Chapter10 Solution

465

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.

Chapter 10: Elasticity and Oscillations College Physics

466

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.

College Physics Chapter 10: Elasticity and Oscillations

467

Problems

1. Strategy The stress is proportional to the strain. Use Eq. (10-4).

Solution Find the vertical compression of the beam. 4

9 3 2(5.8 10 N)(2.5 m), so 0.097 mm .

(200 10 Pa)(7.5 10 m )L F FLY L

L A YA −∆ ×= ∆ = = =

× ×

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. 12

10 2 4 2(91 kg)(9.80 N kg)(0.50 m)

29 m(1.1 10 N m )(7.0 10 m )

FLLYA −∆ = = = µ

× ×


Solution Find how much the wire stretches. 3

10 2 3 2(5.0 10 N)(2.0 m) 2.2 cm

(9.2 10 Pa)(5.0 mm )(10 m mm)FLLYA −

×∆ = = =×


Solution Find Young’s modulus for the wire. 3

102 2 2 3(1.00 10 N)(5.00 m) 7.69 10 Pa

(0.100 cm )(10 m cm) (6.50 10 m)FLY

A L − −×= = = ×

∆ ×

5. (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 21 2 6 2f i 4f2

av 3( ) (0.45 10 kg)(0.74 m s) 1.2 10 W

2 2(1.0 10 s)

m v vK mvPt t t

−−

−−∆ ×= = = = = ×

∆ ∆ ∆ ×

(b) Strategy and Solution Compute the power output of the flea. 6 6 4(60 W kg)(0.45 10 kg)(0.20) 5.4 10 W 1.2 10 W− − −× = × < ×

No, the flea’s muscle cannot provide the power needed.

(c) Strategy There are two pads, so the total energy stored is 2 2122 2[ ( ) ] ( ) .E U k L k L= = ∆ = ∆ Use Eq. (10-5).

Solution Find the energy stored in the resilin pads. 2

2 2 2 3 6 2 5 3 7( ) ( ) (1.7 10 N m )(6.0 10 m) 3.7 10 JA LE k L Y L Y L YLL L

− −= ∆ = ∆ = = = × × = ×

(d) Strategy Use the definition of average power.

Solution Compute the power provided by the resilin pads. 7

4 4av 3

3.7 10 J 3.7 10 W 1.2 10 W1.0 10 s

EPt

−− −

−∆ ×= = = × > ×∆ ×

Yes, enough power is provided for the jump.


468

6. Strategy The stress is proportional to the strain. Use Eq. (10-4). Solution Find the required stretch of the string.

9 2 6 2(20 N)(0.50 m) 5.0 mm

(2.0 10 N m )(1.0 10 m )FLLYA −∆ = = =

× ×

7. Strategy Form a proportion of the elongations of the left and right wires. Use Eq. (10-4). Solution Find how far the midpoint moves.

L

R

2L R

L R L R2R L

(2 ) 4, so 4 and FL

YAFL

YA

L A r L L L L LL A r

ππ

∆ = = = = ∆ = ∆ ∆ = ∆ + ∆∆

is the total elongation, 1.0 mm.

L R L L L L1 5 4 4, so (1.0 mm) 0.80 mm .4 4 5 5

L L L L L L L L∆ = ∆ + ∆ = ∆ + ∆ = ∆ ∆ = ∆ = =

8. Strategy The stress is proportional to the strain. Use Eq. (10-4). Solution Compute the compression of the abductin ligament.

( )3

26 2 2 1 m1000 mm

(1.5 N)(1.0 10 m) 0.48 mm(4.0 10 N m )(0.78 mm )

FLLYA

−×∆ = = =×

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

7 210 2stress 5.0 10 N m 1.5 10 N m .

strain 0.0033Y ×= = = ×

Compression:

Similarly, for small values of compressive stress and strain, 7 2

9 24.5 10 N m 9.0 10 N m .0.0050

Y − ×= = ×−

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.

2 814

4 4(55 kg)(9.80 N kg)elastic limit , so 1.7 mm .(elastic limit) (2.5 10 Pa)

F F FdA d ππ π

= = = = =×

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.

52 81

4

4 4(1.2 N)tensile strength , so 8.7 10 m .(tensile strength) (2.0 10 Pa)

F F FdA d ππ π

−= = = = = ××


469

12. Strategy Compare the substances in each case using ratios.

Solution

(a) Tendon: 6

4 33

tensile strength 80.0 10 Pa 7.3 10 Pa m /kgdensity 1100 kg m

×= = × ⋅

Steel: 9

4 33

0.50 10 Pa 6.5 10 Pa m /kg7700 kg m

× = × ⋅

Tendon is stronger than steel.

(b) Bone: 6

5 33

160 10 Pa 1.0 10 Pa m /kg1600 kg m

× = × ⋅

Concrete: 9

5 33

0.40 10 Pa 1.5 10 Pa m /kg2700 kg m

× = × ⋅

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.

2 2 8elastic limit, so (elastic limit) (0.0010 m) (2.0 10 Pa) 630 N .F F rA

π π< < = × =

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.

2 2 8tensile strength, so (tensile strength) (0.0010 m) (4.0 10 Pa) 1300 N .F F rA

π π< < = × =

15. Strategy Use Eqs. (10-1), (10-2), and (10-4).

Solution

(a) 4

74 2

7.0 10 Nstress 2.8 10 Pa25 10 m

FA −

×= = = ××

(b) 4

410 4 27.0 10 Nstrain 4.7 10

(6.0 10 Pa)(25 10 m )L F

L YA−

−∆ ×= = = = ×

× ×

(c) 4

410 4 2

(7.0 10 N)(2.0 m) 9.3 10 m(6.0 10 Pa)(25 10 m )

FLLYA

−−

×∆ = = = ×× ×

(d) Set the compressive strength equal to the stress to find the maximum weight the column can support. 8 4 2 5compressive strength, so (compressive strength) (2.0 10 Pa)(25 10 m ) 5.0 10 N .F F A

A−= = = × × = ×


470

16. (a) Strategy The stress is proportional to the strain. Use Eq. (10-4). Solution Find the diameter and tensile stress of the wire. Diameter:

1 9 34

4 4(120 N)(3.0 m), so 1.3 mm .(120 10 Pa)(2.1 10 m)

F F L FLY dA L Y Ld ππ π −

∆= = = = =∆ × ×

Tensile stress: 3

9 7 22.1 10 m(120 10 Pa) 8.4 10 N/m3.0 m

F LYA L

−∆ ×= = × = ×

(b) Strategy Set the maximum stress equal to the tensile strength. Solution Find the maximum weight the wire can support.

