Download Chapter 10 ELASTICITY AND OSCILLATIONS

Chapter 10
ELASTICITY AND OSCILLATIONS
Conceptual Questions
1. Young’s modulus does not tell us which is stronger. Instead, it tells us which is more resistant to deformation for a
given stress. The ultimate strength would tell us which is stronger—i.e., which can withstand the greatest stress.
2. The pendulum should be lengthened to increase its period and slow down the clock.
3. When the block is struck, it initially begins to bend downward before actually breaking. Thus, the top of the block
is compressed, while the bottom is stretched. Since concrete has much less tensile strength than compressive
strength, it will break at the bottom first.
4. (a) Answer: 2F. Since stress is proportional to strain, the same force would produce the same strain on the bar of
half the length. However, strain is a relative change in length, defined as ∆L/L. Therefore, the change in
length would be half as much. To compress the bar by the same amount, then, would require a force twice as
great.
(b) Answer: F/4. To compress the bar by the same amount, with the same length as before, would require the
same stress. Stress is defined as F/A. With half the radius, the area is reduced to 1/4 of its initial value.
Therefore, to produce the same stress would require a force 1/4 as great as before.
5. The compressive force experienced by the columns is greater at the bottom than at the top, because the bottom
must support the weight of the column itself in addition to whatever the column is holding up. By increasing the
cross-sectional area of the bottom of the column, the stress it experiences is reduced. Tapering columns so that
they are thicker at the base prevents the stress at the bottom from being too large.
6. Although the distance traveled by the mass during each cycle is proportional to the amplitude of the oscillation,
the maximum velocity of the mass is as well. If the mass has farther to go, for example, it travels correspondingly
faster. This is how the period of the mass-spring system can be independent of amplitude.
7. Yes, the motion of the saw blade is SHM. The Scotch yoke effectively makes the horizontal displacement of the
saw blade equal to the x-component of the position of the knob, which is moving in a circle. When an object
moves in uniform circular motion, its x- (or y-) component exhibits SHM.
8. For the mass and spring system, the period will remain 1 s, because the period depends only on the mass and the
spring constant. For a pendulum, the period depends on the length and the gravitational field strength. With a
stronger gravitational field, the period of the pendulum would be less than 1 s.
9. The tension in the bungee cord at the lowest point would be greater than the person’s weight, because there is an
upward acceleration. In fact, the tension would have its maximum value at the bottom, because that is where the
upward acceleration is the greatest.
10. The breaking point of a rope is determined by the maximum strain it can withstand. The strain is the ratio of the
change in the length of the rope to the original length—the maximum strain is therefore independent of the rope’s
length. The strain is directly proportional to the stress—defined as the force per unit area. Thus, ropes of varying
length that are otherwise identical require the same force to reach the breaking point. The actual distance the rope
stretches before breaking is greater for a longer rope—more work, and thus more energy, is required to break a
longer rope.
482
College Physics
Chapter 10: Elasticity and Oscillations
11. The velocity of the plane is lower at the top of its loop than at the bottom. Thus, the speed of the plane’s shadow
at the midpoint of its flight depends on the direction of travel—the shadow does not therefore exhibit simple
harmonic motion.
12. Concrete is much stronger under compressive stress than tensile stress. As a result of this, concrete is very strong
in vertical columns where most of the force is compressive. Concrete is weaker in horizontal columns because it
must withstand additional tensile stresses. Steel rods with a high tensile strength are therefore inserted into the
concrete to reinforce it against these tensile stresses.
13. To produce the same strain, the ratio of the force to the cross-sectional area must remain unchanged. The total
cross-sectional area of the two wires together is twice the original area. Thus, the force applied to the two wires
must be doubled as well. Modeling a thick wire as a bundle of thin wires, the preceding argument explains why
the force to produce a given strain must be proportional to the cross-sectional area—and thus why the strain
depends on the stress.
14. A given tensile stress produces a given stretch in the length of a single spring. Wires of different lengths can be
modeled as being comprised of varying numbers of springs in series. The elongation of any wire under an
identical tensile stress is therefore equal to the sum of the distances that each of the springs it is built from has
stretched. The elongation is thus proportional to the number of springs—but so too is the original length.
Therefore, a given tensile stress produces an elongation of the wire proportional to the wire’s initial length.
15. Using stress and strain to describe deformations provides the means to describe properties of materials in a way
that is independent of their dimensions. These concepts also allow physical laws to be expressed in terms of
general properties of materials—rather than properties particular to a given piece of a material.
16. The shock absorbing system in the car can be thought of as a set of springs that dampen out oscillations induced
by bumps in the road. These springs have a natural oscillation frequency just as a simple spring does. When the
frequency of encounters with bumps in the road matches the resonant frequency of the springs, the car’s
oscillations are amplified. The passengers in the car are experiencing resonance.
17. In the mass-spring system, the restoring force supplied by the spring is independent of the object’s mass. Thus, the
larger inertia of a more massive object produces a longer period. The restoring force for small amplitude
oscillations of the pendulum is the horizontal component of the tension in the string. In this case, the magnitude of
the tension is approximately equal to the weight of the bob. Although a more massive bob has more inertia, it also
has a proportionally larger restoring force. Thus, the period of oscillation of the pendulum is independent of the
mass.
18. The percent of kinetic energy (short dashes), potential energy (medium dashes) and total energy (long dashes) of a
mass connected to an ideal spring oscillating on a frictionless horizontal surface are plotted below as a function of
time.
483
Chapter 10: Elasticity and Oscillations
College Physics
Problems
1. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find the vertical compression of the beam.
∆L F
FL
(5.8 × 104 N)(2.5 m)
Y
= , so ∆L =
=
= 0.097 mm .
L
A
YA (200 × 109 Pa)(7.5 × 10−3 m 2 )
2. Strategy The man is standing on two feet, so the force for one leg is equal to half his weight. The stress is
proportional to the strain. Use Eq. (10-4).
Solution Find the compression of the thighbone.
1 (91 kg)(9.80 N kg)(0.50 m)
FL
∆L =
= 2
= 29 µm
YA (1.1× 1010 N m 2 )(7.0 × 10−4 m 2 )
3. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find how much the wire stretches.
FL
(5.0 × 103 N)(2.0 m)
∆L =
=
= 2.2 cm
10
YA (9.2 × 10 Pa)(5.0 mm 2 )(10−3 m mm)2
4. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find Young’s modulus for the wire.
FL
(1.00 × 103 N)(5.00 m)
Y=
=
= 7.69 × 1010 Pa
A∆L (0.100 cm 2 )(10−2 m cm)2 (6.50 × 10−3 m)
5. Strategy Form a proportion of the elongations of the left and right wires. Use Eq. (10-4).
Solution Find how far the midpoint moves.
FL
∆LL YAL
A
π (2r )2
= FL = R =
= 4, so ∆LL = 4∆LR and ∆L = ∆LL + ∆LR is the total elongation, 1.0 mm.
∆LR YA
AL
π r2
R
∆L = ∆LL + ∆LR = ∆LL +
1
5
4
4
∆LL = ∆LL , so ∆LL = ∆L = (1.0 mm) = 0.80 mm .
4
4
5
5
6. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Compute the compression of the abductin ligament.
FL
(1.5 N)(1.0 × 10−3 m)
∆L =
=
= 0.48 mm
YA (4.0 × 106 N m 2 )(0.78 mm 2 ) 1 m 2
1000 mm
(
)
7. Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find the required stretch of the string.
FL
(20 N)(0.50 m)
∆L =
=
= 5.0 mm
YA (2.0 × 109 N m 2 )(1.0 × 10−6 m 2 )
484
College Physics
Chapter 10: Elasticity and Oscillations
8. (a) Strategy The average power required is equal to the kinetic energy change divided by the elapsed time it
takes for the flea to reach its peak velocity.
Solution Find the average power required.
2
2
1
∆K 2 m(vf − vi ) mvf2 (0.45 × 10−6 kg)(0.74 m s)2
Pav =
=
=
=
= 1.2 × 10−4 W
∆t
∆t
2∆t
2(1.0 × 10−3 s)
(b) Strategy and Solution Compute the power output of the flea.
(60 W kg)(0.45 × 10−6 kg)(0.20) = 5.4 × 10−6 W < 1.2 × 10−4 W
No, the flea’s muscle cannot provide the power needed.
(c) Strategy There are two pads, so the total energy stored is E = 2U = 2[ 12 k (∆L)2 ] = k (∆L)2. Use Eq. (10-5).
Solution Find the energy stored in the resilin pads.
