PHY300 Chapter 10 physics 5e

Giambattista Physics
Chapter 10
©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.

©McGraw-Hill Education
Chapter 10: Elasticity and Oscillations
10.1 Elastic Deformations of Solids
10.2 Hooke’s Law for Tensile and Compressive Forces
10.3 Beyond Hooke’s Law
10.4 Shear and Volume Deformations
10.5 Simple Harmonic Motion
10.6 The Period and Frequency for SHM
10.7 Graphical Analysis of SHM
10.8 The Pendulum
10.9 Damped Oscillations
10.10 Forced Oscillations and Resonance

10.1 Elastic Deformations of Solids
A deformation is a change in the size or shape of the object.
Many solids are stiff enough that the deformation cannot be
seen with the human eye; a microscope or other sensitive device
is required to detect the change in size or shape.
When the contact forces are removed, an elastic object returns
to its original shape and size.

10.2 Hooke’s Law for Tensile
and Compressive Forces
Suppose we stretch a wire by applying tensile forces of
magnitude F to each end. The length of the wire increases from L
to L + L . How does the elongation L depend on the original
length L?

Example 10.1
If a given tensile force stretches a tendon by an amount L, how
much would the same force stretch a tendon twice as long but
identical in thickness and composition?
Solution Think of the tendon of length 2L as two tendons of
length L placed end-to-end. Under the same tension, each of the
two tendons stretches by an amount L, so the total
deformation of the long tendon is 2L.

Stress and Strain
The fractional length change is called the strain; it is a
dimensionless measure of the degree of deformation.
Strain (fractional length change)
L
L


The force per unit area is called the stress:
Stress (force per unit cross-sectional area)
F
A


Hooke's Law
According to Hooke’s law, the deformation is proportional to the
deforming forces as long as they are not too large:
The constant k depends on the length and cross-sectional area
of the object. A larger cross-sectional area A makes k larger; a
greater length L makes k smaller.
F L
k
 

Hooke’s Law Continued
stress strain

F L YA
Y k
A L L

 
The constant of proportionality Y is called the elastic modulus,
or Young’s modulus; Y has the same units as those of stress
(Pa or N/m2), since strain is dimensionless.
Young’s modulus can be thought of as the inherent stiffness of a
material; it measures the resistance of the material to elongation
or compression.

Example 10.2
A man whose weight is 0.80 kN is
standing upright.
By approximately how much is his
femur (thighbone) shortened
compared with when he is lying down?
Assume that the compressive force on
each femur is about half his weight.
The average cross-sectional area of the
femur is 8.0 cm2 and the length of the
femur when lying down is 43.0 cm.

Example 10.2 Strategy
A change in length of the femur involves a strain.
After finding the stress and looking up the Young’s modulus, we
can find the strain using Hooke’s law.
We assume that each femur supports half the man’s weight.

Example 10.2 Solution
/
F L F A
Y L L
A L Y

   
9
9.4 10 Pa
Y  
2
2 2
1 m
cm 0.00080 m
100 cm
8.0
A
 
 

 
 
2 2
9
5
/
(4.0 10 N) / (0.00080 m )
43.0 cm
9.4 10 Pa
=5.3 10 43.0 cm 0.0023 cm
L
F A
L
Y







  

10.3 Beyond Hooke’s Law
A ductile material continues to stretch beyond its ultimate
tensile strength without breaking; the stress then decreases
from the ultimate strength.
For a brittle substance, the ultimate strength and the breaking
point are close together.

Beyond Hooke’s Law Continued

Beyond Hooke’s Law Example: Bone

Example 10.3
A crane is required to lift loads of up to 1.0 × 105 N (11 tons).
(a) What is the minimum diameter of the steel cable that must
be used?
(b) If a cable of twice the minimum diameter is used and it is 8.0
m long when no load is present, how much longer is it when
supporting a load of 1.0 × 105 N?
Data for steel:
Y = 2.0 × 1011 Pa; proportional limit = 2.0 × 108 Pa; elastic limit =
3.0 × 108 Pa; tensile strength = 5.0 × 108 Pa.

The elastic limit is the maximum stress so that no permanent
deformation occurs; the tensile strength is the maximum stress
so that the cable does not break.
We choose a minimum diameter in (a) to keep the stress below
the elastic limit.

