAe sinbωt + φ g
bL
FG − b IJ cosbωt + φ g − Ae ω sinbωt + φ gOP
= − M Ae
H 2m K
2N
Q
b
cosbωt + φ g
+ Ae
ω sinbωt + φ g − mω Ae
2
Compare the coefficients of Ae
cosbωt + φ g and Ae
sinbωt + φ g :
F k b I = −k + b
b
bF b I
b
= − G−
− mω =
− mG −
cosine-term: − k +
J
K
H
2m
2
2m
4m
2m
H m 4m JK
− kAe − bt 2 m cos ωt + φ +
− bt 2 m
− bt 2 m
− bt 2 m
− bt 2 m
− bt 2 m
− bt 2 m
2
− bt 2 m
− bt 2 m
2
sine-term:
bω = +
2
2
2
2
2
af af
b
b
ω + ω = bω
2
2
b
g
Since the coefficients are equal, x = Ae − bt 2 m cos ωt + φ is a solution of the equation.
*P15.43
The frequency if undamped would be ω 0 =
(a)
k
=
m
With damping
ω = ω 02 −
FG b IJ
H 2m K
2
=
2.05 × 10 4 N m
= 44.0 s.
10.6 kg
FG 44 1 IJ − FG 3 kg IJ
H s K H s 2 10.6 kg K
= 1 933.96 − 0.02 = 44.0
f=
(b)
b
ω 44.0
=
= 7.00 Hz
2π 2π s
2
1
s
g
In x = A 0 e − bt 2 m cos ωt + φ over one cycle, a time T =
A0 e − b 2π
2 mω
2
2π
ω
, the amplitude changes from A0 to
for a fractional decrease of
A 0 − A 0 e − πb mω
= 1 − e −π 3 a10.6⋅44.0 f = 1 − e −0 .020 2 = 1 − 0.979 98 = 0.020 0 = 2.00% .
A0
(c)
The energy is proportional to the square of the amplitude, so its fractional rate of decrease is
twice as fast:
E=
We specify
1 2 1 2 − 2 bt 2 m
= E0 e − bt m .
kA = kA 0 e
2
2
0.05E0 = E0 e − 3 t 10.6
0.05 = e − 3 t 10.6
e + 3 t 10 .6 = 20
3t
= ln 20 = 3.00
10.6
t = 10.6 s
Chapter 15
Section 15.7
P15.44
(a)
Forced Oscillations
For resonance, her frequency must match
f0 =
(b)
ω0
1
=
2π 2π
4.30 × 10 3 N m
= 2.95 Hz .
12.5 kg
k
1
=
m 2π
dx
dv
= − Aω sin ωt , and a =
= − Aω 2 cos ωt , the maximum acceleration
dt
dt
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
P15.45
or
b g
F = 3.00 cos 2π t N
2π
= 2π rad s
T
(a)
ω=
(b)
In this case,
A=
g
ω2
=
g
k
m
e9.80 m s jb12.5 kgg =
A=
2
gm
=
k
4.30 × 10 3 N m
and
k = 20.0 N m
so
T = 1.00 s
ω0 =
k
=
m
2.85 cm
20.0
= 3.16 rad s
2.00
The equation for the amplitude of a driven oscillator,
P15.46
455
FG F IJ eω
H mK
0
2
− ω 02
j
−1
a f
3
4π 2 − 3.16
2
with b = 0, gives
A=
Thus
A = 0.050 9 m = 5.09 cm .