A
L2 2
E = k (∆L)2 = Y (∆L)2 = Y
L = YL3 = (1.7 × 106 N m 2 )(6.0 × 10−5 m)3 = 3.7 × 10−7 J
L
L
(d) Strategy Use the definition of average power.
Solution Compute the power provided by the resilin pads.
∆E 3.7 × 10−7 J
Pav =
=
= 3.7 × 10−4 W > 1.2 × 10−4 W
∆t 1.0 × 10−3 s
Yes, enough power is provided for the jump.
9. Strategy Refer to Fig. 10.4c. The stress is proportional to the strain.
Solution Calculate Young’s moduli for tension and compression of bone.
Tension:
For tensile stress and strain, the graph is far from being linear, but for relatively small values of stress and strain, it
is approximately linear. So, for small values of tensile stress and strain, Young’s Modulus is
stress 5.0 × 107 N m 2
Y=
=
= 1.5 × 1010 N m 2 .
strain
0.0033
Compression:
−4.5 × 107 N m 2
= 9.0 × 109 N m 2 .
Similarly, for small values of compressive stress and strain, Y =
−0.0050
10. Strategy Set the stress equal to the elastic limit to find the minimum diameter.
Solution Find the minimum diameter of the wire required to support the acrobat.
4F
4(55 kg)(9.80 N kg)
F
F
=
= 1.7 mm .
elastic limit = =
, so d =
2
1
π (elastic limit)
A
πd
π (2.5 × 108 Pa)
4
11. Strategy Set the stress equal to the tensile strength of the hair to find the diameter of the hair.
Solution Find the diameter of the hair.
F
F
4F
4(1.2 N)
, so d =
=
= 8.7 × 10−5 m .
tensile strength = =
8
A 1 πd2
π (tensile strength)
(2.0
×
10
Pa)
π
4
485
Chapter 10: Elasticity and Oscillations
College Physics
12. Strategy Compare the substances in each case using ratios.
Solution
(a) Tendon:
tensile strength 80.0 × 106 Pa
=
= 7.3 × 104 Pa ⋅ m3/kg
density
1100 kg m3
Steel:
0.50 × 109 Pa
= 6.5 × 104 Pa ⋅ m3/kg
7700 kg m3
Tendon is stronger than steel.
(b) Bone:
160 × 106 Pa
1600 kg m3
Concrete:
0.40 × 109 Pa
2700 kg m3
= 1.0 × 105 Pa ⋅ m3/kg
= 1.5 × 105 Pa ⋅ m3/kg
Concrete is stronger than bone.
13. Strategy The stress on the copper wire must be less than its elastic limit.
Solution Find the maximum load that can be suspended from the copper wire.
F
< elastic limit, so F < π r 2 (elastic limit) = π (0.0010 m)2 (2.0 × 108 Pa) = 630 N .
A
14. Strategy The stress on the copper wire must be less than its tensile strength.
Solution Find the maximum load that can be suspended from the copper wire.
F
< tensile strength, so F < π r 2 (tensile strength) = π (0.0010 m)2 (4.0 × 108 Pa) = 1300 N .
A
15. Strategy Set the stresses equal to the compressive strengths to determine the effective cross-sectional areas.
Solution Find the effective cross-sectional areas.
5 × 104 N
F
= 1.6 × 108 Pa, so A =
= 3 cm 2 .
Human:
8
A
1.6 × 10 Pa
Horse: A =
10 × 104 N
8
1.4 × 10 Pa
= 7.1 cm 2
16. Strategy Assume that the stress is proportional to the strain up to the breaking point. Use Eq. (10-4).
Solution Find the stress at the breaking point of the steel wire.
∆L
⎛ 0.20 ⎞
8
= (2.0 × 1011 N m 2 ) ⎜
stress at breaking point = Y
⎟ = 4.0 × 10 Pa
L
100
⎝
⎠
486
College Physics
Chapter 10: Elasticity and Oscillations
17. Strategy Use Eqs. (10-1), (10-2), and (10-4).
Solution
(a) stress =
7.0 × 104 N
F
=
= 2.8 × 107 Pa
A 25 × 10−4 m 2
(b) strain =
∆L F
7.0 × 104 N
=
=
= 4.7 × 10−4
−
10
4
2
L YA (6.0 × 10 Pa)(25 × 10 m )
(c) ∆L =
FL
(7.0 × 104 N)(2.0 m)
=
= 9.3 × 10−4 m
YA (6.0 × 1010 Pa)(25 × 10−4 m 2 )
(d) Set the compressive strength equal to the stress to find the maximum weight the column can support.
F
= compressive strength, so F = (compressive strength)A = (2.0 ×108 Pa)(25 ×10−4 m 2 ) = 5.0 × 105 N .
A
18. (a) Strategy The stress is proportional to the strain. Use Eq. (10-4).
Solution Find the diameter and tensile stress of the wire.
Diameter:
∆L
4 FL
4(120 N)(3.0 m)
F
F
= 1
=Y
, so d =
=
= 1.3 mm .
π Y ∆L
A 4πd
L
π (120 × 109 Pa)(2.1× 10−3 m)
Tensile stress:
F
∆L
2.1× 10−3 m
=Y
= (120 × 109 Pa)
= 8.4 × 107 N/m 2
A
L
3.0 m
(b) Strategy Set the maximum stress equal to the tensile strength.
Solution Find the maximum weight the wire can support.
Wmax
1
= tensile strength, so Wmax = π (0.00135 m) 2 (4.0 × 108 N m 2 ) = 570 N .
A
4
19. Strategy Use Hooke’s law for volume deformations.
Solution Compute the fractional changes of the volume and radius of the sphere.
∆V
∆P 1.0 × 108 N m 2
Volume:
=
=
= 7.7 × 10−4
V
B
130 × 109 Pa
Radius:
4 3
3V
.
π r = V , so r = 3
3
4π
3V ′
3
V − ∆V
∆r r − r ′
r′
V′
∆V
3
=
= 1 − = 1 − 4π = 1 − 3
= 1− 3
= 1− 3 1−
= 1 − 1 − 7.7 × 10−4 = 2.6 × 10−4
3
V
r
r
r
V
V
V
3
4π
487
Chapter 10: Elasticity and Oscillations
College Physics
20. Strategy Use ∆P = ρ gd and the relationship between volume, mass, and density.
Solution Find the percent increase in the density of water at a depth of 1.0 km.
∆ρ
ρ
=
ρ′ − ρ ρ′
= −1 =
ρ
ρ
Since V ′ ≈ V ,
100% ×
∆ρ
ρ
≈
∆ρ
ρ
≈−
m
V′
m
V
−1 =
∆V
V
V −V ′
−1 =
=−
V′
V′
V′
∆V
. Use Hooke’s law for volume deformations.
V
ρ gd
∆P
(1.0 × 103 kg m3 )(9.80 N kg)(1.0 × 103 m)
× 100% = w × 100% =
× 100% = 0.45%
B
B
2.2 × 109 Pa
21. Strategy Use Hooke’s law for volume deformations.
Solution Find the change in volume of the sphere.
∆V
V ∆P
(1.00 cm3 )(9.12 × 106 Pa)
∆P = − B
, so ∆V = −
=−
= −57 × 10−6 cm3 .
V
B
160 × 109 Pa
The volume of the steel sphere would decrease by 57 × 10−6 cm3 .
22. Strategy Use Hooke’s law for volume deformations. The pressure of the Moon is roughly 10−9 Pa.
Solution Find the change in volume of the aluminum.
∆V
V ∆P
(1.00 cm3 )(10−9 Pa − 1.013 × 105 Pa)
∆P = − B
, so ∆V = −
=−
= 1.4 × 10−6 cm3 .
V
B
70 × 109 Pa
The volume of the aluminum sphere would increase by 1.4 × 10−6 cm3 .
23. Strategy Set the shear stress equal to the total shear strength to find the maximum shearing force.
Solution Find the maximum shearing force F on the plates that the four bolts can withstand.
F
= shear strength, so F = 4π r 2 (shear strength) = 4π (0.010 m)2 (6.0 × 108 Pa) = 7.5 × 105 N .
4 Abolt
24. Strategy Use Hooke’s law for volume deformations.
Solution Find the change in the volume of the anchor.
∆V
∆P
V ∆P
(0.230 m3 )(1.75 × 106 Pa)
=−
=−
= − 6.71 cm3 .
, so ∆V = −
V
B
B
60.0 × 109 Pa
25. Strategy Use Hooke’s law for shear deformations.
Solution Find the magnitude of the tangential force.