Example 10.3 Solution 1
(a) 8
elastic limit 3.0 10 Pa
F
A
  
5
N
1.0 10
F  
5
4 2
8
1.0 10 N
3.33 10 m
elastic limit 3.0 10 Pa
A
F 
 




4 2
4 4 3.33 10 m
2.1 cm
d
A
 

 

 
The minimum diameter is therefore 2.1 cm.

(b) 8
7
3.0 10 Pa
7.5 10 Pa
4
F
A

  
7
11
/ 7.5 10 Pa
0.000375
2.0 10 Pa
L F A
L Y
 
  

0.000375 0.000375 8.0 m 0.0030 m 3.0 mm
L L
     

10.4 Shear and Volume Deformations
A shear deformation is the result of a pair of equal and opposite
forces that act parallel to two opposite surfaces.
The shear stress is the magnitude of the shear force divided by
the area of the surface on which the force acts:
shear force
shear stress
area of surface
F
A
 

Shear Strain
Shear strain is the ratio of the relative displacement x to the
separation L of the two surfaces:
displacement of surfaces
shear strain
separation of surfaces
x
L

 

Hooke’s Law for Shear Deformation
shear stress shear strain

F x
S
A L


The shear strain is proportional to the shear stress as long as the
stress is not too large.
The constant of proportionality is the shear modulus S.
The units of shear stress and the shear modulus are the same as
for tensile or compressive stress and Young’s modulus: Pa or
N/m2 . The strain is once again dimensionless.

Shear Deformation: Scissors

Example 10.4
A sheet of paper of thickness 0.20 mm is cut with scissors that
have blades of length 10.0 cm and width 0.20 cm.
While cutting, the scissors blades each exert a force of 3.0 N on
the paper; the length of each blade that makes contact with the
paper is approximately 0.5 mm.
What is the shear stress on the paper?

Shear stress is a force divided by an area. In this problem,
identifying the correct area is tricky.
The blades push two cross-sectional paper surfaces in opposite
directions so they are displaced with respect to one another. The
shear stress is the force exerted by each blade divided by this
cross-sectional area—the thickness of the paper times the length
of blade in contact with the paper.

4 4 7 2
thickness contact length
2.0 10 m 5 10 m 1 10 m
A
  

    


7 2
7 2
3.0 N
3 10 N/m
1 10 m
F
A 
  


Volume Deformation
The fluid pressure P is the
force per unit surface area;
it can be thought of as the
volume stress on the solid
object.
volume stress pressure
F
P
A
  

Volume Deformation Continued
The resulting deformation of
the object is characterized by
the volume strain, which is the
fractional change in volume:
change in volume
volume strain
original volume
V
V




Hooke’s Law for Volume Deformation
V
P B
V

  
Unless the stress is too large, the stress and strain are
proportional within a constant of proportionality called the bulk
modulus B.

Example 10.5
A marble statue of volume 1.5 m3 is being transported by ship
from Athens to Cyprus. The statue topples into the ocean when
an earthquake-caused tidal wave sinks the ship; the statue ends
up on the ocean floor, 1.0 km below the surface.
Find the change in volume of the statue in cm3 due to the
pressure of the water. The density of seawater is 1025 kg/m3.
Strategy
The water pressure is the volume stress; it is the force per unit
area pressing inward and perpendicular to all the surfaces of the
statue.

3
7
1025 kg/m 9.8 N/kg 1000 m
=1.005 10 Pa
P gd

 
  

V
P B
V

  
7
3
9
3
4 3 3
1.005 10 Pa
1.5 m
70 10
0
2 =
. 2
2
Pa
10 cm
10 m 2 0 cm
1 m
P
V V
B


 
     

 
  

 
 
The statue’s volume decreases by approximately 220 cm3.

10.5 Simple Harmonic Motion
Vibrations occur in the vicinity of a point
of stable equilibrium. An equilibrium
point is stable if the net force on an
object when it is displaced a small
distance from equilibrium points back
toward the equilibrium point.
Such a force is called a restoring force
since it tends to restore equilibrium.
A special kind of vibratory motion—called simple harmonic
motion (or SHM )—occurs whenever the restoring force is
proportional to the displacement from equilibrium.

