1. Chapter 11: Elasticity and Periodic Motion
Elastic Deformations
Hooke’s Law
Stress and Strain
Shear Deformations
Volume Deformations
Simple Harmonic Motion
The Pendulum
Damped Oscillations, Forced Oscillations, and Resonance

2. Elastic Deformation of Solids (optional)
A deformation is the change in size or shape of an object.
An elastic object is one that returns to its original size and shape after contact forces have been removed. If the forces acting on the object are too large, the object can be permanently distorted.

3. Hooke’s Law F F
Apply a force to both ends of a long wire. These forces will stretch the wire from length L to L+L.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 10, Questions 1, 2, 3, 4, 5, and 6.

4. Define:
The fractional change in length
Force per unit cross-sectional area

5. Hooke’s Law (Fx) can be written in terms of stress and strain (stress  strain).
The spring constant k is now
Y is called Young’s modulus and is a measure of an object’s stiffness. Hooke’s Law holds for an object to a point called the proportional limit.

6. Example (text problem 10.1): A steel beam is placed vertically in the basement of a building to keep the floor above from sagging. The load on the beam is 5.8104 N and the length of the beam is 2.5 m, and the cross-sectional area of the beam is 7.5103 m2. Find the vertical compression of the beam.
Force of ceiling on beam
Force of floor on beam
For steel Y = 200109 Pa.

7. Example (text problem 10. 7): A 0
Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional area 1.0106 m2, has a Young’s modulus of 2.0109 Pa. By how much must you stretch a guitar string to obtain a tension of 20 N?

8. §11.3 Beyond Hooke’s Law
If the stress on an object exceeds the elastic limit, then the object will not return to its original length.
An object will fracture if the stress exceeds the breaking point. The ratio of maximum load to the original cross-sectional area is called tensile strength.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 10, Question 7.

9. The ultimate strength of a material is the maximum stress that it can withstand before breaking.

10. Example (text problem 10.10): An acrobat of mass 55 kg is going to hang by her teeth from a steel wire and she does not want the wire to stretch beyond its elastic limit. The elastic limit for the wire is 2.5108 Pa. What is the minimum diameter the wire should have to support her?
Want

11. §10.4 Shear and Volume Deformations
A shear deformation occurs when two forces are applied on opposite surfaces of an object.

12. Define:
Hooke’s law (stressstrain) for shear deformations is
where S is the shear modulus

13. Example (text problem 10.25): The upper surface of a cube of gelatin, 5.0 cm on a side, is displaced by 0.64 cm by a tangential force. If the shear modulus of the gelatin is 940 Pa, what is the magnitude of the tangential force? F
From Hooke’s Law:

14. An object completely submerged in a fluid will be squeezed on all sides.
The result is a volume strain;

15. For a volume deformation, Hooke’s Law is (stressstrain):
where B is called the bulk modulus. The bulk modulus is a measure of how easy a material is to compress.

16. Example (text problem 10.24): An anchor, made of cast iron of bulk modulus 60.0109 Pa and a volume of m3, is lowered over the side of a ship to the bottom of the harbor where the pressure is greater than sea level pressure by 1.75106 Pa. Find the change in the volume of the anchor.

17. Tensile or compressive
Deformations summary table
Tensile or compressive
Shear
Volume
Stress
Force per unit cross-sectional area
Shear force divided by the area of the surface on which it acts
Pressure
Strain
Fractional change in length
Ratio of the relative displacement to the separation of the two parallel surfaces
Fractional change in volume
Constant of proportionality
Young’s modulus (Y)
Shear modulus (S)
Bulk Modulus (B)
This table appears in this chapter’s Master the Concepts.