2 8 2maxmax

1tensile strength, so (0.00135 m) (4.0 10 N m ) 570 N .4

W WA

π= = × =

17. Strategy Set the stresses equal to the compressive strengths to determine the effective cross-sectional areas. Solution Find the effective cross-sectional areas.

Human: 4

8 28

5 10 N1.6 10 Pa, so 3 cm .1.6 10 Pa

F AA

×= × = =×

Horse: 4

28

10 10 N 7.1 cm1.4 10 Pa

A ×= =×

18. 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.

11 2 80.20stress at breaking point (2.0 10 N m ) 4.0 10 Pa100

LYL

∆ ⎛ ⎞= = × = ×⎜ ⎟⎝ ⎠

19. Strategy Use Hooke’s law for volume deformations. Solution Compute the fractional changes of the volume and radius of the sphere.

Volume: 8 2

49

1.0 10 N m 7.7 10130 10 Pa

V PV B

−∆ ∆ ×= = = ××

Radius: 3 34 3, so .3 4

Vr V rππ

= =

3334 4 433 3

334

1 1 1 1 1 1 1 1 7.7 10 2.6 10V

V

V Vr r r r V Vr r r V V V

π

π

′− −− ∆′ ′ ′∆ − ∆= = − = − = − = − = − − = − − × = ×

20. Strategy Use P gdρ∆ = 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.

1 1 1mVmV

V V V VV V V

ρ ρ ρ ρρ ρ ρ

′′ ′ ′∆ − − ∆= = − = − = − = = −′ ′ ′

Since , .VV VV

ρρ

∆ ∆′ ≈ ≈ − Use Hooke’s law for volume deformations.

3 3 3w

9(1.0 10 kg m )(9.80 N kg)(1.0 10 m)100% 100% 100% 100% 0.45%

2.2 10 PagdP

B Bρρ

ρ∆ ∆ × ×× ≈ × = × = × =

×


471

21. Strategy Use Hooke’s law for volume deformations. Solution Find the change in volume of the sphere.

3 66 3

9(1.00 cm )(9.12 10 Pa), so 57 10 cm .

160 10 PaV V PP B V

V B−∆ ∆ ×∆ = − ∆ = − = − = − ×

×

6 3The volume of the steel sphere would decrease by 57 10 cm .−×

22. Strategy Use Hooke’s law for volume deformations. The pressure of the Moon is roughly 910 Pa.− Solution Find the change in volume of the aluminum.

3 9 56 3

9(1.00 cm )(10 Pa 1.013 10 Pa), so 1.4 10 cm .

70 10 PaV V PP B V

V B

−−∆ ∆ − ×∆ = − ∆ = − = − = ×

×

6 3The volume of the aluminum sphere would increase by 1.4 10 cm .−×

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.

2 2 8 5

boltshear strength, so 4 (shear strength) 4 (0.010 m) (6.0 10 Pa) 7.5 10 N .

4F F r

Aπ π= = = × = ×

24. Strategy Use Hooke’s law for volume deformations. Solution Find the change in the volume of the anchor.

3 63

9(0.230 m )(1.75 10 Pa), so 6.71 cm .

60.0 10 PaV P V PV

V B B∆ ∆ ∆ ×= − ∆ = − = − = −

×

25. Strategy Use Hooke’s law for shear deformations. Solution Find the magnitude of the tangential force.

22 , so (940 Pa)(0.64 10 m)(0.050 m) 0.30 N .F F xS F S xL

A LL−∆= = = ∆ = × =

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∆

tan (2.0 cm) tan8.0° 2.8 mmx L γ∆ = = =

(b) Strategy Use Hooke’s law for shear deformations. Solution Compute the shear modulus of the sponge.

44 212 N 2.0 10 Pa

tan tan (42 10 m ) tan 8.0°FL FL FSA x AL Aγ γ −= = = = = ×

∆ ×


472

27. Strategy Use Eqs. (10-20a) and (10-20b) and Newton’s second law.

Solution Find the spring constant. 2

childchild

(24 kg)(9.80 m s )0, so 840 N m.0.28 my

m gF kx m g k

xΣ = − = = = =

Find the mass of the wooden horse.

child horse child2child horse child

, so .k km mm m

ωω

= = −+

mchildg

Fsy

Find the oscillation frequency of the spring when no one is sitting on the horse.

2 2 2child

horsehorse 840 N m

horse child(0.88 Hz) (2 rad cycle)

1 1 1 840 N m 2.5 Hz2 2 2 2 24 kgk

k kfm m

ω π

ωπ π π π

= = = = =− −

28. Strategy At the equilibrium point, the speed is at its maximum. Use Eq. (10-21). Solution Find the speed at the equilibrium point.

m2 2 (0.050 m) 0.63 m s

0.50 sv A A

Tπ πω= = = =

29. 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.

2 22 2

m 2 24 4 (0.050 m) 7.9 m s

(0.50 s)Aa A

Tπ πω= = = =

30. Strategy The amplitude is half the maximum distance. Use Eq. (10-21). Solution Find the maximum needle speed.

( )38.42

m 9.0 s24

2 10 m2 7.0 cm sAv AT

ππω−×

= = = =

31. 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.

3m

2.242 2 (440.0 Hz) 10 m 3.10 m s2

v A fAω π π −⎛ ⎞= = = × =⎜ ⎟⎝ ⎠

2 2 2 2 2 3 2m

2.244 4 (440.0 Hz) 10 m 8560 m s2

a A f Aω π π −⎛ ⎞= = = × =⎜ ⎟⎝ ⎠

32. (a) Strategy Solve for the spring constant in Eq. (10-20a). Solution Find the spring constant.