∆x
F F
= 2 =S
, so F = S ∆xL = (940 Pa)(0.64 × 10−2 m)(0.050 m) = 0.30 N .
A L
L
488
College Physics
Chapter 10: Elasticity and Oscillations
26. (a) Strategy The tangent of the deformation angle is equal to the relative displacement of the sponge divided by
its thickness.
Solution Compute the relative displacement, ∆x.
∆x = L tan γ = (2.0 cm) tan 8.0° = 2.8 mm
(b) Strategy Use Hooke’s law for shear deformations.
Solution Compute the shear modulus of the sponge.
FL
FL
F
12 N
S=
=
=
=
= 2.0 × 104 Pa
A∆x AL tan γ
A tan γ (42 × 10−4 m 2 ) tan 8.0°
27. Strategy At the maximum extension of the spring, x = A and the magnitude of the acceleration is maximum.
Use Eq. (10-22).
Solution Find the magnitude of the acceleration at the point of maximum extension of the spring.
4π 2 A 4π 2 (0.050 m)
am = ω 2 A =
=
= 7.9 m s 2
T2
(0.50 s) 2
28. Strategy The amplitude is half the maximum distance. Use Eq. (10-21).
Solution Find the maximum needle speed.
vm = ω A =
2π A 2π
=
T
( 8.42 ×10−3 m ) =
9.0 s
24
7.0 cm s
29. Strategy The amplitude is half the maximum distance. Use Eqs. (10-21) and (10-22).
Solution Find the maximum velocity and maximum acceleration of the prong.
⎛ 2.24
⎞
× 10−3 m ⎟ = 3.10 m s
vm = ω A = 2π fA = 2π (440.0 Hz) ⎜
2
⎝
⎠
⎞
2
2 2
2
2 ⎛ 2.24
am = ω A = 4π f A = 4π (440.0 Hz) ⎜
× 10−3 m ⎟ = 8560 m s 2
2
⎝
⎠
30. Strategy At the equilibrium point, the speed is at its maximum. Use Eq. (10-21).
Solution Find the speed at the equilibrium point.
2π
2π (0.050 m)
vm = ω A =
A=
= 0.63 m s
T
0.50 s
31. Strategy Replace each quantity with its SI units.
Solution a has units m s 2 . − ω 2 x has units s −2 ⋅ m = m s 2 . So, a = −ω 2 x is consistent for units.
ω has units s −1.
k m has units
Nm
=
kg
kg ⋅ m s 2
1
= 2 = s −1. So,
m ⋅ kg
s
489
k m has the same units as ω .
Chapter 10: Elasticity and Oscillations
College Physics
32. (a) Strategy Solve for the spring constant in Eq. (10-20a).
Solution Find the spring constant.
k
ω=
, so k = ω 2 m = (3.00 Hz) 2 (2π rad cycle)2 (0.17 kg) = 60 N m .
m
(b) Strategy Use Eq. (10-17).
Solution The amplitude is A = 12.0 cm. Find the angular frequency.
ω = (2π rad cycle)(3.00 Hz) = 6.00π rad s
Thus, the equation that describes the position of the object as a function of time is
x(t ) = (12.0 cm) cos[(6.00π s −1 )t ] .
33. Strategy Use Eqs. (10-21) and (10-22).
Solution
(a) Find vm and am in terms of f. Then compare high- and low-frequency sounds.
vm = ω A = 2π fA ∝ f and am = ω 2 A = 4π 2 f 2 A ∝ f 2 , so vm and am are greatest for high frequency .
(b) vm = 2π (20.0 Hz)(1.0 × 10−8 m) = 1.3 × 10−6 m s
am = 4π 2 (20.0 Hz)2 (1.0 × 10−8 m) = 1.6 × 10−4 m s 2
(c) vm = 2π (20.0 × 103 Hz)(1.0 × 10−8 m) = 0.0013 m s
am = 4π 2 (20.0 × 103 Hz)2 (1.0 × 10−8 m) = 160 m s 2
34. Strategy Use Eqs. (10-21) and (10-22).
2
Solution For SHM, show that vm
= am A.
vm = ω A, so ω =
vm
v2
v2
2
. am = ω 2 A = m2 A = m , so vm
= am A.
A
A
A
35. Strategy The angular frequency of oscillation is inversely proportional to the square root of the mass. Form a
proportion.
Solution Find the new value of ω.
ω∝
1 mf
ω
1
=
, so f =
ωi
m
1 mi
mi
=
mf
1
1
ω
10.0 rad s
=
= 5.0 rad s .
. Therefore, ωf = i =
4.0 2.0
2.0
2.0
490
College Physics
Chapter 10: Elasticity and Oscillations
36. (a) Strategy Use Hooke’s law.
Solution When the cart moves to the right, the spring on the
right compresses an amount x and the spring on the left stretches
an amount x. The spring on the right pushes on the cart with a
force of kx to the left, while the spring on the left pulls on the
cart with a force of kx to the left. When the cart moves to the
left, the situation is reversed. Thus, the forces are identical and
the magnitude of the net force on the cart is 2kx .
x
x
kx
kx
(b) Strategy Use Eq. (10-20a).
Solution The effective spring constant for the two springs (as if they were one) is 2k, so the angular
frequency of the cart is ω =
2k m .
37. Strategy Use Eqs. (10-20a) and (10-20b) and Newton’s second law.
Solution Find the spring constant.
m
g (24 kg)(9.80 m s 2 )
ΣFy = kx − mchild g = 0, so k = child =
= 840 N m.
0.28 m
x
Find the mass of the wooden horse.
k
k
ωchild =
, so mhorse =
− mchild .
mchild + mhorse
ωchild 2
Find the oscillation frequency of the spring when no one is sitting on the horse.
ω
1
k
1
k
1
840 N m
=
=
f horse = horse =
840 N m
k
π
2π
2π mhorse 2π
2
−
m
child
2
ωchild
(0.88 Hz) 2 (2π rad cycle)2
Fs
y
mchildg
− 24 kg
= 2.5 Hz
38. Strategy Use Eq. (10-22).
Solution Find the oscillation frequency of the bird’s wingtips.
am
am = ω 2 A = 4π 2 f 2 A, so f =
4π 2 A
=
12 m s 2
4π 2 (0.050 m)
= 2.5 Hz .
39. Strategy Use Eqs. (10-21) and (10-22) and Newton’s second law.
Solution Find the radio’s maximum displacement and maximum speed, and the maximum net force exerted on it.
a
(a) am = ω 2 A, so A = m2 =
ω
(b) vm = ω A = ω
am
ω2
=
am
ω
98 m s 2
2
4π (120 Hz)
=
2
= 1.7 × 10−4 m .
98 m s 2
= 0.13 m s
2π (120 Hz)
(c) According to Newton’s second law, Fm = mam = (5.24 kg)(98 m s 2 ) = 510 N .
491
Chapter 10: Elasticity and Oscillations
College Physics
40. Strategy Use Eqs. (10-21) and (10-22).
Solution Compute the maximum speed and acceleration to which the pilot is subjected.
vm = ω A = 2π (25.0 Hz)(0.00100 m) = 0.157 m s
am = ω 2 A = 4π 2 (25.0 Hz)2 (0.00100 m) = 24.7 m s 2
41. (a) Strategy Use Newton’s second law and Eq. (10-22).
Solution Find the maximum force acting on the diaphragm.
Fm = mam = mω 2 A = 4π 2 (0.0500 kg)(2.0 × 103 Hz)2 (1.8 × 10−4 m) = 1.4 kN .
(b) Strategy The maximum elastic potential energy of the diaphragm is equal to the total mechanical energy.
Solution Find the mechanical energy of the diaphragm.
1
E = U = mω 2 A2 = 2π 2 (0.0500 kg)(2.0 × 103 Hz)2 (1.8 × 10−4 m)2 = 0.13 J .
2
42. Strategy Use Hooke’s law and Newton’s second law.
Solution
(a) According to Newton’s second law, at equilibrium, ΣFy = 0 = kd − mg , so
k = mg d . When the extension of the spring is a maximum,
2kd
kd
ΣFy = ma = k (2d ) − mg = 2(mg d )d − mg = mg. Therefore, a = g .
a
mg
mg
(b) At maximum extension, ΣFy = kx − mg = ma. Solve for x.
m
(1.0 kg)(9.80 N kg + 9.80 N kg)
= 0.78 m .
kx = mg + ma = m( g + a), so x = ( g + a) =
k
25 N m
43. (a) Strategy The speed is maximum when the spring and mass system is at its equilibrium point. Use Newton’s
second law.