18. Simple Harmonic Motion
SHM (Mandatory)
Simple Harmonic Motion

19. The Ideal Spring & Hooke’s Law
Springs are objects that exhibit elastic behavior
An ideal spring is:
Massless (the mass of the spring is negligible)
The applied force (Fapplied) required to compress/stretch is proportional to the displacement of the spring from its unstrained length (x) or
Fapplied = kx
where k is called the spring constant (or stiffness of the spring)
To stretch/compress a spring, the spring exerts a restoring force of equal & opposite magnitude (reaction force, F) against the stretching/compressing force or
F = -kx
{this is referred to as Hooke’s Law!}

20. The (Elastic) Restoring Force (& Newton’s 3rd Law)
Action:
Applied force is proportional to displacement of the spring:
Fapplied = kx
Reaction:
Restoring force is equal/opposite to applied force:
F = -Fapplied = -kx

21. Displacing an Ideal Spring results in Simple Harmonic Motion
Releasing a strained a spring results in oscillating motion due to the spring restoring force
This type of motion is called Simple Harmonic Motion

22. ConcepTest 13.3a Spring Combination I
A spring can be stretched a distance of 60 cm with an applied force of 1 N. If an identical spring is connected in parallel with the first spring, and both are pulled together, how much force will be required to stretch this parallel combination a distance of 60 cm?
1) 1/4 N
2) 1/2 N
3) 1 N
4) 2 N
5) 4 N

23. ConcepTest 13.3a Spring Combination I
A spring can be stretched a distance of 60 cm with an applied force of 1 N. If an identical spring is connected in parallel with the first spring, and both are pulled together, how much force will be required to stretch this parallel combination a distance of 60 cm?
1) 1/4 N
2) 1/2 N
3) 1 N
4) 2 N
5) 4 N
Each spring is still stretched 60 cm, so each spring requires 1 N of force. But since there are two springs, there must be a total of 2 N of force! Thus, the combination of two parallel springs behaves like a stronger spring!!

24. ConcepTest 13.3b Spring Combination II
A spring can be stretched a distance of 60 cm with an applied force of 1 N. If an identical spring is connected in series with the first spring, how much force will be required to stretch this series combination a distance of 60 cm?
1) 1/4 N
2) 1/2 N
3) 1 N
4) 2 N
5) 4 N

25. ConcepTest 13.3b Spring Combination II
A spring can be stretched a distance of 60 cm with an applied force of 1 N. If an identical spring is connected in series with the first spring, how much force will be required to stretch this series combination a distance of 60 cm?
1) 1/4 N
2) 1/2 N
3) 1 N
4) 2 N
5) 4 N
Here, the springs are in series, so each spring is only stretched 30 cm, and only half the force is needed. But also, since the springs are in a row, the force applied to one spring is transmitted to the other spring (like tension in a rope). So the overall applied force of 1/2 N is all that is needed. The combination of two springs in series behaves like a weaker spring!!

26. Simple Harmonic Motion
Displacement (x) is the distance of the object from equilibrium.
Amplitude (A) is the maximum displacement.
Period (T) is the time for one complete oscillation.
Frequency (f) is the number of complete oscillations per second.
It is measured in Hz or s-1.

27. Simple Harmonic Motion (SHM)
If n oscillations are happening in the time t, then:
T = t/n, and f = n/t.
Therefore T= 1/f, f = 1/T, or fT =1 k m k m k m

28. Simple Harmonic Motion

29. Example T, f
A mass oscillates 20 times in 5s. What is the period and the frequency of the oscillation?
T = 5s/20 = 0.25 s
f = 20/5s = 4 Hz
Check Tf = 1, 4 Hz x .25 s = 1

30. ConcepTest 13.1a Harmonic Motion I
1) 0
2) A/2
3) A
4) 2A
5) 4A
A mass on a spring in SHM has amplitude A and period T. What is the total distance traveled by the mass after a time interval T?

31. ConcepTest 13.1a Harmonic Motion I
1) 0
2) A/2
3) A
4) 2A
5) 4A
A mass on a spring in SHM has amplitude A and period T. What is the total distance traveled by the mass after a time interval T?
In the time interval T (the period), the mass goes through one complete oscillation back to the starting point. The distance it covers is: A + A + A + A (4A).

32. ConcepTest 13.1b Harmonic Motion II
1) 0
2) A/2
3) A
4) 2A
5) 4A
A mass on a spring in SHM has amplitude A and period T. What is the net displacement of the mass after a time interval T?