2 2 2, so (3.00 Hz) (2 rad cycle) (0.17 kg) 60 N m .k k mm

ω ω π= = = =


473

(b) Strategy Use Eq. (10-17). Solution The amplitude is 12.0 cm.A = Find the angular frequency.

(2 rad cycle)(3.00 Hz) 6.00 rad sω π π= = Thus, the equation that describes the position of the object as a function of time is

1( ) (12.0 cm)cos[(6.00 s ) ] .x t tπ −=

33. Strategy Replace each quantity with its SI units. Solution a has units 2 2m s . xω− has units 2 2s m m s .− ⋅ = So, 2a xω= − is consistent for units.

ω has units 1s .− k m has units 2

12

N m kg m s 1 s .kg m kg s

−⋅= = =⋅

So, k m has the same units as .ω

34. Strategy Use Eq. (10-22). Solution Find the oscillation frequency of the bird’s wingtips.

22 2 2 m

m 2 212 m s4 , so 2.5 Hz .

4 4 (0.050 m)aa A f A f

Aω π

π π= = = = =

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

ff i

i fi

11 1 1, so .4.0 2.01

m mm mm

ωωω

∝ = = = = Therefore, if

10.0 rad s 5.0 rad s .2.0 2.0ωω = = =

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 2 .k mω =


474

37. Strategy Use Eqs. (10-21) and (10-22).

Solution

(a) Find mv and ma in terms of f. Then compare high- and low-frequency sounds.

m 2v A fA fω π= = ∝ and 2 2 2 2m 4 ,a A f A fω π= = ∝ so mv and ma are greatest for high frequency .

(b) 8 6m 2 (20.0 Hz)(1.0 10 m) 1.3 10 m sv π − −= × = ×

2 2 8 4 2m 4 (20.0 Hz) (1.0 10 m) 1.6 10 m sa π − −= × = ×

(c) 3 8m 2 (20.0 10 Hz)(1.0 10 m) 0.0013 m sv π −= × × =

2 3 2 8 2m 4 (20.0 10 Hz) (1.0 10 m) 160 m sa π −= × × =

38. Strategy Use Eqs. (10-21) and (10-22). Solution For SHM, show that 2

m m .v a A=

m ,v Aω= so m .vA

ω = 2 2

2 m mm 2 ,v va A A

AAω= = = so 2

m m .v a A=

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) 2

2 4mm 2 2 2

98 m s, so 1.7 10 m .4 (120 Hz)

aa A Aωω π

−= = = = ×

(b) 2

m mm 2

98 m s 0.13 m s2 (120 Hz)

a av Aω ωω πω

= = = = =

(c) According to Newton’s second law, 2m m (5.24 kg)(98 m s ) 510 N .F ma= = =

40. Strategy Use Eqs. (10-21) and (10-22). Solution Compute the maximum speed and acceleration to which the pilot is subjected.

m 2 (25.0 Hz)(0.00100 m) 0.157 m sv Aω π= = =

2 2 2 2m 4 (25.0 Hz) (0.00100 m) 24.7 m sa Aω π= = =

41. (a) Strategy Use Newton’s second law and Eq. (10-22). Solution Find the maximum force acting on the diaphragm.

2 2 3 2 4m m 4 (0.0500 kg)(2.0 10 Hz) (1.8 10 m) 1.4 kN .F ma m Aω π −= = = × × =


475

(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.

2 2 2 3 2 4 21 2 (0.0500 kg)(2.0 10 Hz) (1.8 10 m) 0.13 J .2

E U m Aω π −= = = × × =

42. Strategy Use Hooke’s law and Newton’s second law.

Solution

(a) According to Newton’s second law, at equilibrium, 0 ,yF kd mgΣ = = − so

.k mg d= When the extension of the spring is a maximum,

(2 ) 2( ) .yF ma k d mg mg d d mg mgΣ = = − = − = Therefore, .a g=

mg

kd

mg

2kd

a

(b) At maximum extension, .yF kx mg maΣ = − = Solve for x.

(1.0 kg)(9.80 N kg 9.80 N kg)( ), so ( ) 0.78 m .25 N m

mkx mg ma m g a x g ak

+= + = + = + = =

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. (0.60 kg)(9.80 N kg)0, so 0.39 m .

15 N mymgF kx mg xk

Σ = − = = = =

mg

kx

(b) Strategy Use Eqs. (10-20a) and (10-21). Solution Find the maximum speed of the body.

m15 N m (0.39 m) 2.0 m s0.60 kg

kv A xm

ω= = = =

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.

2 21 1 (25 N m)(0.050 m) 0.031 J2 2

K kx∆ = − = − = −

45. Strategy Use Hooke’s law, Newton’s second law, and Eq. (10-20c).

Solution Find the “spring constant” of the boat. At equilibrium, person

person 0, so .ym g

F kx m g kx

Σ = − = =

Compute the period of oscillation.

total total total2

person person

(47 kg 92 kg)(0.080 m)2 2 2 2 0.70 s(92 kg)(9.80 m s )

m m m xT

k m g x m gπ π π π += = = = =

mperson g

kx


476

46. (a) Strategy Use Newton’s second law and Eq. (10-20a).

Solution Determine the spring constant of the cord.

At equilibrium, 0 .yF kd mgΣ = = − So, ,mgkd

= where d = 0.20 m.

Calculate the period.

22 2 0.20 m2 2 2 2 0.90 s

9.80 m sm m dTk mg d gk m

π π π π π πω

= = = = = = = mg

kd

(b) Strategy Use Eqs. (10-21) and (10-20a). Solution Find the maximum speed of the baby.

2

m9.80 m s(0.080 m) 0.56 m s

0.20 mk mg gv A A A Am dm d

ω= = = = = =

47. Strategy and Solution Since ( ) sin ,y t A tω= 1.57 rad s 0.250 Hz .2 2

f ωπ π

= = =

48. (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.

2(0.306 kg)(9.80 m s )0, so 12 cm.25 N my

mgF kA mg Ak

Σ = − = = = = The object will

move up and down a total vertical distance of 2 24 cm.A = Thus, the 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

a positive cosine plot of amplitude 12 cm .