Solution Find the extension of the spring.
mg (0.60 kg)(9.80 N kg)
ΣFy = kx − mg = 0, so x =
=
= 0.39 m .
k
15 N m
kx
mg
(b) Strategy Use Eqs. (10-20a) and (10-21).
Solution Find the maximum speed of the body.
k
15 N m
vm = ω A =
x=
(0.39 m) = 2.0 m s
m
0.60 kg
492
College Physics
Chapter 10: Elasticity and Oscillations
44. Strategy Assuming SHM, the kinetic energy decreases by an amount equal to the increase in elastic potential
energy.
Solution Find the change in kinetic energy.
1
1
∆K = − kx 2 = − (25 N m)(0.050 m)2 = −0.031 J
2
2
45. Strategy Use Hooke’s law, Newton’s second law, and Eq. (10-20c).
Solution Find the “spring constant” of the boat. At equilibrium,
mperson g
.
ΣFy = kx − mperson g = 0, so k =
x
Compute the period of oscillation.
mtotal
mtotal
mtotal x
(47 kg + 92 kg)(0.080 m)
= 2π
= 2π
= 2π
= 0.70 s
T = 2π
k
mperson g x
mperson g
(92 kg)(9.80 m s 2 )
kx
mperson g
46. (a) Strategy Use Newton’s second law and Eq. (10-20a).
Solution Determine the spring constant of the cord.
mg
At equilibrium, ΣFy = 0 = kd − mg. So, k =
, where d = 0.20 m.
d
Calculate the period.
2π
2π
0.20 m
m
m
d
=
= 2π
= 2π
= 2π
= 2π
= 0.90 s
T=
ω
k
mg d
g
k m
9.80 m s 2
kd
mg
(b) Strategy Use Eqs. (10-21) and (10-20a).
Solution Find the maximum speed of the baby.
vm = ω A = A
k
mg
g
9.80 m s 2
=A
=A
= (0.080 m)
= 0.56 m s
m
dm
d
0.20 m
47. Strategy and Solution Since y(t ) = A sin ω t , f =
ω
2π
=
1.57 rad s
= 0.250 Hz .
2π
48. Strategy The maximum kinetic energy occurs at the equilibrium point where v = vm = ω A. For SHM,
ω = k m.
Solution Find the maximum kinetic energy of the body.
1 2 1
1 ⎛k⎞
1
1
K max = mvm
= mω 2 A2 = m ⎜ ⎟ A2 = kA2 = (2.5 N m)(0.040 m) 2 = 2.0 mJ
2
2
2 ⎝m⎠
2
2
493
Chapter 10: Elasticity and Oscillations
College Physics
49. (a) Strategy The object will oscillate up and down with an amplitude determined by the spring constant and the
mass of the spring.
Solution Find the amplitude of the motion. At the equilibrium point, the net force on
the object is zero.
ΣFy = kA − mg = 0, so A = mg k = (0.306 kg)(9.80 m s 2 ) (25 N m) = 12 cm.
The object will move up and down a total vertical distance of 2 A = 24 cm. Thus, the
kA
pattern traced on the paper by the pen is a vertical straight line of length 24 cm .
(b) Strategy and Solution As the paper moves to the left at constant speed while the pen
oscillates vertically in SHM, the pen traces a pattern of
mg
a positive cosine plot of amplitude 12 cm .
24 cm
50. Strategy Graph (a) x, (b) vx , and (c) a x on the vertical axis and t on the horizontal axis. Analyze the slopes.
Solution
(a) x(t ) = A sin ω t
x
A
0
2π
ω
t
π
ω
(b) The slope of the x(t) graph is maximum at t = 0, so vx (t ) is max at t = 0. Since vx (t ) starts at its maximum
value and then is positive and decreasing, it is a cosine function, vx (t ) = vm cos ωt.
vx
vm
0
π
ω
2π
ω
t
(c) The slope of the vx (t ) graph is zero at t = 0 and becomes negative just after, so a x (t ) starts at zero and
decreases. Since a x (t ) starts at zero and then is negative and decreasing, it is a negative sine function,
a x (t ) = −am sin ω t.
ax
am
0
π
ω
2π
ω
t
(d) Compare minimums. x(t ) has its minimum at t =
3π
π
π
.
, vx (t ) has its at t = , and a x (t ) has its at t =
ω
2ω
2ω
π 3π π
π
π π
π
and −
.
=
, so vx (t ) is one quarter cycle ahead of x(t) and
− =
ω 2ω 2ω
2ω 2ω ω 2ω
a x (t ) is one quarter cycle ahead of vx (t ).
One quarter cycle is
494
College Physics
Chapter 10: Elasticity and Oscillations
51. Strategy Use the definition of average speed and Eq. (10-21). In (d), graph vx on the vertical axis and t on the
horizontal axis.
Solution
(a) The average speed is the total distance traveled divided by the time of travel.
∆x 4 A
4A
2
ωA
vav =
=
=
=
π
∆t
2π ω
T
(b) The maximum speed for SHM is vm = ω A .
2
vav π ω A
=
=
ωA
π
vm
2
(c)
(d) Graph vx (t ) and a line from the origin to vm .
vx
ωA
2π
ω
0
t
π
ω
−ω A
If the acceleration were constant so that the speed varied linearly, the average speed would be 1/2
of the maximum velocity. Since the actual speed is always larger than what it would be for constant
acceleration, the average speed must be larger.
52. Strategy Use the equations of motion for constant acceleration. Graph y on the vertical axis and t on the
horizontal axis.
Solution Analyze the motion of the ball.
During the fall:
1
1
1
y − yi = y − h = viy ∆t − g (∆t ) 2 = (0)∆t − g (t − 0)2 , so y = h − gt 2. At y = 0, t =
2
2
2
2h
.
g
During the rise:
The speed of the ball just before and after it hits the ground is v = gt = g
ti =
2h
= 2 gh . So, at
g
1
2h
, viy = 2 gh , and y = yi = 0. If ti = 0 when y = yi = 0, then y = viy ∆t − g (∆t ) 2. The graph is shifted
2
g
2
⎛
2h
2h ⎞ 1 ⎛
2h ⎞
2h
− g ⎜t −
, so y = 2 gh ⎜ t −
.
⎟
⎟⎟ , and y = h when t = 2
⎜
⎟ 2 ⎜
g
g
g
g
⎝
⎠
⎝
⎠
to the right by
y
h
0
2h
g
2h
2 g
t
The motion is not SHM, since it is not a sine or cosine function . The graph is always nonnegative and has a
parabolic shape.
495
Chapter 10: Elasticity and Oscillations
College Physics
53. (a) Strategy Use Eq. (10-20a) to find the spring constant. Then, find the elastic potential energy using 12 kx 2 .
Solution Find the spring constant.
k
ω=
, so k = ω 2 m = (2.00 Hz) 2 (2π rad cycle)2 (0.2300 kg) = 36.3 N m.
m
The equation for the elastic potential energy is
1
U (t ) = (36.3 N m)(0.0800 m)2 sin 2 [ (2.00 Hz)(2π rad cycle)t ] = (116 mJ) sin 2 ⎡ (4.00π s −1 )t ⎤ .
⎣
⎦
2
Since the sine function is squared, the period of U (t ) is half that of a sine function or
T=
π
π
=
= 250 ms. Graph U (t ).
ω 4.00π s −1
U (mJ)
116
0
0
500 t (ms)
250
(b) Strategy Find the kinetic energy using 12 mvx 2 .
Solution The equation for the kinetic energy is
2
K (t ) =
⎛ 2π rad ⎞
⎛ 2π rad ⎞ ⎤
1
2
2⎡
(0.2300 kg)(2.00 Hz) 2 ⎜
⎟ (0.0800 m) cos ⎢ (2.00 Hz) ⎜
⎟t⎥
2
cycle
⎝
⎠
⎝ cycle ⎠ ⎦
⎣
= (116 mJ) cos 2 ⎡(4.00π s −1 )t ⎤ .
⎣
⎦
Since the cosine function is squared, the period of K (t ) is half that of a cosine function or
T = π ω = π (4.00π s −1 ) = 250 ms, which is the same as U (t ). Graph K (t ).
K (mJ)
116
0
0
500 t (ms)
250
(c) Strategy Add U (t ) and K (t ) and graph the result.