F0 cos ωt − kx = m
b
x = A cos ωt + φ
d2x
dt 2
ω0 =
=
2 −1
k
m
(1)
g
(2)
b
dx
= − Aω sin ωt + φ
dt
g
d2x
= − Aω 2 cos ωt + φ
dt 2
b
(3)
g
(4)
b g e j b
j cosbωt + φ g = F cos ωt
Substitute (2) and (4) into (1):
F0 cos ωt − kA cos ωt + φ = m − Aω 2 cos ωt + φ
Solve for the amplitude:
ekA − mAω
2
0
These will be equal, provided only that φ must be zero and kA − mAω 2 = F0
Thus, A =
F0
m
c h−ω
k
m
2
g
456
P15.47
Oscillatory Motion
From the equation for the amplitude of a driven oscillator with no damping,
F0 m
A=
eω
2
− ω 02
j
2
e
ω = 2π f = 20.0π s −1
e
F0 = mA ω
F0 =
P15.48
j
FG 40.0 IJ e2.00 × 10 jb3 950 − 49.0g =
H 9.80 K
−2
eω
2
− ω 02
j + b bω m g
2
k
200
=
= 49.0 s −2
40 .0
m
9.80
c h
318 N
A=
2
Fext m
eω
2
− ω 02
j
2
=
e
Fext m
±ω
2
− ω 02
j
=±
ω 2 = ω 02 ±
This yields
ω = 8.23 rad s or ω = 4.03 rad s
Then,
f=
ω
2π
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
Thus,
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=
*P15.50
− ω 02
ω 02 =
Fext m
A=
With b = 0,
P15.49
2
j
1
2π
g
1
=
L 2π
9.80 m s 2
= 1.74 Hz .
0.082 1 m
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
2
Now as the occupants exit
k
1.82 × 10
5
kg s
2
−2
5
kg s 2
Chapter 15
457
Additional Problems
P15.51
Let F represent the tension in the rod.
pivot
(a)
At the pivot, F = Mg + Mg = 2 Mg
A fraction of the rod’s weight Mg
FG y IJ as well as the
H LK
P
L
weight of the ball pulls down on point P. Thus, the
tension in the rod at point P is
F = Mg
FG y IJ + Mg =
H LK
FG
H
Mg 1 +
y
L
IJ
K
y
.
M
FIG. P15.51
(b)
Relative to the pivot, I = I rod + I ball =
1
4
ML2 + ML2 = ML2
3
3
I
where m = 2 M and d is the distance from the
mgd
pivot to the center of mass of the rod and ball combination. Therefore,
For the physical pendulum, T = 2π
d=
For L = 2.00 m, T =
P15.52
(a)
Total energy =
M
4π
3
c h + ML = 3L and T = 2π
L
2
M+M
a
4
f=
2 2.00 m
9.80 m s
2
b
4
3
ML2
a 2 M f gc h
3L
4
4π
3
=
2L
.
g
2.68 s .
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.
continued on next page
458
Oscillatory Motion
(b)
The energy of the m1 -spring system at equilibrium is:
b
gb
1
1
m1 v 2 = 9.00 kg 0.500 m s
2
2
This is also equal to
Therefore,
g
2
= 1.125 J .
a f
1
2
k A′ , where A ′ is the amplitude of the m1 -spring system.
2
a fa f
1
100 A ′
2
2
= 1.125 or A ′ = 0.150 m.
m1
= 1.885 s
k
The period of the m1 -spring system is T = 2π
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
P15.53
F d xI
GH dt JK
2
FG T IJ − A ′ = 0.500 m s a0.471 sf − 0.150 m = 0.085 6 =
H 4K
µs
= Aω 2
2
B
max
P
fmax = µ sn = µ s mg = mAω 2
A=
8.56 cm .
µsg
= 6.62 cm
ω2
n
f
B
mg
FIG. P15.53
P15.54
The maximum acceleration of the oscillating system is a max = Aω 2 = 4π 2 Af 2 . The friction force
exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B
is about to slip,
e
f = fmax = µ sn = µ s mg = m 4π 2 Af 2
P15.55
j
or
µs g
.
4π 2 f 2
A=
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.