24 cm

mg

kA

49. Strategy The maximum kinetic energy occurs at the equilibrium point where m .v v Aω= = For SHM,

.k mω = Solution Find the maximum kinetic energy of the body.

2 2 2 2 2 2max m

1 1 1 1 1 (2.5 N m)(0.040 m) 2.0 mJ2 2 2 2 2

kK mv m A m A kAm

ω ⎛ ⎞= = = = = =⎜ ⎟⎝ ⎠


477

50. Strategy Graph (a) x, (b) , and (c) x xv a on the vertical axis and t on the horizontal axis. Analyze the slopes.

Solution

(a) ( ) sinx t A tω=

A

0t

x

π

ω

2πω

(b) The slope of the x(t) graph is maximum at 0,t = so ( )xv t is max at 0.t = Since ( )xv t starts at its maximum value and then is positive and decreasing, it is a cosine function, m( ) cos .xv t v tω=

0t

vm

π

ω2πω

vx

(c) The slope of the ( )xv t graph is zero at 0t = and becomes negative just after, so ( )xa t starts at zero and decreases. Since ( )xa t starts at zero and then is negative and decreasing, it is a negative sine function,

m( ) sin .xa t a tω= −

0t

am

π

ω

2πω

ax

(d) Compare minimums. ( )x t has its minimum at 3 , ( )2 xt v tπω

= has its at ,t πω

= and ( )xa t has its at .2

t πω

=

One quarter cycle is . 2 2 2π π π πω ω ω ω

3 − = and ,2 2

π π πω ω ω

− = so ( )xv t is one quarter cycle ahead of x(t) and

( )xa t is one quarter cycle ahead of ( ).xv t

51. Strategy Use the definition of average speed and Eq. (10-21). In (d), graph xv 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.

av4 4 2

2x A Av At T

ωπ ω π

∆= = = =∆

(b) The maximum speed for SHM is m .v Aω=

(c) 2

av

m

Avv A

π ωω π

2= =


478

(d) Graph ( )xv t and a line from the origin to m.v

0t

vx

π

ω

2πω

ωA

−ωA

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:

2 2 2i i

1 1 1( ) (0) ( 0) , so .2 2 2yy y y h v t g t t g t y h gt− = − = ∆ − ∆ = ∆ − − = − At 20, .hy t

g= =

During the rise:

The speed of the ball just before and after it hits the ground is 2 2 .hv gt g ghg

= = = So, at

i i2 , 2 ,y

ht v ghg

= = and i 0.y y= = If i 0t = when i 0,y y= = then 2i

1 ( ) .2yy v t g t= ∆ − ∆ The graph is shifted

to the right by 2 ,hg

so 2

2 1 22 ,2

h hy gh t g tg g

⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ and y = h when 22 .ht

g=

0 t2hg

yh

22hg

The motion is not SHM, since it is not a sine or cosine function . The graph is always nonnegative and has a parabolic shape.

53. (a) Strategy Use Eq. (10-20a) to find the spring constant. Then, find the elastic potential energy using 212 .kx

Solution Find the spring constant.

2 2 2, so (2.00 Hz) (2 rad cycle) (0.2300 kg) 36.3 N m.k k mm

ω ω π= = = =

The equation for the elastic potential energy is

[ ]2 2 2 11( ) (36.3 N m)(0.0800 m) sin (2.00 Hz)(2 rad cycle) (116 mJ)sin (4.00 s ) .2

U t t tπ π −⎡ ⎤= = ⎣ ⎦


479

Since the sine function is squared, the period of ( )U t is half that of a sine function or

1 250 ms.4.00 s

T π πω π −= = = Graph ( ).U t

0t (ms)

U (mJ)

0 250 500

116

(b) Strategy Find the kinetic energy using 212 .xmv

Solution The equation for the kinetic energy is

22 2 2

2 1

1 2 rad 2 rad( ) (0.2300 kg)(2.00 Hz) (0.0800 m) cos (2.00 Hz)2 cycle cycle

(116 mJ)cos (4.00 s ) .

K t t

t

π π

π −

⎡ ⎤⎛ ⎞ ⎛ ⎞= ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤= ⎣ ⎦

Since the cosine function is squared, the period of ( )K t is half that of a cosine function or

1 250 ms,4.00 s

T π πω π −= = = which is the same as ( ).U t Graph ( ).K t

0t (ms)

K (mJ)

0 250 500

116

(c) Strategy Add ( ) and ( )U t K t and graph the result. Solution

{ }2 1 2 1

2 1 2 1

( ) ( ) ( ) (116 mJ)sin (4.00 s ) (116 mJ)cos (4.00 s )

(116 mJ) sin (4.00 s ) cos (4.00 s ) (116 mJ)(1) 116 mJ

E t U t K t t t

t t

π π

π π

− −

− −

⎡ ⎤ ⎡ ⎤= + = +⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤= + = =⎣ ⎦ ⎣ ⎦

Graph ( ) ( ) ( ).E t U t K t= +

0t (ms)

E (mJ)

0 250 500

116

(d) Strategy and Solution Friction does nonconservative work on the object, thus, , , and would gradually be reduced to zero .U K E


480

54. (a) Strategy Draw the velocity vector for point P in Fig. 10.17b and then find its x-component. Solution Show that ( ) sin .xv t A tω ω= −

vy

vx

θ

vm y

x

From the figure, m sin .xv v θ= − For SHM, mv Aω= and ,tθ ω= so ( ) sin .xv t A tω ω= −

(b) Strategy Use conservation of energy and Eq. (10-20a). Solution Verify that the expressions for x(t) and ( )xv t are consistent with energy conservation.

212

E kA= = 2 2 2 2 2 2 2

22 2 2 2 2 2 2 2 2

1 1 1 1sin cos2 2 2 2

1 1 1 1 1sin cos (sin cos ) (1)2 2 2 2 2

K U mv kx m A t kA t

km A t kA t kA t t kA kAm

ω ω ω

ω ω ω ω

+ = + = +

⎛ ⎞= + = + = =⎜ ⎟⎜ ⎟

⎝ ⎠

The expressions for x(t) and ( )xv t are consistent with energy conservation.