33. ConcepTest 13.1b Harmonic Motion II
1) 0
2) A/2
3) A
4) 2A
5) 4A
A mass on a spring in SHM has amplitude A and period T. What is the net displacement of the mass after a time interval T?
The displacement is Dx = x2–x1. Since the initial and final positions of the mass are the same (it ends up back at its original position), then the displacement is zero.
Follow-up: What is the net displacement after a half of a period?

34. ConcepTest 13.1c Harmonic Motion III
A mass on a spring in SHM has amplitude A and period T. How long does it take for the mass to travel a total distance of 6A ?
1) 1/2 T
2) 3/4 T
3) 1 1/4 T
4) 1 1/2 T
5) 2 T

35. ConcepTest 13.1c Harmonic Motion III
A mass on a spring in SHM has amplitude A and period T. How long does it take for the mass to travel a total distance of 6A ?
1) 1/2 T
2) 3/4 T
3) 1 1/4 T
4) 1 1/2 T
5) 2 T
We have already seen that it takes one period T to travel a total distance of 4A. An additional 2A requires half a period, so the total time needed for a total distance of 6A is 1 1/2 T.
Follow-up: What is the net displacement at this particular time?

36. Simple Harmonic Motion and Uniform Circular Motion
A ball is attached to the rim of a turntable of radius A
The focus is on the shadow that the ball casts on the screen
When the turntable rotates with a constant angular speed, the shadow moves in simple harmonic motion

37. SHM and Circular Motion
A spinning DVD is another example of periodic, repeating motion
Observe a particle on the edge of the DVD
The particle moves with a constant speed vc
Its position is given by θ = ω t
ω = 2πf is the angular velocity
Section 11.1

38. Simple Harmonic Motion
Systems that oscillate in a sinusoidal matter are called simple harmonic oscillators
They exhibit simple harmonic motion
Abbreviated SHM
The position can be described by y = A sin (2πƒt)
A is the amplitude of the motion
The object moves back and forth between the positions
y =  A
ƒ is the frequency of the motion
Section 11.1

39. SHM: Velocity as a Function of Time
Although the speed of the particle is constant, its y-component is not constant
vy = vc cosθ = vc cos(ωt)
= vc cos (2 π ƒ t)
From the definition of velocity, we can express the particle’s speed as v = 2 π ƒ A cos (2 π ƒ t) = v = vmax cos (2 π ƒ t),
where vmax = 2πfA
Section 11.1

40. Displacement (y) y = Asin(wt) or Displacement (x) x = Acos(wt)
Derivation of SHM formulae from uniform circular motion
Displacement (y)
y = Asin(wt) or
Displacement (x)
x = Acos(wt)
Velocity (v)
v = vmaxcos(wt)
vmax = ωA
Acceleration (a)
a = -amaxsin(wt)
amax=ω2A

41. SHM graphically

42. A simple harmonic oscillator can be described mathematically by:
where A is the amplitude of the motion, the maximum displacement from equilibrium, A = vmax, and A2 = amax.
Or by:

43. The period of oscillation is
where  is the angular frequency of the oscillations, k is the spring constant and m is the mass of the block.

44. At the equilibrium point x = 0 so a = 0 too.
When the stretch is a maximum, a will be a maximum too.
The velocity at the end points will be zero, and it is a maximum at the equilibrium point.

45. Equations of Motion
This figure shows the displacement as a function of time.

46. Displacement (y) y = Asin(ωt) Velocity (v) v = vmaxcos(wt)
Acceleration (a)
a = -amaxsin(wt)
T/4
3T/4 T
T/2

47. Simple Harmonic Motion
When the restoring force of a spring obeys Hooke’s Law (F=-kx), the resulting motion is called Simple Harmonic Motion
Consider a mass attached to a stretched spring that is released at to=0:
The displacement (x) of the mass due to the spring’s restoring force will be
y = A sin(wt)
where A is the amplitude of the strained spring
The velocity (v) of the mass will be
v = Aw cos(wt) = -vmax sin wt
where vmax = Aw
The acceleration (a) of the mass will be
a = -Aw2 sin(wt) = -amax cos wt
where amax = Aw2