MD = 2MH
ωD
=
ωH
k
MD
k
MH
=
MH
=
MD
1
2
fD =
fH
2
= 0.919 × 10 14 Hz
Chapter 15
P15.56
1
1
mv 2 + IΩ 2 ,
2
2
where Ω is the rotation rate of the ball about its
center of mass. Since the center of the ball moves
along a circle of radius 4R, its displacement from
equilibrium is s = 4R θ and its speed is
ds
dθ
= 4R
v=
. Also, since the ball rolls without
dt
dt
slipping,
459
The kinetic energy of the ball is K =
5R
a f
FG IJ
H K
v=
θ
ds
= RΩ
dt
Ω=
so
R
h
FG IJ
H K
v
dθ
=4
R
dt
s
FIG. P15.56
The kinetic energy is then
K=
=
FG
H
1
dθ
m 4R
2
dt
IJ
K
2
FG IJ
H K
112mR 2 dθ
10
dt
+
FG
H
1 2
mR 2
2 5
IJ FG 4 dθ IJ
K H dt K
2
2
a
f
When the ball has an angular displacement θ, its center is distance h = 4R 1 − cos θ higher than
when at the equilibrium position. Thus, the potential energy is U g = mgh = 4mgR 1 − cos θ . For small
a
f
angles, 1 − cos θ ≈
a
f
2
θ
(see Appendix B). Hence, U g ≈ 2mgRθ 2 , and the total energy is
2
E = K +Ug =
FG IJ
H K
112mR 2 dθ
10
dt
2
+ 2mgRθ 2 .
FG IJ
H K
28 R d θ
d θ
F 5 g IJθ .
+ gθ = 0 , or
This reduces to
= −G
H 28R K
5 dt
dt
Since E = constant in time,
2
FG IJ
H K
112 mR 2 dθ d 2θ
dE
dθ
.
+ 4mgRθ
=0=
5
dt
dt dt 2
dt
2
2
2
With the angular acceleration equal to a negative constant times the angular position, this is in the
5g
.
defining form of a simple harmonic motion equation with ω =
28 R
The period of the simple harmonic motion is then T =
2π
ω
= 2π
28 R
.
5g
460
P15.57
Oscillatory Motion
(a)
Li
L
a
h
a
FIG. P15.57(a)
(b)
T = 2π
π 1 dL
dT
=
dt
g L dt
L
g
af
We need to find L t and
dL
. From the diagram in (a),
dt
FG IJ
H K
1 dh
a h dL
=−
− ;
.
2 dt
2 2 dt
L = Li +
But
(1)
dM
dV
dh
=ρ
= − ρA . Therefore,
dt
dt
dt
IJ
K
(2)
FG 1 IJ FG dM IJ t = L − L
H 2 ρA K H dt K
(3)
FG
H
1 dM dL
dh
1 dM
=−
=
;
ρA dt dt
dt
2 ρA dt
z
L
Also,
dL =
Li
i
Substituting Equation (2) and Equation (3) into Equation (1):
F
GH
π
1
g 2 ρa 2
dT
=
dt
(c)
I FG dM IJ
JK H dt K
1
Li +
1
2 ρa 2
c ht
dM
dt
.
Substitute Equation (3) into the equation for the period.
2π
T=
g
Li +
FG IJ
H K
1
dM
t
2
dt
2 ρa
Or one can obtain T by integrating (b):
F
GH
I FG dM IJ
z
JK H dt K z L + dt c ht
L 2 OPL
O
1 F dM I
π F 1 I F dM I M
T −T =
L +
t− L P
M
G
J
G
J
G
J
g H 2 ρa K H dt K M
PQ
N c h PQMN 2 ρa H dt K
L
2π
1 F dM I
, so T =
L +
G Jt .
g
2 ρa H dt K
g
Ti
dT =
T
i
But Ti = 2π
i
t
π
1
g 2 ρa 2
0
2
i
2
i
1
2 ρa 2
1
2 ρa 2
dM
dt
dM
dt
i
2
i
Chapter 15
P15.58
ω=
(a)
P15.59
k
2π
=
m
T
2
k =ω m =
4π 2 m
m′ =
(b)
T2
a f
k T′
= m
4π 2
FG T ′ IJ
HTK
2
Hy
We draw a free-body diagram of the pendulum.
The force H exerted by the hinge causes no torque
about the axis of rotation.
τ = Iα
Hx
h
d 2θ
= −α
dt 2
and
τ = MgL sin θ + kxh cos θ = − I
kx
Lθ
x
2
d θ
dt 2
k
m
mg
For small amplitude vibrations, use the
approximations: sin θ ≈ θ , cos θ ≈ 1, and x ≈ s = hθ .