55. Strategy Use Eq. (10-26b). Solution Compute the period of the pendulum.

24.0 m2 2 4.0 s

9.80 m sLTg

π π= = =

56. Strategy The total mechanical energy for a simple pendulum is 2 212 .E m Aω= Use Eq. (10-26a).

Solution Find the amplitude of the pendulum.

22 2 2 21 1 2 2(0.015 J)(0.75 m), so 3.0 cm .

2 2 2 (2.5 kg)(9.80 N kg)g mg ELE m A m A A AL L mg

ω⎛ ⎞

= = = = = =⎜ ⎟⎜ ⎟⎝ ⎠

57. Strategy and Solution According to Eq. (10-26b), 2 ,LTg

π= which does not depend upon the mass.

Therefore, 1.5 s .T =

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.

f

i

f ff i

i i

2 2 2, so 2 (2.0 s) 2 2.8 s .2

Lg

Lg

T L L T TT L L

π

π= = = = = = =


481

59. (a) Strategy Graph xv on the vertical axis and t on the horizontal axis. Use Eq. (10-21). Solution xv leads x by one quarter cycle [ (2 )]π ω and m ,v Aω= so if sin , cos .xx A t v A tω ω ω= =

0t

vx

ωA

−ωA

π

ω2πω

(b) Strategy Use Eqs. (6-6) and (10-21). Solution Find the maximum kinetic energy.

2m m

1 , so2

K mv= 2 2m

1 .2

K m Aω=

60. Strategy Use Eq. (10-26b). Solution Find the length of the pendulum.

2 2 2

2 2(9.80 m s )(1.0 s)2 , so 0.25 m .

4 4L gTT Lg

ππ π

= = = =

61. Strategy Use Eq. (10-26b) to find the length of the pendulum. Then, form a ratio of the lengths. Solution Solve for L.

2

22 , so .4

L gTT Lg

ππ

= =

Form a proportion. 22

2 22

1 1

1.00 s 1.110.950 s

L TL T

⎛ ⎞= = =⎜ ⎟⎝ ⎠

62. Strategy The total mechanical energy of a pendulum is 2 212 .E m Aω= Form a proportion.

Solution Find the mechanical energy of the pendulum.

2 222 2 2

2 121 11

3.0 cm, so (5.0 mJ) 11 mJ .2.0 cm

E A AE EE AA

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠


482

63. (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.

1E X

E XE X X E X E

2 and 2 , so 2 2 .T gL L L LT T

g g T g g gπ π π π

−⎛ ⎞

= = = =⎜ ⎟⎜ ⎟⎝ ⎠

Thus, 2 2

E 2 2X E

X

0.650 s(9.80 m s ) 5.57 m s .0.862 s

Tg g

T⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

64. (a) Strategy Use Eq. (10-26b). Solution Compute the period of the pendulum.

21.000 m2 2 2.01 s

9.80 m sLTg

π π= = =

(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 r .m Find the net torque acting on the physical pendulum.

r r rsin sin sin ( 2 ) ( 2 ),2 2 2L gL gLF r m g mg L m m m mτ θ θ θ θ⊥Σ = = − − = − + ≈ − +

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

L

Axis

mrg

θ

mg

L/2

r( 2 ) ,2

gL m m Iτ θ αΣ = − + = and the angular acceleration is r r

r

( 2 ) ( 2 ),

2 2( )gL m m gL m m

I I Iθ

α θ+ +

= − = −+

where

rI 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

spring, 2 ,xa xω= − where the angular frequency in our case is r

r

( 2 ).

2( )gL m m

I Iω

+=

+ Thus, the period of the

pendulum is r

r

2( )2 2 .( 2 )

I IT

gL m mπ π

ω+

= =+

Now, the rotational inertias of the rod and point mass are

2 21r r3 and ,I m L I mL= = respectively, so the period becomes r

r

2 ( 3 )2 .

3 ( 2 )L m m

Tg m m

π+

=+

So, now we need to

compare the mass of the rod to the total mass.


483

( )( )

( )( )

r

r

r

r

r

r

3r

2r

32

2

3 2 2

2 2

2 2 2

2 2 2r r r

2 2

2 2r

r

2 12 ( 3 )2 2

3 ( 2 ) 3 1

2 1

2 3 1

1 3 32 21 8

3 3 2 3 3 21 18 8 8

3 33 14 8

1

mm

mm

mm

mm

mmm

m

LL m mT

g m m g

LT

g

T g T gL L

m T g m T g T g mm m mL L L

m T g T gm L L

mm

π π

π

π π

π π π

π π

++= =

+ +

+⎛ ⎞ =⎜ ⎟⎝ ⎠ +

+ ⎛ ⎞= =⎜ ⎟+ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ = + = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞

− = −⎜ ⎟⎜ ⎟⎝ ⎠

+

2

23

2 28r

2 2 2r

2

1 16 313 24 634

T gLm m L T g

m T g L T gL

π ππ

π

−+ −= = + =−−

2 2 2 2 2r

2 2 2 2 2r

24 6 24 (1.000 m) 6(0.99 2.007 s) (9.80 m s )100% 100% 100% 11.3%16 3 16 (1.000 m) 3(0.99 2.007 s) (9.80 m s )

m L T gm m L T g

π ππ π

− − ×× = × = × =+ − − ×

The percentage of the total mass of the pendulum in the uniform thin rod is 11.3% .

65. 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:

3

m2 2 (10.0 10 m) 3.14 cm s

2.00 sv A A

Tπ πω

−×= = = =

2nd method: Find the length L of the pendulum.

L

θmax

A

2 2 2

2 2(2.00 s) (9.80 m s )2 , so 0.993 m.

4 4L T gT Lg

ππ π

= = = =

Find max.θ

Since the displacement is small, 3

2max max

10.0 10 msin 1.007 10 rad.0.993 m

AL

θ θ−

−×≈ = = = ×

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, cos .L θ So, (1 cos ).h L θ= − Find m.v

2m

2 2m

1 , so 22 2 (1 cos ) 2(9.80 m s )(0.993 m)(1 cos1.007 10 ) 3.14 cm s .