48. Example The equation of and SHM is y = 5sin20πt. Find:
a) The frequency of the motion f=ω/2π= 20π/2π=10 Hz
b) The period of the motion:
T = 1/f = 0.1 s
c) The amplitude: A = 5 m
d) The maximum velocity
v= ωA = 20πx5 = 100π m/s

49. ConcepTest 13.2 Speed and Acceleration
A mass on a spring in SHM has amplitude A and period T. At what point in the motion is v = 0 and a = 0 simultaneously?
1) x = A
2) x > 0 but x < A
3) x = 0
4) x < 0
5) none of the above

50. ConcepTest 13.2 Speed and Acceleration
A mass on a spring in SHM has amplitude A and period T. At what point in the motion is v = 0 and a = 0 simultaneously?
1) x = A
2) x > 0 but x < A
3) x = 0
4) x < 0
5) none of the above
If both v and a were zero at the same time, the mass would be at rest and stay at rest! Thus, there is NO point at which both v and a are both zero at the same time.
Follow-up: Where is acceleration a maximum?

51. ConcepTest 13.4 To the Center of the Earth
A hole is drilled through the center of Earth and emerges on the other side. You jump into the hole. What happens to you ?
1) you fall to the center and stop
2) you go all the way through and continue off into space
3) you fall to the other side of Earth and then return
4) you won’t fall at all

52. ConcepTest 13.4 To the Center of the Earth
A hole is drilled through the center of Earth and emerges on the other side. You jump into the hole. What happens to you ?
1) you fall to the center and stop
2) you go all the way through and continue off into space
3) you fall to the other side of Earth and then return
4) you won’t fall at all
You fall through the hole. When you reach the center, you keep going because of your inertia. When you reach the other side, gravity pulls you back toward the center. This is Simple Harmonic Motion!
Follow-up: Where is your acceleration zero?

53. Newton’s 2nd Law & Ideal Springs
Applying Newton’s 2nd Law to a stretched ideal spring:
SF = ma = -kx
The acceleration of the spring is
a = - (k/m).x
The acceleration of the spring at any point in the motion is proportional to the displacement of the spring
For motions of this type, the angular frequency (w) of the motion is
w =(k/m)½
The period of the motion (T) is
T = 1/f = 2p/w = 2p(m/k)½
The larger the mass the greater the period (T) or T ~ m
General form:
a = w2 x (when a ~ x)

54. Period of The Spring and Mass System
m: mass (kg), k spring (elastic) constant (N/m)
Frequency: f = 1/T

55. What is the period of the system?
The mass in a spring and mass system (k = 8N/m, m = 0.5 Kg) is at 5cm away from the equilibrium position. What is the acceleration of the mass?
a= -kx/m
a = -8 x 0.05/0.5 = m/s2
What is the period of the system?

56. Conservation of Energy & Simple Harmonic Motion
When work is performed on a spring due to stretch/compression by an applied force the spring gains potential energy equal to
Uelastic = ½ kx2
As the spring is released and the restoring force w/in the spring drives the motion of the spring (assuming no friction)
PEelastic is converted to KE as the spring force does work
When the spring’s length equals its unstretched length, all of the PEelastic is converted to KE
Applying conservation of Energy to the spring:
(Uelastic)stretched = Kunstretched or
½ kA2 = ½ mv2
Therefore, the speed of the spring at its unstretched length is related to the length of the original displacement of the spring: v = (k/m)½ A

57. Energy Transformations
vmax
The block is moving on a frictionless surface
The total mechanical energy of the system is the kinetic energy of the block

58. Energy Transformations, 2
The spring is partially compressed
The energy is shared between kinetic energy and elastic potential energy
The total mechanical energy is the sum of the kinetic energy and the elastic potential energy

59. Energy Transformations, 3
The spring is now fully compressed
The block momentarily stops
The total mechanical energy is stored as elastic potential energy of the spring

60. Energy Transformations 4
vmax
When the block is again in the equilibrium position, the total mechanical energy is in the kinetic energy of the block
The spring force is conservative and the total energy of the system remains constant