Therefore,
F
GH
(a)
I
JK
MgL + kh 2
d 2θ
θ = −ω 2θ
=
−
I
dt 2
b
g
ω=
ML2
1
2π
we have at t = 0
v = −ωA sin φ = − v max
This requires φ = 90° , so
x = A cos ωt + 90°
And this is equivalent to
x = − A sin ωt
Numerically we have
ω=
and v max = ωA
20 m s = 10 s −1 A
In
a
MgL + kh 2
ML2
f
50 N m
= 10 s −1
0.5 kg
e
a
= 2π f
g
v = −ωA sin ωt + φ
j
f e
A=2m
j
x = −2 m sin 10 s −1 t
So
(b)
b
MgL + kh 2
In x = A cos ωt + φ ,
k
=
m
L sinθ
FIG. P15.59
f=
*P15.60
2
1
1
1
mv 2 + kx 2 = kA 2 ,
2
2
2
implies
FG
H
1 2
1
kx = 3 mv 2
2
2
11 2 1 2 1 2
kx + kx = kA
32
2
2
x=±
continued on next page
IJ
K
3
A = ±0.866 A = ±1.73 m
4
4 2
x = A2
3
hcosθ
461
462
Oscillatory Motion
(c)
ω=
(d)
In
g
L
g
L=
a
ω2
=
9.8 m s 2
e10 s j
−1 2
f e
= 0.098 0 m
j
x = −2 m sin 10 s −1 t
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.
FπI
The minimum time for the motion is ∆t in 10 ∆t = G J s
H 6K
FπI
∆t = G J s = 0.052 4 s
H 60 K
−1
P15.61
(a)
FIG. P15.60(d)
At equilibrium, we have
F LI
∑ τ = 0 − mg GH 2 JK + kx0 L
where x 0 is the equilibrium compression.
After displacement by a small angle,
FIG. P15.61
F LI
F LI
∑ τ = − mg GH 2 JK + kxL = −mg GH 2 JK + kbx0 − Lθ gL = − kθL2
But,
1
∑ τ = Iα = 3 mL2
d 2θ
.
dt 2
So
d 2θ
3k
= − θ.
2
m
dt
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
Chapter 15
*P15.62
463
As it passes through equilibrium, the 4-kg object has speed
v max = ωA =
100 N m
k
A=
2 m = 10.0 m s.
m
4 kg
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
= kA 2
mv max
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
k
100 N m
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
= 4 kg 10 m s
mv max
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 .
P15.63
(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°
464
P15.64
Oscillatory Motion
One can write the following equations of motion:
T − kx = 0
(describes the spring)
mg − T ′ = ma = m
a
f
R T′ − T = I
2
d x
dt 2
(for the hanging object)
d 2θ I d 2 x
=
dt 2 R dt 2
(for the pulley)
with I =
FIG. P15.64
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=
P15.65
ω
2π
=
k
1
=
m + 12 M 2π
1
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
100 N m
.
0.200 kg + 12 M
Suppose a 100-kg biker compresses the suspension 2.00 cm.
Then,
k=
F
980 N
=
= 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.
Chapter 15
P15.66
(a)
For each segment of the spring
dK =
Also,
vx =
x
v
A
a f
1
dm v x2 .
2
dm =
and
m
dx .
A
FIG. P15.66
Therefore, the total kinetic energy of the block-spring system is
K=
(b)
ω=
k
m eff
Therefore,
P15.67
(a)
∑ F = −2T sin θ j
z FGH
A
0
I
JK
FG
H
IJ
K
x2v2 m
m 2
1
dx =
M+
v .
2
2
3
A
A
FG
H
IJ
K
1
1
m 2
m eff v 2 =
M+
v
2
2
3
and
T=
1
1
Mv 2 +
2
2
2π
ω
= 2π
M + m3
k
.
where θ = tan −1
FG y IJ
H LK
Therefore, for a small displacement
y
sin θ ≈ tan θ =
L
(b)
and
FIG. P15.67
−2Ty
∑ F = L j
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,
∑ F = −kx
becomes here
Therefore, the effective spring constant is
2T
and
L
∑F = −
ω=
2T
y.