K mv U mgh

v gh gL θ −

∆ = = −∆ =

= = − = − × =

L

Axis

θ

h

L cosθ


484

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:

22 22 2 2 2I mr r DTmgd mgr g g

π π π π= = = =

So, H1 1 ,

2gf

T Dπ= = where D is the diameter of the hoop.

Simple pendulum:

P H1

2 2gf fL

ωπ π

= = = if L = D.

67. (a) Strategy The total mechanical energy of a pendulum is 2 212 .E m Aω= Use Eq. (10-26a).

Solution Find the energy of the pendulum.

2 2 2 22 2 21 1 (0.50 kg)(9.80 m s )(0.050 m) 6.1 mJ

2 2 2 2(1.0 m)g mgAE m A m AL L

ω⎛ ⎞

= = = = =⎜ ⎟⎜ ⎟⎝ ⎠

(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.

2 LTg

π= and 2U m gh= where 2 2.0 kg.m =

1 wk 604,800 scycles per week2

gT Lπ

= =

( )3

cycles/wk 2 22604,800 s 21 112 2

3

2 2

4100% 100% 100%(604,800 s)

4 (2.0 kg)(1.0 m) (1.0 m) 100% 1.1%(604,800 s)(0.50 kg)(0.050 m) 9.80 m s

U

g gL L

m gh m h LE gm Am Aπ

π

π

× = × = ×

= × =

68. Strategy 2 2 212E m A Aω= ∝ for a pendulum. Compare the fractional decrease per cycle. Let n = the number of

cycles for E to decrease 5.0%. Solution Find the number of cycles it takes for the energy of the oscillator to decrease 5.0%.

2 22

2 20.950 1

12

0.050 0.050, so (10 cycles) 5.1 cycles .10 cycles 10 cycles 0.950 1

AEE A nn n

−∆∆ − −= = = = =−


485

69. Strategy 2 2 212E m A Aω= ∝ for a pendulum. Form a proportion.

Solution Find by what factor the energy has decayed.

2 22 2 1

2 2 21 1 1

( 20.0) 1 140020.0

E A AE A A

= = = =

The energy has decayed by a factor of 400 .

70. Strategy Use Eq. (10-20b). Solution Find the spring constant.

2 21 1

1 , so 4 ,2

kf k f mm

ππ

= = where 1m is the combined mass of all four people and the car.

Compute the frequency when only the 45-kg person is present. 2 2 2

1 12

2 2

41 1 (2.00 Hz) (1020 kg 45 kg 52 kg 67 kg 61 kg) 2.16 Hz2 2 1020 kg 45 kg

f mkfm m

ππ π

+ + + += = = =+

71. Strategy Use Eq. (10-21). Assume the amplitude of the pendulum is small. Solution Find the period T of the pendulum.

m 0.50 m sv Aω= = and 0.50 m s 2.5 rad s.0.20 m

ω = = Thus, 2 2 2.5 s .2.5 rad s

T π πω

= = =

72. Strategy Use Eq. (10-26a). Solution We must assume that the pendulum is located on Earth. Find its length.

2

2 29.80 m s, so 0.994 m .

(3.14 rad s)g gLL

ωω

= = = =

73. Strategy Use Equation (10-5) and conservation of energy. Solution Find the “spring constant” of the bar.

2 11 2 27(2.0 10 Pa) (0.75 10 m) 7.1 10 N/m

0.50 mA Y rk YL L

π π −× ×= = = = ×

The gravitational potential energy is converted into elastic potential energy. 2 2

2 22 11 2 2

1 1 2 2(0.50 m)(0.70 kg)(9.80 m s )(1.0 m), so 0.44 mm .2 2 (2.0 10 Pa) (0.75 10 m)

Y r Lmghk L L mgh LL Y r

ππ π −∆ = ∆ = ∆ = = =

× ×

74. Strategy 2 ( )T I mgdπ= for a physical pendulum, where d is the distance from the axis to the center of mass. 21

3I mL= for a uniform rod with the axis through its end.

Solution Find the period of the pendulum for each horizontal axis.

(a) 21

32

1 2 (1.00 m)2 2 1.64 s3 3(9.80 m s )(0.500 m)

mLT L

mgd gdππ π= = = =


486

(b) Treat the meterstick as two rods with lengths 75 cm and 25 cm. 2 2

2 2 21 1 3 3 1 1 27 1 28 73 4 4 3 4 4 3 64 64 3 64 48

m L m LI mL mL mL⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = + = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

27 248

27 7 7(1.00 m)2 2 2 2 1.53 s

48 ( 4) 12 12(9.80 m s )

mL L LTmgd g L g

π π π π= = = = =

(c) Treat the meterstick as two rods with lengths 60 cm and 40 cm. 2 2

2 21 4 4 1 6 6 1 64 216 73 10 10 3 10 10 3 1000 1000 75

m L m LI mL mL⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

2775

214 14(1.00 m)2 2 2 1.94 s

( 10) 15 15(9.80 m s )

mL LTmg L g

π π π= = = =

75. 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. 2 2

fish fish fishfish weight

weight weight weight

2202 , so . Thus, (4.90 N) 56 N .65

T m TmT m mk T m T

π⎛ ⎞ ⎛ ⎞⎜ ⎟= = = = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

76. 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.

245 m2 2 13 s

9.80 m sLTg

π π= = =

77. Strategy Refer to Eqs. (10-20). Use conservation of energy and the fact that 2.E A∝

Solution Analyze the mass and spring system.

(a)

The frequency and period don’t vary with amplitude, they only vary with and . Since these twovalues remain constant, so do the frequency and period.

m k

(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 is four times that for an amplitude of .D 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 21

i2 ,mv due to the initial kinetic energy.


487

78. 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, 0, so .yF kd mg k m g dΣ = − = = Calculate the frequency of oscillation.

2

31 1 1 9.80 m s 91 Hz .

2 2 2 0.030 10 mk gfm dπ π π −= = = =

× mg

kd

79. (a) Strategy Use Eq. (10-4) and the geometry of the web.