61. Energy in the Spring-Mass System
The energy = kinetic + potential energy
U = (k/2)(displacement)2
K = ½ m v2
where x= displacement and v = the velocity.
According to conservation of energy the sum of the energies K and U is constant at all times.
Since when the spring is fully compressed (or stretched) the energy is potential only and equal to kA2/2, we can write:

62. Energy continuation: (*)
The previous equation allows the calculation of the velocity in SHM
Speed is a maximum at x = 0
Speed is zero at x = ±A
The ± indicates the object can be traveling in either direction

63. Energy Graphs

64. Example air track

65. Solution Example air track
a. F = kx, k = F/x = (6 N)/(0.03m) = 200 N/m
v = x = 0 b.
vmin = 0 m/s c.
amax = kA/m = (200 N/m x 0.04 m)/(0.5 Kg) = 16 m/s2
amin = 0 d.
a = -kx/m = - (200N/m x 0.02m)/0.5Kg =- 8m/s2 e.
K = ½ mv2 = ½ (0.5 kg) x(0.69m/s)2 = 0.12J
U = ½ kx2 = ½ 200N/m x (0.02m)2 = 0.04 J
E = K + U = 0.16 J
E = Umax = ½ KA2 = ½ 200N/m x (0.04m)2 = 0.16 J

66. ConcepTest 13.5a Energy in SHM I
A mass oscillates in simple harmonic motion with amplitude A. If the mass is doubled, but the amplitude is not changed, what will happen to the total energy of the system?
1) total energy will increase
2) total energy will not change
3) total energy will decrease

67. ConcepTest 13.5a Energy in SHM I
A mass oscillates in simple harmonic motion with amplitude A. If the mass is doubled, but the amplitude is not changed, what will happen to the total energy of the system?
1) total energy will increase
2) total energy will not change
3) total energy will decrease
The total energy is equal to the initial value of the elastic potential energy, which is PEs = 1/2 kA2. This does not depend on mass, so a change in mass will not affect the energy of the system.
Follow-up: What happens if you double the amplitude?

68. ConcepTest 13.5b Energy in SHM II
If the amplitude of a simple harmonic oscillator is doubled, which of the following quantities will change the most?
1) frequency
2) period
3) maximum speed
4) maximum acceleration
5) total mechanical energy

69. ConcepTest 13.5b Energy in SHM II
If the amplitude of a simple harmonic oscillator is doubled, which of the following quantities will change the most?
1) frequency
2) period
3) maximum speed
4) maximum acceleration
5) total mechanical energy
Frequency and period do not depend on amplitude at all, so they will not change. Maximum acceleration and maximum speed do depend on amplitude, and both of these quantities will double (you should think about why this is so). The total energy equals the initial potential energy, which depends on the square of the amplitude, so that will quadruple.
Follow-up: Why do maximum acceleration and speed double?

70. force = -mg(x/l)= -(mg/l)x
The Simple Pendulum
Consider the motion of a mass (m) attached to a string (length, l):
The gravitational force (mg) moving the mass (at all but the bottom point of the swing) is: - mg sinq
since sin q ~ q = x/l
force = -mg(x/l)= -(mg/l)x
Comparing to –kx yields k=mg/l.
Substitute in :
The period of the motion (T) is therefore q l m x mg

71. ConcepTest 13.6a Period of a Spring I
A glider with a spring attached to each end oscillates with a certain period. If the mass of the glider is doubled, what will happen to the period?
1) period will increase
2) period will not change
3) period will decrease

72. ConcepTest 13.6a Period of a Spring I
A glider with a spring attached to each end oscillates with a certain period. If the mass of the glider is doubled, what will happen to the period?
1) period will increase
2) period will not change
3) period will decrease
The period is proportional to the square root of the mass. So an increase in mass will lead to an increase in the period of motion.
T = 2p (m/k)
Follow-up: What happens if the amplitude is doubled?