Solution
E (t ) = U (t ) + K (t ) = (116 mJ) sin 2 ⎡(4.00π s −1 )t ⎤ + (116 mJ) cos 2 ⎡ (4.00π s −1 )t ⎤
⎣
⎦
⎣
⎦
2
−
1
2
−
1
⎡
⎤
⎡
⎤
= (116 mJ) sin (4.00π s )t + cos (4.00π s )t = (116 mJ)(1) = 116 mJ
⎣
⎦
⎣
⎦
Graph E (t ) = U (t ) + K (t ).
{
}
E (mJ)
116
0
0
250
500 t (ms)
(d) Strategy and Solution Friction does nonconservative work on the object, thus,
U , K , and E would gradually be reduced to zero .
496
College Physics
Chapter 10: Elasticity and Oscillations
54. (a) Strategy Draw the velocity vector for point P in Fig. 10.17b and then find its x-component.
Solution Show that vx (t ) = −ω A sin ω t.
vm
y
θ
vy
x
vx
From the figure, vx = −vm sin θ . For SHM, vm = ω A and θ = ω t , so vx (t ) = −ω A sin ω t.
(b) Strategy Use conservation of energy and Eq. (10-20a).
Solution Verify that the expressions for x(t) and vx (t ) are consistent with energy conservation.
1
1
1
1
1
E = kA2 = K + U = mv 2 + kx 2 = mω 2 A2 sin 2 ωt + kA2 cos 2 ωt
2
2
2
2
2
2
1 ⎛ k ⎞ 2 2
1 2
1 2
1 2
1 2
2
2
2
= m⎜
⎟ A sin ωt + kA cos ωt = kA (sin ωt + cos ωt ) = kA (1) = kA
2 ⎜⎝ m ⎟⎠
2
2
2
2
The expressions for x(t) and vx (t ) are consistent with energy conservation.
55. Strategy Use Eq. (10-26b).
Solution Compute the period of the pendulum.
L
4.0 m
= 2π
= 4.0 s
T = 2π
g
9.80 m s 2
56. Strategy The total mechanical energy for a simple pendulum is E = 12 mω 2 A2. Use Eq. (10-26a).
Solution Find the amplitude of the pendulum.
2
1
1 ⎛ g ⎞ 2 mg 2
E = mω 2 A2 = m ⎜
A , so A =
⎟ A =
2
2 ⎜⎝ L ⎟⎠
2L
2 EL
=
mg
57. Strategy and Solution According to Eq. (10-26b), T = 2π
2(0.015 J)(0.75 m)
= 3.0 cm .
(2.5 kg)(9.80 N kg)
L
, which does not depend upon the mass.
g
Therefore, T = 1.5 s .
58. Strategy Use Eq. (10-26b) and form a proportion with the final and initial periods.
Solution Find the period of oscillation of the pendulum.
Tf 2π
=
Ti 2π
Lf
g
Li
g
=
Lf
=
Li
2L
= 2, so Tf = Ti 2 = (2.0 s) 2 = 2.8 s .
L
497
Chapter 10: Elasticity and Oscillations
College Physics
59. (a) Strategy Graph vx on the vertical axis and t on the horizontal axis. Use Eq. (10-21).
Solution vx leads x by one quarter cycle [π (2ω )] and vm = ω A, so if x = A sin ωt , vx = ω A cos ωt .
vx
ωA
0
π
ω
2π
ω
t
−ω A
(b) Strategy Use Eqs. (6-6) and (10-21).
Solution Find the maximum kinetic energy.
1 2
1
K m = mvm
, so K m = mω 2 A2 .
2
2
60. Strategy Use Eq. (10-26b).
Solution Find the length of the pendulum.
L
gT 2 (9.80 m s 2 )(1.0 s) 2
T = 2π
, so L =
=
= 0.25 m .
g
4π 2
4π 2
61. Strategy Use Eq. (10-26b) to find the length of the pendulum. Then, form a ratio of the lengths.
Solution Solve for L.
L
gT 2
T = 2π
, so L =
.
g
4π 2
Form a proportion.
2
L2 T22 ⎛ 1.00 s ⎞
=
=
= 1.11
L1 T12 ⎜⎝ 0.950 s ⎟⎠
62. (a) Strategy and Solution Since the period of a pendulum is inversely proportional to gravitational field
strength, the greater period implies that the gravitational field strength on the other planet is less than
that on Earth.
(b) Strategy Use Eq. (10-26b). Form a proportion.
Solution Refer to the mystery planet as X.
T
L
L
L ⎛
L
TE = 2π
and TX = 2π
, so E = 2π
⎜⎜ 2π
gE
gX
TX
gE ⎝
gX
2
⎞
⎟⎟
⎠
−1
=
2
⎛T ⎞
⎛ 0.650 s ⎞
2
Thus, g X = g E ⎜⎜ E ⎟⎟ = (9.80 m s 2 ) ⎜
⎟ = 5.57 m s .
0.862
s
T
⎝
⎠
⎝ X⎠
498
gX
gE
.
College Physics
Chapter 10: Elasticity and Oscillations
63. Strategy The amplitude is (20.0 mm) 2 = 10.0 mm and the period is 2.00 s. Use Eqs. (10-21) and (10-26b) and
conservation of energy.
Solution Find the maximum speed of the pendulum bob.
1st method:
2π
2π (10.0 × 10−3 m)
vm = ω A =
A=
= 3.14 cm s
T
2.00 s
2nd method:
Find the length L of the pendulum.
L
T 2 g (2.00 s) 2 (9.80 m s 2 )
T = 2π
, so L =
=
= 0.993 m.
g
4π 2
4π 2
θ max
L
A
Find θ max .
A 10.0 × 10−3 m
=
= 1.007 × 10−2 rad.
L
0.993 m
The height h of the pendulum bob above its lowest point is the difference between the
length of the pendulum L and the vertical distance from the axis to the bob when it is at its
maximum height, L cos θ . So, h = L(1 − cos θ ).
Find vm .
Since the displacement is small, θ max ≈ sin θ max =
∆K =
Axis
θ
L cos θ
L
1 2
mvm = −∆U = mgh, so
2
vm = 2 gh = 2 gL(1 − cos θ ) = 2(9.80 m s 2 )(0.993 m)(1 − cos1.007 × 10−2 ) = 3.14 cm s .
h
64. (a) Strategy Use Eq. (10-26b).
Solution Compute the period of the pendulum.
L
1.000 m
= 2π
= 2.01 s
T = 2π
g
9.80 m s 2
(b) Strategy Refer to the derivation of the physical pendulum in Section 10.8 of
College Physics.
Solution Let the mass of the point mass by m and the mass of the uniform thin rod
be mr . Find the net torque acting on the physical pendulum.
Axis
θ
L
L/2
m rg
L
gL
gL
Στ = F⊥ r = − mr g sin θ − mg sin θ L = −
sin θ ( mr + 2m) ≈ − θ ( mr + 2m),
2
2
2
mg
where the approximation for small amplitudes (sin θ ≈ θ ) was used. The net torque
is equal to the rotational inertia times the angular acceleration, so
gLθ (mr + 2m)
gL(mr + 2m)
gL
Στ = − θ (mr + 2m) = I α , and the angular acceleration is α = −
=−
θ , where
2
2I
2( I r + I )
I r and I are the rotational inertias for the rod and the point mass, respectively. Since we have SHM, our
equation for the angular acceleration is analogous to the equation for the linear acceleration of the oscillating
gL(mr + 2m)
spring, a x = −ω 2 x, where the angular frequency in our case is ω =
. Thus, the period of the
2( I r + I )
pendulum is T =
2π
ω
= 2π
2( I r + I )
. Now, the rotational inertias of the rod and point mass are
gL(mr + 2m)
499
Chapter 10: Elasticity and Oscillations
College Physics
I r = 13 mr L2 and I = mL2 , respectively, so the period becomes T = 2π
compare the mass of the rod to the total mass.