Approximate Nonlinear Restoring Force
A nonlinear restoring force (curved line) can be approximated as
a linear restoring force (straight line) for small displacements.

Spring Relaxed Position
A spring in its relaxed position. We choose the object’s
equilibrium position as the origin (x = 0).
x
F kx
 

Energy Analysis in SHM
The oscillator slows as it approaches the endpoints and gains
speed as it approaches the equilibrium point.
The total mechanical energy of the mass and spring is constant.
constant
E K U
  

Energy Analysis in SHM Continued
2
1
2
2 2
1 1
2 2
x
U kx
mv kx
E



The maximum displacement of the body is the amplitude A. The
total energy E at the endpoints is:
2
1
total 2
2
1
total m
2
2 2
1 1
m
2 2
m
At 0:
A
E kA
x E mv
mv kA
k
m
v

 



Acceleration in SHM
m
( ) ( )
x x
x
kx ma
k
t x t
m
k
a A
F
m
a
  
 


Example 10.6
A model rocket of 1.0-kg mass is attached to a horizontal spring
with a spring constant of 6.0 N/cm. The spring is compressed by
18.0 cm and then released.
The intent is to shoot the rocket horizontally, but the release
mechanism fails to disengage, so the rocket starts to oscillate
horizontally.

Example 10.6 Continued
Ignore friction and assume the spring to be ideal.
(a) What is the amplitude of the oscillation?
(b) What is the maximum speed?
(c) What are the rocket’s speed and acceleration when it is 12.0
cm from the equilibrium point?
Strategy
The speed at any position can be found using energy
conservation:
2 2 2
1 1 1
2 2 2
x
kx mv kA
 

(a) The amplitude of the oscillation is the maximum
displacement, so A = 18.0 cm.
(b)
2 2
1 1
m m
2 2
mv E kA
K   
2
m
6.0 10 N/m
0.180 m 4.4 m/s
1.0 kg
k
v A
m

   

(c) 2 2 2
1 1 1
2 2 2
x
kx mv kA
 
2 2
2 2
2
2 2
( )
6.0 10 N/m
[(0.180 m) (0.120 m) ] 3.3 m/s
1.0 kg
v
kA kx k
A x
m m

 
  


x x
kx ma
F   
At 0.120 m,
x  
2
2
6.0 10 N/m
( 0.120 m) 72 m/s
1.0 kg
x x
a
k
m

      

10.6 The Period and Frequency for SHM
The period T is the time taken by one complete cycle.
The frequency f is the number of cycles per unit time:
1
(SI unit: Hz = cycles per second)
T
f 

Circular Motion and SHM
( )
t t
 

( ) cos
cos
x t A
A t





Circular Motion and SHM Continued
2 2
2
2
cos cos
cos cos
( ) ( )
( )
x
x
x
x
m
A t
a r A
a a A t
a t x
t
t
k k
a x
m
A  
 
  



 
   




 

Period and Frequency for an
Ideal Mass-Spring System
Since the object in SHM and the pin in circular motion have the
same frequency and period, the relationships between , f, and
T still apply.
Therefore, the frequency and period of a mass-spring system
are
1
2 2
1
2
k
m
m
T
f k
f

 





In the context of SHM, the quantity ω is called the angular
frequency.

Maximum Speed and Acceleration in SHM
With the identification of  for a mass-spring system,
m
2
m
A
a A
v 




Finding Angular Frequency for SHM
• Write down the restoring force as a function of the
displacement from equilibrium. Since the restoring force is
linear, it always takes the form F = −kx, where k is a constant.
• Use Newton’s second law to relate the restoring force to the
acceleration.
• Solve for  using ax = −2x.

A Vertical Mass and Spring

A Vertical Mass and Spring Continued
spring,
0 (at equilibrium)
( )
( )
y
y
y
F kd mg
F k d y
F k d y mg kd ky mg
   
 
     


(at displacement y from equilibrium)
y
F ky
kd mg

 


Example 10.7
A spring with spring constant k is suspended vertically. A model
goose of mass m is attached to the unstretched spring and then
released so that the bird oscillates up and down. (Ignore friction
and air resistance; assume an ideal massless spring.) Calculate
the kinetic energy, the elastic potential energy, the gravitational
potential energy, and the total mechanical energy at
(a) the point of release and
(b) The equilibrium point. Take the gravitational potential
energy to be zero at the equilibrium point.
(c) How long does it take the bird to move from its highest to its
lowest position?