73. ConcepTest 13.6b Period of a Spring II
A glider with a spring attached to each end oscillates with a certain period. If identical springs are added in parallel to the original glider, what will happen to the period?
1) period will increase
2) period will not change
3) period will decrease

74. ConcepTest 13.6b Period of a Spring II
A glider with a spring attached to each end oscillates with a certain period. If identical springs are added in parallel to the original glider, what will happen to the period?
1) period will increase
2) period will not change
3) period will decrease
We saw in the last section that two springs in parallel act like a stronger spring. So the spring constant has been effectively increased, and the period is inversely proportional to the square root of the spring constant, which leads to a decrease in the period of motion.
T = 2p (m/k)

75. ConcepTest 13.7a Spring in an Elevator I
A mass is suspended from the ceiling of an elevator by a spring. When the elevator is at rest, the period is T. What happens to the period when the elevator is moving upward at constant speed?
1) period will increase
2) period will not change
3) period will decrease

76. ConcepTest 13.7a Spring in an Elevator I
A mass is suspended from the ceiling of an elevator by a spring. When the elevator is at rest, the period is T. What happens to the period when the elevator is moving upward at constant speed?
1) period will increase
2) period will not change
3) period will decrease
Nothing at all changes when the elevator moves at constant speed. The equilibrium elongation of the spring is the same, and the period of simple harmonic motion is the same.

77. ConcepTest 13.7b Spring in an Elevator II
A mass is suspended from the ceiling of an elevator by a spring. When the elevator is at rest, the period is T. What happens to the period when the elevator is accelerating upward?
1) period will increase
2) period will not change
3) period will decrease

78. ConcepTest 13.7b Spring in an Elevator II
A mass is suspended from the ceiling of an elevator by a spring. When the elevator is at rest, the period is T. What happens to the period when the elevator is accelerating upward?
1) period will increase
2) period will not change
3) period will decrease
When the elevator accelerates upward, the hanging mass feels “heavier” and the spring will stretch a bit more. Thus, the equilibrium elongation of the spring will increase. However, the period of simple harmonic motion does not depend upon the elongation of the spring–it only depends on the mass and the spring constant, and neither one of them has changed.

79. ConcepTest 13.7c Spring on the Moon
A mass oscillates on a vertical spring with period T. If the whole setup is taken to the Moon, how does the period change?
1) period will increase
2) period will not change
3) period will decrease

80. ConcepTest 13.7c Spring on the Moon
A mass oscillates on a vertical spring with period T. If the whole setup is taken to the Moon, how does the period change?
1) period will increase
2) period will not change
3) period will decrease
The period of simple harmonic motion only depends on the mass and the spring constant and does not depend on the acceleration due to gravity. By going to the Moon, the value of g has been reduced, but that does not affect the period of the oscillating mass-spring system.
Follow-up: Will the period be the same on any planet?

81. ConcepTest 13.8a Period of a Pendulum I
Two pendula have the same length, but different masses attached to the string. How do their periods compare?
1) period is greater for the greater mass
2) period is the same for both cases
3) period is greater for the smaller mass

82. ConcepTest 13.8a Period of a Pendulum I
Two pendula have the same length, but different masses attached to the string. How do their periods compare?
1) period is greater for the greater mass
2) period is the same for both cases
3) period is greater for the smaller mass
The period of a pendulum depends on the length and the acceleration due to gravity, but it does not depend on the mass of the bob.
T = 2p (L/g)
Follow-up: What happens if the amplitude is doubled?

83. ConcepTest 13.8b Period of a Pendulum II
Two pendula have different lengths: one has length L and the other has length 4L. How do their periods compare?
1) period of 4L is four times that of L
2) period of 4L is two times that of L
3) period of 4L is the same as that of L
4) period of 4L is one-half that of L
5) period of 4L is one-quarter that of L

84. ConcepTest 13.8b Period of a Pendulum II
Two pendula have different lengths: one has length L and the other has length 4L. How do their periods compare?
1) period of 4L is four times that of L
2) period of 4L is two times that of L
3) period of 4L is the same as that of L
4) period of 4L is one-half that of L
5) period of 4L is one-quarter that of L
The period of a pendulum depends on the length and the acceleration due to gravity. The length dependence goes as the square root of L, so a pendulum 4 times longer will have a period that is 2 times larger.
T = 2p (L/g)