Solution Find the angle the web makes with the horizontal. 9 2

9 21.4 10 N mmaximum stress, so 0.35.4.0 10 N m

L LYL L

∆ ∆ ×= = =×

,L r= so 0.35.L r∆ = The new length of a stretched web strand is the hypotenuse of a right triangle.

r + ∆L

r

d

θ

1 11 1 1cos , so cos cos 42.2° .1 0.351 1L L

r r

rr L

θ θ − −∆ ∆= = = = =

+ ∆ ++ +

(b) Strategy Use Newton’s second law and Eq. (10-2).

Solution Find the tension.

sin 0,yF T mgθΣ = − = so .sinmgT

θ=

Determine the mass of the bug. 9 21.4 10 N m , so

sinF T mgA A A θ

= = = ×

T

mg

θ

9 2 11 2 9 2

2sin (1.4 10 N m ) 50(1.0 10 m )sin 42.2 (1.4 10 N m ) 48 g .

9.80 m sAm

gθ −× × ° ×= = =

(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.

( )sin 1 sin (0.10 m)(1.35)sin 42.2 9.1 cmLd r L rr

θ θ∆⎛ ⎞= + ∆ = + = ° =⎜ ⎟⎝ ⎠


488

80. 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 1( ) (10 cm)cos[(1.57 s ) ].x t t−= −

10

t (s)

x (cm)

01.0

10

−

3.0 4.02.0

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.

81. Strategy The inertia of the system is 2 211 23 .I m L m L= + 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 1

1 2 22 2

1 2.

mLm m L md L

m m m+ +

= =+

Find the period of this physical pendulum.

( )( )

( )( )

1 1

11

22 21 2 23 31 23 1 2

1 222 22

2 ( 3 )2 2 2 2 23 ( 2 )

m m

mm

L m L mm L m LI L m mTmgd mgd g m mg mmg m m L

π π π π π+ ++ += = = = =

⎡ ⎤ +++⎢ ⎥⎣ ⎦

(b) For each case, replace the smaller mass with zero. Then

1 2 1 22for , 2 , and for , 2 .3

L Lm m T m m Tg g

π π>> = << =

The former is the period for the uniform rod alone and the latter is the period for the block alone.

82. 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.

m 2 2 (28 Hz)(0.012 m) 2.1 m sv A fAω π π= = = = and 2 2 2 2m 4 (28 Hz) (0.012 m) 370 m s .a Aω π= = =


489

83. (a) Strategy Use conservation of energy. Do not assume SHM. Solution Find the speed of the pendulum bob at the bottom of its swing.

21 , so 2 .2

K mv U mgL v gL∆ = = −∆ = =

(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 2 .4 2

L Lπ π=

Assuming SHM, m .2 2

gv A L gLL

π πω ⎛ ⎞= = =⎜ ⎟⎝ ⎠

Since 1 ,vT

ω∝ ∝ a smaller speed implies a larger

period. Since m 2 1,2 2 2

gLvv gL

π π= = > the period of a pendulum for large amplitudes is larger than that

given by Eq. (10-26b).

84. (a) Strategy Draw a diagram of a pendulum. Solution From the figure below, we see that cos ,L y Lθ + = or (1 cos ).y L θ= −

L

L cos

L

y

θ

θ

(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

22 21 1 1(1 cos )

2 2 2x mgU mgy mgL mgL x kxL L

θ⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= = − ≈ = =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

with .mgkL

=

85. Strategy 2I mL= for a simple pendulum of length L and mass m. Use Eq. (10-28a). Solution

2 2 2( ) ,mgd mgd gd g L g

I LmL L Lω = = = = = which is the angular frequency of a simple pendulum.


490

86. (a) Strategy Use Hooke’s law and Newton’s second law.

Solution Find the spring constant. At equilibrium, the spring has stretched a distance 0.310 m 0.200 m 0.110 m.d = − =

2(1.10 kg)(9.80 m s )0, so 98.0 N m .0.110 my

mgF kd mg kd

Σ = − = = = = mg

kd

(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 0.0500 m.d =

2 22 2

max max max max1 1 (98.0 N m)(0.0500 m), so 0.472 m s .2 2 1.10 kg

kdK mv U kd vm

= = = = = =

(c) Strategy Use conservation of energy. Solution Find the speed of the brick.

2 2 2

2 2 2 2

1 1 1 , so2 2 2

98.0 N m( ) [(0.0500 m) (0.0250 m) ] 0.409 m s .1.10 kg

E kd K U mv ky

kv d ym

= = + = +

= − = − =

(d) Strategy Use Eq. (10-20c). Solution The time is will take for the brick to oscillate five times is five times the period.

1.10 kg5 5 2 10 3.33 s98.0 N m

mTk

π π⎛ ⎞

= = =⎜ ⎟⎜ ⎟⎝ ⎠

87. Strategy Since the body begins with its maximum amplitude at 0,t = 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. 4.0 N0, so 1.6 cm.

250 N mymgF kA mg Ak

Σ = − = = = =

Find the angular frequency .ω 2(250 N m)(9.80 m s ) 25 rad s

4.0 Nk kgm mg

ω = = = =

mg

kA

y

Thus, the equation describing the motion of the body is [ ](1.6 cm)cos (25 rad s ) .y t=

88. Strategy Use dimensional analysis. Solution 2[ ] m, [ ] kg s , and [ ] kg;A k m= = = we need 1[ ] s .f −= Only k has units which include seconds, so f

must be proportional to the square root of k to get 1s .− kg must be eliminated and only m has units of kg; therefore, .f k m∝


491

89. (a) Strategy Use conservation of energy. Solution Let the maximum displacement be d. Find d.

2 2 2

2 22 2

1 1 1 , so2 2 2

(1.24 kg)(0.543 m s) (0.345 m) 0.395 m .9.82 N m

E K U mv kx kd

mvd xk

= + = + =

= + = + =

(b) Strategy The maximum speed of the block occurs when the block is at it equilibrium position, 0.x = Use conservation of energy. Solution Find the maximum speed of the block.