(a)
0
y
F kd mg
mg
A d
k
   
 

2
g
2
e g
( )
( )
mg
mgy mgA
k
mg
E K U U
k
U   
   

(b) 2 2 2
1 1
m
2 2
K mv m A

 
Substituting A = mg/k and 2 = k/m,
2 2
2
2 2
2
e 2
2 2 2
e g
1 ( ) 1 ( )
2 2
1 1 ( ) 1 ( )
2 2 2
1 ( ) 1 ( ) ( )
0
2 2
k mg mg
m
m k k
mg mg
U kA k
k k
mg mg mg
E K U U
k k k
K 
  
     



(c) The period is 2 / .
m k

Moving from y = +A to y = -A is half of a complete cycle, so the
time it takes is
/ .
t m k

 

10.7 Graphical Analysis of SHM

Graphical Analysis of SHM Continued

Example 10.8
A loudspeaker has a movable diaphragm (the cone) that vibrates
back and forth to produce sound waves. The displacement of a
loudspeaker cone playing a sinusoidal test tone is graphed
below.

Example 10.8 Continued
Find
(a) the amplitude of the motion,
(b) the period of the motion, and
(c) the frequency of the motion.
(d) Write equations for x(t) and vx(t).

(a) The amplitude is the maximum displacement shown on the
graph: A = 0.015 m.
(b) The period is the time for one complete cycle. From the
graph: T = 0.040 s.
(c) 1 1
25 Hz
0.040 s
T
f  


(d)
m
m
cos
0.015 m
2 160 rad/s
( )
(
sin
160 rad/s 0.015 m 2. m s
)
4 /
x
t
A
f
v t v t
x t
v A
A 
 



 
 
   


10.8 The Pendulum
Simple Pendulum
When a pendulum swings back and forth, a string or thin rod
constrains the bob to move along a circular arc.
However, for oscillations with small amplitude , we assume that
the bob moves back and forth along the x-axis; the vertical
motion of the bob is negligible.

Simple Pendulum
The restoring force is the tangential
component of the weight
tan sin
mg
F 
 
For small angles, sin   , and  = s/L,
where s in the circular arc length.
Therefore,
tan ( / )
mg s
F L
 
The effective spring constant—the constant of proportionality
between the restoring force and the displacement, is
eff /L
k mg


Simple Pendulum Continued
The angular frequency is
eff
k g
m L
  
and the period is
2
2
L
g
T






Example 10.9
A grandfather clock uses a pendulum with period 2.0 s to keep
time.
In one such clock, the pendulum bob has mass 150 g; the
pendulum is set into oscillation by displacing it 33 mm to one
side.
(a) What is the length of the pendulum?
(b) Does the initial displacement satisfy the small angle
approximation?

Strategy
The period depends on the length of the pendulum and on the
gravitational field strength g. It does not depend on the mass of
the bob.
It also does not depend on the initial displacement, as long as it
is small compared to the length.
Solution
(a)
2
T
L
g


2
2
2 2
2
(2 )
(2.0 s) 9.80 m/s
0.99 m
(2 )
T g
L




 

Example 10.9 Solution Continued
(b) 33 mm
0.033
990 mm
x
L
 
1
sin 0.033 0.033006
 
 
sin  and  differ by less than 0.02 %. Since we only know T to
two significant figures, the approximation is good.

Physical Pendulum 1
The center of mass of the bar
is located at the midpoint, a
distance d = 1/2 L from the
axis. Since the mass is on
average closer to the axis, the
period is shorter than that of
the simple pendulum.
Would this bar have a period equal to that of a simple
pendulum of length d = 1/2 L?