85. ConcepTest 13.9 Grandfather Clock
A grandfather clock has a weight at the bottom of the pendulum that can be moved up or down. If the clock is running slow, what should you do to adjust the time properly?
1) move the weight up
2) move the weight down
3) moving the weight will not matter
4) call the repairman

86. ConcepTest 13.9 Grandfather Clock
A grandfather clock has a weight at the bottom of the pendulum that can be moved up or down. If the clock is running slow, what should you do to adjust the time properly?
1) move the weight up
2) move the weight down
3) moving the weight will not matter
4) call the repairman
The period of the grandfather clock is too long, so we need to decrease the period (increase the frequency). To do this, the length must be decreased, so the adjustable weight should be moved up in order to shorten the pendulum length.
T = 2p (L/g)

87. ConcepTest 13.10a Pendulum in Elevator I
A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period is T. What happens to the period when the elevator is moving upward at constant speed?
1) period will increase
2) period will not change
3) period will decrease

88. ConcepTest 13.10a Pendulum in Elevator I
A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period is T. What happens to the period when the elevator is moving upward at constant speed?
1) period will increase
2) period will not change
3) period will decrease
Nothing at all changes when the elevator moves at constant speed. Neither the length nor the effective value of g has changed, so the period of the pendulum is the same.

89. ConcepTest 13.10b Pendulum in Elevator II
A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period is T. What happens to the period when the elevator is accelerating upward?
1) period will increase
2) period will not change
3) period will decrease

90. ConcepTest 13.10b Pendulum in Elevator II
A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period is T. What happens to the period when the elevator is accelerating upward?
1) period will increase
2) period will not change
3) period will decrease
When the elevator accelerates upward, the hanging mass feels “heavier” – this means that the effective value of g has increased due to the acceleration of the elevator. Since the period depends inversely on g, and the effective value of g increased, then the period of the pendulum will decrease (i.e., its frequency will increase and it will swing faster).

91. ConcepTest 13.10c Pendulum in Elevator III
A swinging pendulum has period T on Earth. If the same pendulum were moved to the Moon, how does the new period compare to the old period?
1) period increases
2) period does not change
3) period decreases

92. ConcepTest 13.10c Pendulum in Elevator III
A swinging pendulum has period T on Earth. If the same pendulum were moved to the Moon, how does the new period compare to the old period?
1) period increases
2) period does not change
3) period decreases
The acceleration due to gravity is smaller on the Moon. The relationship between the period and g is given by:
therefore, if g gets smaller, T will increase. g L T = 2p
Follow-up: What can you do to return the pendulum to its original period?

93. ConcepTest 13.11 Damped Pendulum
After a pendulum starts swinging, its amplitude gradually decreases with time because of friction.
What happens to the period of the pendulum during this time ?
1) period increases
2) period does not change
3) period decreases

94. ConcepTest 13.11 Damped Pendulum
After a pendulum starts swinging, its amplitude gradually decreases with time because of friction.
What happens to the period of the pendulum during this time ?
1) period increases
2) period does not change
3) period decreases
The period of a pendulum does not depend on its amplitude, but only on its length and the acceleration due to gravity. g L T = 2p
Follow-up: What is happening to the energy of the pendulum?

95. ConcepTest 13.12 Swinging in the Rain
You are sitting on a swing. A friend gives you a push, and you start swinging with period T1. Suppose you were standing on the swing rather than sitting. When given the same push, you start swinging with period T2. Which of the following is true?
1) T1 = T2
2) T1 > T2
3) T1 < T2 T1
[CORRECT 5 ANSWER]

96. ConcepTest 13.12 Swinging in the Rain
You are sitting on a swing. A friend gives you a push, and you start swinging with period T1. Suppose you were standing on the swing rather than sitting. When given the same push, you start swinging with period T2. Which of the following is true?
1) T1 = T2
2) T1 > T2
3) T1 < T2 L1 T1 L2 T2
Standing up raises the Center of Mass of the swing, making it shorter !! Since L1 > L2 then T1 > T2 g L T = 2p