L
k
=
m
2T
.
mL
465
466
P15.68
Oscillatory Motion
(a)
Assuming a Hooke’s Law type spring,
F = Mg = kx
and empirically
Mg = 1.74x − 0.113
k = 1.74 N m ± 6% .
so
(b)
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
0.471
0.686
0.080 0
0.493
0.784
We may write the equation as theoretically
T2 =
4π 2
4π 2
M+
ms
3k
k
FIG. P15.68
and empirically
T 2 = 21.7 M + 0.058 9
k=
so
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
The k values 1.74 N m ± 6%
and
so
(c)
1.82 N m ± 3% differ by 4%
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.
Chapter 15
P15.69
(a)
∆K + ∆U = 0
Thus, K top + U top = K bot + U bot
M
R
where K top = U bot = 0
1 2
Iω , but
2
h = R − R cos θ = R 1 − cos θ
Therefore, mgh =
a
θ
f
v
R
MR 2 mr 2
and I =
+
+ mR 2
2
2
Substituting we find
ω=
2
2
2
(b)
M
m
M
m
2
+
r2
R2
r2
R2
+2
1
2
dCM =
af
mR + M 0
m+M
MR 2 + 12 mr 2 + mR 2
mgR
We require Ae − bt 2 m =
A
2
bt
= ln 2
or
2m
The spring constant is irrelevant.
or
(b)
2
2
I
mT gd CM
T = 2π
T = 2π
(a)
2
2
mT = m + M
P15.70
e + bt 2 m = 2
0.100 kg s
t = 0.693
2 0.375 kg
b
∴ t = 5.20 s
g
We can evaluate the energy at successive turning points, where
1
1
1
1 1 2
cos ωt + φ = ±1 and the energy is kx 2 = kA 2 e − bt 2 m . We require kA 2 e − bt 2 m =
kA
2
2
2
2 2
m ln 2 0.375 kg 0.693
∴t =
=
= 2.60 s .
or
e + bt m = 2
b
0.100 kg s
b
g
a
(c)
m
FIG. P15.69
2
so
θ
v
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
a
mgR 1 − cos θ =
and
467
f
1 2
kA , the fractional rate of change of energy over time is
2
2
d 1
dE
dA
dA
1
dt 2 kA
dt
dt
2 k 2 A dt
= 1
=
=
2
2
1 kA 2
E
A
kA
2
2
From E =
e
j
a f
two times faster than the fractional rate of change in amplitude.
FG
H
IJ
K
468
P15.71
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. P15.71
LM k OPx
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
(b)
2
1
2
This is in the form
F = k eff x = ma
and
T = 2π
b
m k1 + k 2
m
= 2π
k eff
k1 k 2
g
In this case each spring is distorted by the distance x which the mass is displaced. Therefore,
the restoring force is
b
g
F = − k1 + k 2 x
so that
P15.72
1
1 2
1
where a is the acceleration of the mass m.
2
T = 2π
b
k eff = k1 + k 2
and
m
k1 + k 2
g
.
Let A represent the length below water at equilibrium and M the tube’s mass:
∑ Fy = 0 ⇒ − Mg + ρπ r 2 Ag = 0 .
Now with any excursion x from equilibrium
a f
− Mg + ρπ r 2 A − x g = Ma .
Subtracting the equilibrium equation gives
− ρπ r 2 gx = Ma
a=−
F ρπ r g I x = −ω x
GH M JK
2
2
The opposite direction and direct proportionality of a to x imply SHM with angular frequency
ω=
T=
ρπ r 2 g
M
2π
ω
=
FG 2 IJ
HrK
πM
ρg
Chapter 15
P15.73
For θ max = 5.00° , the motion calculated by the Euler method
agrees quite precisely with the prediction of θ max cos ωt . The
period is T = 2.20 s .
Time,
t (s)
0.000
0.004
0.008
…
0.544
0.548
0.552
…
1.092
1.096
1.100
1.104
…
1.644
1.648
1.652
…
2.192
2.196
2.200
2.204
Angle,
θ (°)
5.000 0
4.999 3
4.998 0
Ang. speed
(°/s)
0.000 0
–0.163 1
–0.326 2
Ang. Accel.