(
(
2 L 1 + 3mm
2 L(mr + 3m)
r
T = 2π
= 2π
3 g (mr + 2m)
2
m
3g 1 + m
(
(
3m
mr
2
2L 1 +
⎛ T ⎞
=
⎜
⎟
⎝ 2π ⎠
3 g 1 + 2mm
1 + 3mm
r
1 + 2mm
r
r
)
)
r
2 L(mr + 3m)
. So, now we need to
3 g (mr + 2m)
)
)
2
⎛ T ⎞ 3 g 3T 2 g
=⎜
=
⎟
⎝ 2π ⎠ 2 L 8π 2 L
3m 3T 2 g ⎛ 2m ⎞ 3T 2 g 3T 2 g ⎛ 2m ⎞
=
+
1+
⎜1 +
⎟=
⎜
⎟
mr 8π 2 L ⎜⎝ mr ⎟⎠ 8π 2 L 8π 2 L ⎜⎝ mr ⎟⎠
m⎛
3T 2 g ⎞ 3T 2 g
⎜⎜ 3 − 2 ⎟⎟ = 2 − 1
mr ⎝
4π L ⎠ 8π L
3T 2 g
− 1 16π 2 L − 3T 2 g
2
m mr + m
1+
=
= 1 + 8π L
=
mr
mr
3T 2 g 24π 2 L − 6T 2 g
3−
4π 2 L
mr
mr + m
× 100% =
24π 2 L − 6T 2 g
16π 2 L − 3T 2 g
× 100% =
24π 2 (1.000 m) − 6(0.99 × 2.007 s) 2 (9.80 m s 2 )
16π 2 (1.000 m) − 3(0.99 × 2.007 s) 2 (9.80 m s 2 )
× 100% = 11.3%
The percentage of the total mass of the pendulum in the uniform thin rod is 11.3% .
65. Strategy The total mechanical energy of a pendulum is E = 12 mω 2 A2. Form a proportion.
Solution Find the mechanical energy of the pendulum.
2
2
⎛A ⎞
E2 A22
⎛ 3.0 cm ⎞
= 2 , so E2 = ⎜ 2 ⎟ E1 = ⎜
⎟ (5.0 mJ) = 11 mJ .
E1 A1
⎝ 2.0 cm ⎠
⎝ A1 ⎠
66. Strategy Use Eqs. (10-26a) and (10-28b).
Solution Show that the thin circular hoop oscillates with the same frequency as a simple pendulum of length
equal to the diameter of the hoop.
Hoop:
T = 2π
I
2mr 2
2r
D
= 2π
= 2π
= 2π
mgd
mgr
g
g
1
1 g
, where D is the diameter of the hoop.
=
T 2π D
Simple pendulum:
ω
1 g
fP =
=
= f H if L = D.
2π 2π L
So, f H =
500
College Physics
Chapter 10: Elasticity and Oscillations
67. (a) Strategy The total mechanical energy of a pendulum is E = 12 mω 2 A2. Use Eq. (10-26a).
Solution Find the energy of the pendulum.
2
E=
1
1 ⎛ g ⎞ 2 mgA2 (0.50 kg)(9.80 m s 2 )(0.050 m)2
mω 2 A2 = m ⎜
=
= 6.1 mJ
⎟ A =
2
2 ⎜⎝ L ⎟⎠
2L
2(1.0 m)
(b) Strategy Use Eq. (10-26b) and the equation for the potential energy of an object in a uniform gravitational
field.
Solution Find the percentage of the pendulum’s energy lost during one cycle.
L
T = 2π
and U = m2 gh where m2 = 2.0 kg.
g
cycles per week =
U
cycles/wk
E
1 wk 604,800 s
=
T
2π
× 100% =
=
g
L
m2 gh
604,800 s
2π
g
L
(
1m g
2 1L
A2
)
× 100% =
4π m2h
(604,800 s)m1 A2
L3
× 100%
g
(1.0 m)3
4π (2.0 kg)(1.0 m)
(604,800 s)(0.50 kg)(0.050 m)2
× 100% = 1.1%
9.80 m s 2
68. Strategy E = 12 mω 2 A2 ∝ A2 for a pendulum. Form a proportion.
Solution Find by what factor the energy has decreased.
E2 A22 ( A1 20.0)2
1
1
=
=
=
=
2
2
E1 A12
400
A1
20.0
The energy has decreased by a factor of 400 .
69. Strategy E = 12 mω 2 A2 ∝ A2 for a pendulum.
Solution Find the percent decrease of the oscillator’s energy in ten cycles.
∆E
∆A 2
(1 − 0.0500)2 − 12
0.95002 − 12
× 100% = 2 × 100% =
×
=
× 100% = − 9.75%
100%
E
A
12
12
70. Strategy Use Eq. (10-20b).
Solution Find the spring constant.
1 k
f =
, so k = 4π 2 f12 m1 , where m1 is the combined mass of all four people and the car.
2π m
Compute the frequency when only the 45-kg person is present.
f2 =
1
2π
k
1
=
m2 2π
4π 2 f12 m1
(2.00 Hz)2 (1020 kg + 45 kg + 52 kg + 67 kg + 61 kg)
=
= 2.16 Hz
m2
1020 kg + 45 kg
501
Chapter 10: Elasticity and Oscillations
College Physics
71. Strategy Use Eq. (10-21). Assume the amplitude of the pendulum is small.
Solution Find the period T of the pendulum.
2π
2π
0.50 m s
=
= 2.5 s .
vm = ω A = 0.50 m s and ω =
= 2.5 rad s. Thus, T =
ω 2.5 rad s
0.20 m
72. Strategy Use Eq. (10-26a).
Solution We must assume that the pendulum is located on Earth. Find its length.
ω=
g
g
9.80 m s 2
, so L =
=
= 0.994 m .
L
ω 2 (3.14 rad s)2
73. Strategy Use Eq. (10-20c).
Solution
(a) The period is directly proportional to the square root of the mass, and the period for the fish is longer than that
for the weight, so the fish weighs more than the weight.
(b) Form a proportion. Let f = fish and w = weight.
T
m
T = 2π
, so f =
k
Tw
mf
mw
⎛T
. Thus, Wf = ⎜⎜ f
=
Ww
⎝ Tw
Wf
2
2
⎞
⎛ 220 ⎞
⎟⎟ Ww = ⎜
⎟ (4.90 N) = 56 N .
⎝ 65 ⎠
⎠
74. Strategy Assume the weight of the cable is negligible compared to the weight of the aviator. Use Eq. (10-26b).
Solution Find the period for a pendulum assuming SHM.
45 m
L
= 2π
= 13 s
T = 2π
g
9.80 m s 2
75. Strategy Refer to Eqs. (10-20). Use conservation of energy and the fact that E ∝ A2 .
Solution Analyze the mass and spring system.
(a)
The frequency and period don’t vary with amplitude, they only vary with m and k . Since these two
values remain constant, so do the frequency and period.
(b) Since the total energy of a spring is directly proportional to the square of the amplitude,
the total energy for an amplitude of 2 D is four times that for an amplitude of D .
(c) The initial speed will essentially result in a greater initial displacement; therefore, it will have a greater
amplitude. Since the frequency and period don’t vary with amplitude, the frequency and period are still
the same.
(d) The energy is greater when given an initial push, since it has an amplitude greater than 2D. The increase in
energy is
1 mv 2 ,
i
2
due to the initial kinetic energy.
502
College Physics
Chapter 10: Elasticity and Oscillations
76. Strategy Assume the web obeys Hooke’s law. Use Eq. (10-20b) and Newton’s second law.
Solution According to Newton’s second law, at equilibrium, ΣFy = kd − mg = 0, so k m = g d .
kd
Calculate the frequency of oscillation.
f =
1
2π
k
1
=
m 2π
g
1
=
d 2π
9.80 m s 2
0.030 × 10−3 m
= 91 Hz .
mg
77. Strategy Graph x on the vertical axis and t on the horizontal axis. Analyze the slope of the graph (the magnitude
of which is the speed) in terms of the distance between the dots to determine the fastest and slowest speeds of the
mass.
Solution Graph x(t ) = −(10 cm) cos[(1.57 s −1 )t ].
x (cm)
10
0
1.0
2.0
4.0 t (s)
3.0
−10
The distance between adjacent dots should be the least at the endpoints and greatest at the center,
so its speed is lowest at the endpoints and fastest at its equilibrium position.
78. Strategy Assume SHM. The amplitude of the motion is half the movement during one complete stroke, or 1.2
cm. Use Eqs. (10-20b), (10-21), and (10-22).
Solution Compute the maximum speed and maximum acceleration of the blade.
vm = ω A = 2π fA = 2π (28 Hz)(0.012 m) = 2.1 m s and am = ω 2 A = 4π 2 (28 Hz)2 (0.012 m) = 370 m s 2 .
79. Strategy I = mL2 for a simple pendulum of length L and mass m. Use Eq. (10-28a).
Solution
mgd
ω=
=
I
mgd
2
mL
=
gd
2
L
=
g ( L)
2
L
=
g
, which is the angular frequency of a simple pendulum.
L
503
Chapter 10: Elasticity and Oscillations
College Physics
80. (a) Strategy Use Hooke’s law and Newton’s second law.