2 2 2

2 2 2 2

max

1 1 1 , so2 2 2

( ) (9.82 N m)[(0.395 m) 0 ] 1.11 m s .1.24 kg

E K U mv kx kd

k d xvm

= + = + =

− −= = =

(c) Strategy Use the result from part (b). Solution

2 2 2 2( ) (9.82 N m)[(0.3953 m) (0.200 m) ] 0.960 m s1.24 kg

k d xvm− −= = =

90. 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 p, , and ,L dθ ∆ the diameter of the tuning pin.

p

2d

s r Lθ θ= = = ∆

Find the angle through which the tuning pin must be turned to tune the piano wire. p

2w

( 2), so

( 2)

dF F LY YA L Ld

θπ

∆= = =

θdp /

2

s = ∆L

2 2 2 11p w p w

2 8 8(402 N 381 N)(0.66 m) 360° 2.0° .2( 2) (0.0080 m)(0.00080 m) (2.0 10 Pa)

FL FLd d Y d d Y

θππ π π

− ⎛ ⎞= = = =⎜ ⎟× ⎝ ⎠

91. 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.

8 82 2 3 21

4

4 4(402 N)tensile stress 8.0 10 Pa 8.26 10 Pa(0.80 10 m)

F T TA d dπ π π −= = = = = × < ×

×

The tensile stress is 88.0 10 Pa; it is just under the elastic limit.×


492

92. Strategy Draw a diagram and use Newton’s second law. Assume the oscillation of the ice cube is small. Solution

N

R

W

y

x

θ

θ

According to the figure, sin tan , for small ; so .x x x

x NxF ma N N N aR mR

θ θ θΣ = = − ≈ − = − ≈ −

cos 0,y yF N mg maθΣ = − = ≈ since ya is negligibly small for a small amplitude of oscillation. So,

cos (1) (cos 1 for small),N N mgθ θ θ≈ ≈ ≈ and .xmgx gxamR R

≈ − = −

Find .ω Recall that 2 .xa xω= −

2 , so .gx gxR R

ω ω− = − = Therefore, 2 2 .RTg

π πω

= =

93. Strategy Treat the swinging gibbon as a physical pendulum. Use Eq. (10-28a). Solution Estimate the frequency of oscillation of the gibbon.

2

21 1 (9.80 m s )(0.40 m) 0.63 Hz

2 2 0.25 mmgdf

Iπ π≈ = =

94. 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. 4cos 1 1strain 1 1 8 10

cos cos cos 0.040L L L L L

L L Lθ

θ θ−′ ′ ′∆ − −= = = = − = − = ×

′

(b) Find the tension in the cable. 640 N2 sin 0, so 8.0 kN .

2sin 2sin 0.040ymgF T mg Tθ

θΣ = − = = = =

L

L′

θ

T

mg

θθ

T


493

(c) Find the cross-sectional area of the cable.

( ) ( )5 2

91 1cos cos 0.040

1 1 , so2 sin cos

640 N 5 10 m .2 sin 1 2(200 10 Pa)sin 0.040 1

T mg LY YA A L

mgAY θ

θ θ

θ−

∆ ⎛ ⎞= = = −⎜ ⎟⎝ ⎠

= = = ×− × −

(d) Compute the stress. 3

8 85 2

8.0 10 N 1.6 10 Pa 2.5 10 Pa5 10 m

TA −

×= = × < ××

No , the cable has not been stretched beyond its elastic limit.

95. 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.

compressive stress F mg Vg hAg ghA A A A

ρ ρ ρ= = = = =

(b) Find the absolute limit to the height of a cylindrical column, regardless of how wide it is. 8

m 3 3 2compressive strength 2.0 10 Pa 7.6 km

(2.7 10 kg m )(9.80 m s )h

gρ×= = =

×

(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.

96. (a) Strategy Approximate the tibia as two concentric circles. Solution Find the average cross-sectional area of the tibia.

2 22 2 2o i

2.5 cm 1.3 cm 3.6 cm2 2

A r rπ π π⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − = − =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

(b) Strategy Use Eq. (10-2). Solution Find the compressive stress in the tibia.

64 2

2800 Ncompressive stress 7.8 10 Pa3.6 10 m

FA −= = = ×

×

(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.

44 2 9

(2800 N)(0.40 m), so 3.3 10 m .(3.6 10 m )(9.4 10 Pa)

L F FLY LL A AY

−−

∆ = ∆ = = = ×× ×


494

97. (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.

22

1 2, so .2

mghky mgh ky

= =

Find the period of oscillations of the bungee cord.

2 22 22 2 (50.0 m 33.0 m) 3.42 s

2 (9.78 m s )(50.0 m)m mT yk ghmgh y

π π π π= = = = − =

(b) Strategy Use conservation of energy and the quadratic formula. Solution Find the extension of the bungee cord 2.y

121

2 222 2 2 2 1 2 12 2 2 2 2 2 2

1 1

1 1 (33.0 m)( 33.0 m), so 0 .2 2

m ghy

m y m yky y m gh m g y y ym h m h

⎛ ⎞= = = + = − −⎜ ⎟⎝ ⎠

Solve for 2.y

2 2 22 1 2 1 2 1

1 1 1

2(33.0 m)

2

22 2 2

4(1)

2(1)

(80.0 kg)(17.0 m) 1 (80.0 kg)(17.0 m) (80.0 kg)(17.0 m) (132 m)2(60.0 kg)(50.0 m) 2 (60.0 kg)(50.0 m) (60.0 kg)(50.0 m)20.3 m or 1

m y m y m ym h m h m h

y

⎛ ⎞ ⎡ ⎤± − − −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦=

⎡ ⎤= ± +⎢ ⎥

⎢ ⎥⎣ ⎦= − 2.6 m

2 0,y > so 2 20.3 m,y = 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.

98. (a) Strategy Use Eq. (10-4). Solution Find the force the wings must exert to extend the resilin.

6 2 6 2(1.7 10 N m )(1.0 10 m )(4.0 cm 1.0 cm), so 5.1 N .1.0 cm

F L YA LY FA L L

−∆ ∆ × × −= = = =

(b) Strategy The energy stored in the resilin is elastic potential energy. Use Hooke’s law. Solution Find the energy stored in the resilin.

2 2 21 1 1 1 (5.1 N)(0.030 m) 7.7 10 J2 2 2 2

FU kx x Fxx

−⎛ ⎞= = = = = ×⎜ ⎟⎝ ⎠

Chapter10 Solution

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