° s2
e j
θ max cos ωt
–40.781 5
–40.776 2
–40.765 6
5.000 0
4.999 7
4.998 7
0.056 0
–0.001 1
–0.058 2
–14.282 3
–14.284 2
–14.284 1
–0.457 6
0.009 0
0.475 6
0.081 0
0.023 9
–0.033 3
–4.999 4
–5.000 0
–5.000 0
–4.999 3
–0.319 9
–0.156 8
0.006 3
0.169 4
40.776 5
40.781 6
40.781 4
40.775 9
–4.998 9
–4.999 8
–5.000 0
–4.999 6
–0.063 8
0.003 3
0.060 4
14.282 4
14.284 2
14.284 1
0.439 7
–0.027 0
–0.493 6
–0.071 6
–0.014 5
0.042 7
4.999 4
5.000 0
5.000 0
4.999 3
0.313 7
0.150 6
–0.012 6
–0.175 7
–40.776 8
–40.781 7
–40.781 3
–40.775 6
4.999 1
4.999 9
5.000 0
4.999 4
For θ max = 100° , the simple harmonic motion approximation
θ max cos ωt diverges greatly from the Euler calculation. The
period is T = 2.71 s , larger than the small-angle period by 23%.
Time,
Angle,
t (s)
θ (°)
0.000 100.000 0
0.004
99.992 6
0.008
99.977 6
…
1.096 –84.744 9
1.100 –85.218 2
1.104 –85.684 0
…
1.348 –99.996 0
1.352 –100.000 8
1.356 –99.998 3
…
2.196
40.150 9
2.200
41.045 5
2.204
41.935 3
…
2.704
99.998 5
2.708 100.000 8
2.712
99.995 7
Ang. speed
(°/s)
0.000 0
–1.843 2
–3.686 5
Ang. Accel.
° s2
e j
θ max cos ωt
–460.606 6
–460.817 3
–460.838 2
100.000 0
99.993 5
99.973 9
–120.191 0
–118.327 2
–116.462 0
465.948 8
466.286 9
466.588 6
–99.995 4
–99.999 8
–99.991 1
–3.053 3
–1.210 0
0.633 2
460.812 5
460.805 7
460.809 3
–75.797 9
–75.047 4
–74.287 0
224.867 7
223.660 9
222.431 8
–301.713 2
–307.260 7
–312.703 5
99.997 1
99.999 3
99.988 5
2.420 0
0.576 8
–1.266 4
–460.809 0
–460.805 7
–460.812 9
12.642 2
11.507 5
10.371 2
FIG. P15.73
469
470
*P15.74
Oscillatory Motion
(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
IJ
K
kµ k mg
µ mg
= −k x − k
= − kx rel
k
k
where x rel is the excursion of the block away from the point
µ k mg
.
k
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 =
b
g
µ s mg µ k mg
−
. It first comes to
k
k
2 µ k − µ s mg
µ k mg
. Almost immediately at this point it
−A=
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.
rest at spring extension
(c)
The graph of
the motion
looks like this:
FIG. P15.74(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
=π
. We ignore the times at the end points of the motion when the speed of
2π
2
k
k
2A
the block changes from v to 0 and from 0 to v. Since v is small compared to
, these
π
times are negligible. Then the period is
T=
continued on next page
b
g
2 µ s − µ k mg
kv
+π
m
.
k
m
k
Chapter 15
(f)
T=
a
fb
b0.024 m sgb12 N mg
f=
Then
*P15.75
ge
2 0.4 − 0.25 0.3 kg 9.8 m s 2
b
g
2 µ s − µ k mg
j +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 .