Solution Find the spring constant. At equilibrium, the spring has stretched a distance
d = 0.310 m − 0.200 m = 0.110 m.
ΣFy = kd − mg = 0, so k =
mg (1.10 kg)(9.80 m s 2 )
=
= 98.0 N m .
d
0.110 m
kd
mg
(b) Strategy Use conservation of energy.
Solution The total energy of the spring is equal to the maximum kinetic energy, as well as the maximum
potential energy. Let d = 0.0500 m.
K max =
1
1
mv 2 = U max = kd 2 , so vmax =
2 max
2
kd 2
(98.0 N m)(0.0500 m) 2
=
= 0.472 m s .
m
1.10 kg
(c) Strategy Use conservation of energy.
Solution Find the speed of the brick.
1
1
1
E = kd 2 = K + U = mv 2 + ky 2 , so
2
2
2
98.0 N m
k 2
2
(d − y ) =
[(0.0500 m)2 − (0.0250 m)2 ] = 0.409 m s .
v=
1.10 kg
m
(d) Strategy Use Eq. (10-20c).
Solution The time is will take for the brick to oscillate five times is five times the period.
⎛
m⎞
1.10 kg
= 3.33 s
5T = 5 ⎜⎜ 2π
⎟⎟ = 10π
k
98.0
N m
⎝
⎠
81. Strategy Since the body begins with its maximum amplitude at t = 0, the body oscillates according to a cosine
function (cos 0 = 1). Use Newton’s second law, Hooke’s law, and Eq. (10-20a).
Solution Find the amplitude A.
mg
4.0 N
ΣFy = kA − mg = 0, so A =
=
= 1.6 cm.
k
250 N m
Find the angular frequency ω .
ω=
k
=
m
kg
=
mg
kA
y
mg
(250 N m)(9.80 m s 2 )
= 25 rad s
4.0 N
Thus, the equation describing the motion of the body is y = (1.6 cm) cos [ (25 rad s )t ] .
82. Strategy Use dimensional analysis.
Solution [ A] = m, [k ] = kg s 2 , and [m] = kg; we need [ f ] = s −1. Only k has units which include seconds, so f
must be proportional to the square root of k to get s −1. kg must be eliminated and only m has units of kg;
therefore, f ∝ k m .
504
College Physics
Chapter 10: Elasticity and Oscillations
83. (a) Strategy Use conservation of energy.
Solution Let the maximum displacement be d. Find d.
1
1
1
E = K + U = mv 2 + kx 2 = kd 2 , so
2
2
2
2
mv
(1.24 kg)(0.543 m s)2
d=
+ x2 =
+ (0.345 m) 2 = 0.395 m .
k
9.82 N m
(b) Strategy The maximum speed of the block occurs when the block is at it equilibrium position, x = 0.
Use conservation of energy.
Solution Find the maximum speed of the block.
1
1
1
E = K + U = mv 2 + kx 2 = kd 2 , so
2
2
2
2
2
k (d − x )
(9.82 N m)[(0.395 m) 2 − 02 ]
vmax =
=
= 1.11 m s .
m
1.24 kg
(c) Strategy Use the result from part (b).
Solution
v=
k (d 2 − x 2 )
(9.82 N m)[(0.3953 m) 2 − (0.200 m)2 ]
=
= 0.960 m s
m
1.24 kg
84. Strategy The angle θ through which the tuning pin must be turned is related to the extension of the wire ∆L by
the formula for arc length. Use Eq. (10-4).
Solution Relate θ , ∆L, and d p , the diameter of the tuning pin.
dp
s = rθ =
θ = ∆L
2
Find the angle through which the tuning pin must be turned to tune the piano wire.
(d p 2)θ
∆L
F
F
=
=Y
=Y
, so
2
A π (d w 2)
L
L
θ=
2 FL
2
d pπ (d w 2) Y
=
8FL
π d p d w2 Y
=
8(402 N − 381 N)(0.66 m)
2
11
π (0.0080 m)(0.00080 m) (2.0 × 10
s = ∆L
dp / 2 θ
⎛ 360° ⎞
⎜
⎟ = 2.0° .
Pa) ⎝ 2π ⎠
85. Strategy Use Eq. (10-2) to find the tensile stress. Then compare the tensile stress to the elastic limit of steel piano
wire.
Solution Find the tensile stress in the piano wire in Problem 90.
F
T
4T
4(402 N)
tensile stress = =
=
=
= 8.0 × 108 Pa < 8.26 × 108 Pa
A 1 π d 2 π d 2 π (0.80 × 10−3 m) 2
4
The tensile stress is 8.0 × 108 Pa; it is just under the elastic limit.
505
Chapter 10: Elasticity and Oscillations
College Physics
86. Strategy Assume that the cable was horizontal prior to being stepped on by the tightrope walker. Neglect the
weight of the cable. Use Eqs. (10-1), (10-2), and (10-4), as well as Newton’s second law.
Solution
(a) Find the strain in the cable.
∆L L′ − L L′ − L′ cos θ
1
1
=
=
=
−1 =
− 1 = 8 × 10−4
strain =
L
L
L′ cos θ
cos θ
cos 0.040
L
θ
L′
(b) Find the tension in the cable.
ΣFy = 2T sin θ − mg = 0, so T =
mg
640 N
=
= 8.0 kN .
2sin θ 2sin 0.040
T
θ
θ
mg
(c) Find the cross-sectional area of the cable.
T
mg
∆L
⎛ 1
⎞
=
=Y
=Y⎜
− 1⎟ , so
A 2 A sin θ
L
⎝ cos θ
⎠
mg
640 N
A=
=
= 5 × 10−5 m 2 .
9
1
1
2Y sin θ cos θ − 1
2(200 × 10 Pa) sin 0.040 cos 0.040
−1
(
)
(
)
(d) Compute the stress.
T 8.0 × 103 N
=
= 1.6 × 108 Pa < 2.5 × 108 Pa
A 5 × 10−5 m 2
No , the cable has not been stretched beyond its elastic limit.
87. Strategy Treat the swinging gibbon as a physical pendulum. Use Eq. (10-28a).
Solution Estimate the frequency of oscillation of the gibbon.
f ≈
1
2π
mgd
1
=
I
2π
(9.80 m s 2 )(0.40 m)
0.25 m 2
= 0.63 Hz
88. (a) Strategy Approximate the tibia as two concentric circles.
Solution Find the average cross-sectional area of the tibia.
⎡⎛ 2.5 cm ⎞2 ⎛ 1.3 cm ⎞ 2 ⎤
2
A = π ro2 − π ri2 = π ⎢⎜
⎟ −⎜
⎟ ⎥ = 3.6 cm
⎢⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦
(b) Strategy Use Eq. (10-2).
Solution Find the compressive stress in the tibia.
F
2800 N
compressive stress = =
= 7.8 × 106 Pa
A 3.6 × 10−4 m 2
(c) Strategy Use Eq. (10-4) and the results of parts (a) and (b). Use Y for a femur in Table 10.1.
Solution Find the change in length for the tibia due to the compressive forces.
FL
∆L F
(2800 N)(0.40 m)
Y
= , so ∆L =
=
= 3.3 × 10−4 m .
4
2
9
−
L
A
AY (3.6 × 10 m )(9.4 × 10 Pa)
506
T
College Physics
Chapter 10: Elasticity and Oscillations
89. Strategy The force on the column is its weight. Use Eq. (10-2) and the relationship between density, mass, and
volume.
Solution
(a) Calculate the compressive stress at the bottom of the column.
F mg ρVg ρ hAg
compressive stress = =
=
=
= ρ gh
A
A
A
A
(b) Find the absolute limit to the height of a cylindrical column, regardless of how wide it is.
compressive strength
2.0 × 108 Pa
hm =
=
= 7.6 km
ρg
(2.7 × 103 kg m3 )(9.80 m s 2 )
(c) It is unlikely that someone would want to build a marble column taller than 7.6 km. So, the answer is no; this
limit is of little practical concern. No beanstalk could ever reach a height of 7.6 km; its height is limited by
other means.
90. (a) Strategy Gravitational potential energy is converted into elastic potential energy in the bungee cord. Assume
the cord obeys Hooke’s law. Assume SHM and use Eq. (10-20c).
Solution Find k for the cord.
1 2
2mgh
.
ky = mgh, so k =
2
y2
Find the period of oscillations of the bungee cord.
m
m
2
2
T = 2π
= 2π
=πy
= π (50.0 m − 33.0 m)
= 3.42 s
2
k
gh
2mgh y
(9.78 m s 2 )(50.0 m)
(b) Strategy Use conservation of energy and the quadratic formula.
Solution Find the extension of the bungee cord y2.
1 2 1 ⎛ 2m1gh ⎞ 2
m y2
m y 2 (33.0 m)
ky2 = ⎜ 2 ⎟ y2 = m2 gh = m2 g ( y2 + 33.0 m), so 0 = y22 − 2 1 y2 − 2 1
.
2
2 ⎝ y1 ⎠
m1h
m1h
Solve for y2.
y2 =
m2 y12
m1h
2
⎛ m y2 ⎞
⎡ m y 2 (33.0 m) ⎤
± ⎜ − m2 h1 ⎟ − 4(1) ⎢ − 2 1 m h
1 ⎠
1
⎝
⎣
⎦⎥
2(1)
(80.0 kg)(17.0 m)2 1
=
±
2(60.0 kg)(50.0 m) 2
2
⎡ (80.0 kg)(17.0 m)2 ⎤
(80.0 kg)(17.0 m)2 (132 m)
⎢
⎥ +
(60.0 kg)(50.0 m)
⎢⎣ (60.0 kg)(50.0 m) ⎥⎦
= 20.3 m or − 12.6 m
y2 > 0, so y2 = 20.3 m, and 33.0 m + 20.3 m = 53.3 m > 50.0 m.
No, he should not use the same cord because his greater mass will stretch if too much and he will hit the
water.
507
Chapter 10: Elasticity and Oscillations
College Physics
91. (a) Strategy Use Eq. (10-4) and the geometry of the web.
Solution Find the angle the web makes with the horizontal.
∆L
∆L 1.4 × 109 N m 2
Y
= maximum stress, so
=
= 0.35.
L
L
4.0 × 109 N m 2
r
θ
d
r + ∆L
L = r , so ∆L r = 0.35. The new length of a stretched web strand is the hypotenuse of
a right triangle.
r
1
1
1
cos θ =
, so θ = cos −1
=
= cos −1
= 42.2° .
L
L
∆
∆
r + ∆L 1 + r
1 + 0.35
1+ r
(b) Strategy Use Newton’s second law and Eq. (10-2).
Solution Find the tension.
θ
mg
.
ΣFy = T sin θ − mg = 0, so T =
sin θ
Determine the mass of the bug.
F T
mg
= =
= 1.4 × 109 N m 2 , so
A A A sin θ
m=
T
mg
A sin θ (1.4 × 109 N m 2 ) 50(1.0 × 10−11 m 2 ) sin 42.2°(1.4 × 109 N m 2 )
=
= 48 g .
g
9.80 m s 2
(c) Strategy The downward extension of the web is the leg of a right triangle opposite θ . The hypotenuse is
r + ∆L.
Solution Find the distance the web extends downward.
d = (r + ∆L) sin θ = r (1 + ∆L r ) sin θ = (0.10 m)(1.35) sin 42.2° = 9.1 cm
92. Strategy T = 2π I (mgd ) for a physical pendulum, where d is the distance from the axis to the center of mass.
I = 13 mL2 for a uniform rod with the axis through its end.
Solution Find the period of the pendulum for each horizontal axis.
(a) T = 2π
1 mL2
3
1
=
3gd
= 2π L
mgd
2π (1.00 m)
3(9.80 m s 2 )(0.500 m)
= 1.64 s
(b) Treat the meterstick as two rods with lengths 75 cm and 25 cm.
2
2
1 ⎛ m ⎞ ⎛ L ⎞ 1 ⎛ 3m ⎞⎛ 3L ⎞
1 2 ⎛ 1 27 ⎞ 1 2 ⎛ 28 ⎞ 7
2
I = ⎜ ⎟⎜ ⎟ + ⎜
⎟ = 3 mL ⎜ 64 + 64 ⎟ = 3 mL ⎜ 64 ⎟ = 48 mL
3 ⎝ 4 ⎠ ⎝ 4 ⎠ 3 ⎝ 4 ⎟⎜
4
⎠⎝ ⎠
⎝
⎠
⎝ ⎠
T = 2π
7 mL2
48
mgd
= 2π
7 L2
7L
7(1.00 m)
= 2π
= 2π
= 1.53 s
48 g ( L 4)
12 g
12(9.80 m s 2 )
(c) Treat the meterstick as two rods with lengths 60 cm and 40 cm.
2
2
1 ⎛ 4m ⎞⎛ 4 L ⎞ 1 ⎛ 6m ⎞ ⎛ 6 L ⎞
1
216 ⎞ 7
⎛ 64
I= ⎜
mL2
+ ⎜
= mL2 ⎜
+
⎟⎜
⎟
⎟
⎜
⎟
⎟=
3 ⎝ 10 ⎠⎝ 10 ⎠ 3 ⎝ 10 ⎠ ⎝ 10 ⎠
3
⎝ 1000 1000 ⎠ 75
T = 2π
7 mL2
75
mg ( L 10)
= 2π
14 L
14(1.00 m)
= 2π
= 1.94 s
15 g
15(9.80 m s 2 )
508
College Physics
Chapter 10: Elasticity and Oscillations
93. (a) Strategy Use conservation of energy. Do not assume SHM.
Solution Find the speed of the pendulum bob at the bottom of its swing.
1
∆K = mv 2 = −∆U = mgL, so v = 2 gL .
2
(b) Strategy Assume (incorrectly, for such a large amplitude) that the motion is SHM. Use Eqs. (10-21) and
(10-26a).
Solution Find the speed of the pendulum bob at the bottom of its swing.
The amplitude A is a quarter of the circumference of a circle with radius L, or
Assuming SHM, vm = ω A =
g ⎛π ⎞ π
L =
L ⎜⎝ 2 ⎟⎠
2
gL . Since v ∝ ω ∝
2π L π
= L.
4
2
1
, a smaller speed implies a larger
T
π
gL
π
v
period. Since m = 2
=
> 1, the period of a pendulum for large amplitudes is larger than that
v
2 gL
2 2
given by Eq. (10-26b).
94. (a) Strategy Draw a diagram of a pendulum.
Solution From the figure below, we see that L cos θ + y = L, or y = L(1 − cos θ ).
θ
L
L cos θ
L
y
(b) Strategy Assume θ is small. Use the gravitational potential energy of a pendulum given in the problem
statement and the result of part (a).
Solution
⎡ 1 ⎛ x ⎞2 ⎤ 1 ⎛ mg ⎞
mg
1
U = mgy = mgL(1 − cos θ ) ≈ mgL ⎢ ⎜ ⎟ ⎥ = ⎜
x 2 = kx 2 with k =
.
⎟
L
2
⎢⎣ 2 ⎝ L ⎠ ⎥⎦ 2 ⎝ L ⎠
509
Chapter 10: Elasticity and Oscillations
College Physics
95. Strategy The inertia of the system is I = 13 m1L2 + m2 L2. Use Eq. (10-28b) and the definition of center of mass.
Solution
(a) The distance to the center of mass from the rotation axis is
m1 L2 + m2 L
m
1
+ m2
= 2
L.
m1 + m2
m
Find the period of this physical pendulum.
d=
I
T = 2π
= 2π
mgd
1 m L2
3 1
+ m2 L2
mgd
= 2π
L2
mg ⎡
⎢⎣
(
(
m1
2
m1
3
+ m2
+ m2
)
) m⎤⎥⎦ L
= 2π
(
g(
L
) = 2π
+m )
m1
3
+ m2
m1
2
2
2 L(m1 + 3m2 )
3g (m1 + 2m2 )
(b) For each case, replace the smaller mass with zero. Then
for m1 >> m2 , T = 2π
2L
L
, and for m1 << m2 , T = 2π
.
3g
g
The former is the period for the uniform rod alone and the latter is the period for the block alone.
96. (a) Strategy Use Eq. (10-4).
Solution Find the force the wings must exert to extend the resilin.
F
∆L
YA∆L (1.7 × 106 N m 2 )(1.0 ×10−6 m 2 )(4.0 cm − 1.0 cm)
=Y
, so F =
=
= 5.1 N .
A
L
L
1.0 cm
(b) Strategy The energy stored in the resilin is elastic potential energy. Use Hooke’s law.
Solution Find the energy stored in the resilin.
1
1⎛F ⎞
1
1
U = kx 2 = ⎜ ⎟ x 2 = Fx = (5.1 N)(0.030 m) = 7.7 × 10−2 J
2
2⎝ x ⎠
2
2
510