(a)
Newton’s law of universal gravitation is
F=−
Thus,
F=−
Which is of Hooke’s law form with
k=
(b)
kv
b
471
g
GMm
r
=−
2
FG
H
IJ
K
Gm 4 3
πr ρ
r2 3
FG 4 πρGmIJ r
H3 K
4
πρGm
3
FG 4 IJ πρGmr = ma
H 3K
F 4I
a = − G J πρGr = −ω r
H 3K
The sack of mail moves without friction according to
−
2
Since acceleration is a negative constant times excursion from equilibrium, it executes SHM
with
ω=
4πρG
3
and period
T=
The time for a one-way trip through the earth is
T
=
2
We have also
g=
so
g
4ρG
=
3
πR e
b g
and
2π
ω
3π
ρG
=
3π
4 ρG
GM e
R e2
=
G 4πR e3 ρ
3 R e2
(a) 4.33 cm; (b) −5.00 cm s ;
P15.6
see the solution
P15.8
12.0 Hz
P15.10
18.8 m s; 7.11 km s 2
2
(c) −17.3 cm s ; (d) 3.14 s; 5.00 cm
P15.4
4
πρGR e
3
Re
6.37 × 10 6 m
T
=π
=π
= 2.53 × 10 3 s = 42.2 min .
2
2
g
9.8 m s
ANSWERS TO EVEN PROBLEMS
P15.2
=
(a) 15.8 cm; (b) −15.9 cm;
(c) see the solution; (d) 51.1 m; (e) 50.7 m
472
Oscillatory Motion
(a) 1.26 s; (b) 0.150 m s; 0.750 m s 2 ;
15 cm
sin 5t ;
(c) x = −3 cmcos 5t ; v =
s
75 cm
a=
cos 5t
s2
P15.42
see the solution
P15.44
(a) 2.95 Hz; (b) 2.85 cm
P15.46
see the solution
P15.48
either 1.31 Hz or 0.641 Hz
P15.14
F vI
(a) ; (b) x = − G J sin ωt
HωK
ω
P15.50
1.56 cm
P15.16
(a) 126 N m; (b) 0.178 m
P15.52
(a) 0.500 m s ; (b) 8.56 cm
P15.18
(a) 0.153 J; (b) 0.784 m s; (c) 17.5 m s 2
P15.54
A=
P15.20
(a) 100 N m; (b) 1.13 Hz;
(c) 1.41 m s at x = 0 ;
P15.56
see the solution
(f) 1.33 m s ; (g) 3.33 m s 2
P15.58
(a) k =
P15.22
(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
P15.60
(a) x = −2 m sin 10t ; (b) at x ± 1.73 m;
(c) 98.0 mm; (d) 52.4 ms
P15.24
(a) 0.218 s and 1.09 s; (b) 14.6 mW
P15.62
P15.26
The position of the piston is given by
x = A cos ωt .
(a) decreased by 0.735 m;
(b) increased by 0.730 s;
(c) decreased by 120 J; (d) see the solution
P15.64
(a) 3.56 Hz ; (b) 2.79 Hz; (c) 2.10 Hz
P15.66
(a)
P15.68
see the solution; (a) k = 1.74 N m ± 6% ;
(b) 1.82 N m ± 3%; they agree;
(c) 8 g ± 12%; it agrees
P15.70
(a) 5.20 s; (b) 2.60 s; (c) see the solution
P15.72
see the solution; T =
P15.74
see the solution; (f) 0.281 Hz ;
(g) decreases; (h) increases; (i) increases;
(j) decreases
P15.12
FG
H
FG
H
IJ
K
IJ
K
v
µsg
4π 2 f 2
2
(d) 10.0 m s at x = ± A ; (e) 2.00 J;
P15.28
gC
= 1.001 5
gT
P15.30
1.42 s; 0.499 m
P15.32
(a) 3.65 s; (b) 6.41 s; (c) 4.24 s
P15.34
(a) see the solution;
(b), (c) 9.85 m s 2 ; agreeing with the
accepted value within 0.5%
P15.36
(a) 2.09 s; (b) 4.08%
P15.38
203 µN ⋅ m
P15.40
see the solution
FG IJ
H K
4π 2 m
T′
; (b) m ′ = m
2
T
T
a
FG
H
2
f a f
IJ
K
M + m3
1
m 2
M+
v ; (b) T = 2π
2
3
k
FG 2 IJ
HrK
πM
ρg