Physical Pendulum 2
The gravitational force acts at
the center of mass, but we
cannot think of all the mass as
being concentrated at that
point—that would give the
wrong rotational inertia. When
set into oscillation, the bar, or
any other rigid object free to
rotate about a fixed axis, is
called a physical pendulum.
The period of a physical pendulum is:
2
T
I
mgd



Physical Pendulum 3
2
T
I
mgd


The rotational inertia of a uniform bar rotating about an axis
through an endpoint is I = (1/3)mL2.
2
1
3
1
2
2
2 2
( ) 3
2
T
mL
I L
mgd mg L g
  
  
The bar has the same period as a simple pendulum of length
2/3L.

Example 10.10
During a relaxed walking pace, an animal’s leg can be thought of
as a physical pendulum of length L that pivots about the hip.
(a) What is the relaxed walking frequency for a cat (L = 30 cm),
dog (60 cm), human (1 m), giraffe (2 m), and a mythological
titan (10 m)?
(b) Derive an equation that gives the walking speed for a given
walking frequency f. [Hint: Draw a picture of the leg position
at the start of the swing (leg back) and the end of the swing
(leg forward) and assume a comfortable angle of about 30°
between these two positions. To how many steps does a
complete period of the pendulum correspond?]
(c) Find the walking speed for each animal listed in part (a).

We have to use an idealized model of the leg, since we don’t
know the exact location of the center of mass or the rotational
inertia.
The simple pendulum is not a good model, since it would assume
all the mass of the leg at the foot!
A much better model is to think of the leg as a uniform cylinder
pivoting about one end.

(a) For a uniform cylinder, the center of mass is a distance d =
(1/2)L from the pivot and the rotational inertia about an
axis at one end is I = (1/3)mL2. Then the period is
2
1
3
1
2
2
2 2
( ) 3
2
T
mL
I L
mgd mg L g
  
  
1 1 3
0.2
2 2
g g
T L L
f


 
Substituting the numerical values of L for each animal, we find
the frequencies to be 1 Hz (cat), 0.8 Hz (dog), 0.6 Hz (human),
0.4 Hz (giraffe), and 0.2 Hz (titan).

(b) One period of the “pendulum” corresponds to two steps.
(c) The speeds are
0.3 m/s (cat),
0.5 m/s (dog),
0.6 m/s (human),
0.9 m/s (giraffe), and
2 m/s (titan).
1
3
2 2
12
0.2
L
D L L
D
v Lf gL
T


   
  


10.9 Damped Oscillations
The oscillations of a swinging
pendulum or a vibrating tuning
fork gradually die out as energy is
dissipated. The amplitude of each
cycle is a little smaller than that of
the previous cycle.
This kind of motion is called
damped oscillation, where the
word damped is used in the sense
of extinguished or restrained.
For a small amount of damping, oscillations occur at
approximately the same frequency as if there were no damping.

10.10 Forced Oscillations and Resonance
Forced oscillations (or driven oscillations) occur when a periodic
external driving force acts on a system that can oscillate.
The frequency of the driving force does not have to match the
natural frequency of the system.
When the driving frequency is equal to the natural frequency of
the system, the amplitude of the motion is a maximum. This
condition is called resonance.

Beyond Hooke’s Law Continued Appendix
Stress-strain curves showing limits for a ductile material and a
brittle material. The ultimate strength and breaking point are
well separated for ductile materials, but close together for brittle
materials. For a ductile material, after the ultimate strength is
reached, as strain increases further, the stress decreases before
the breaking point is reached. For brittle materials, the ultimate
strength and breaking point are virtually identical.

Circular Motion and SHM Appendix
An experiment to show the relation between uniform circular
motion and SHM. A pin P moves counterclockwise around a
circle as a disk rotates with constant angular velocity . A light
source shines on the pin casting a shadow on the wall. The
motion of the shadow is the same as the x-component of the
circular motion of the pin. The angle of the pin in the circle
increases at a constant rate  = t. The amplitude A of the SHM
is the same as the radius of the circle.

Example 10.7 Strategy Appendix
The spring is unstretched before the model bird is released (with
initial speed v = 0) at position y = +A; the model bird passes
through the equilibrium position y = 0 with maximum speed v =
vm; the spring’s maximum extension occurs when the bird is at y
= −A, and v = 0 at that point.

10.7 Graphical Analysis of SHM Appendix
Graphs of position and velocity for SHM. At any time, the value
of vx is the slope of the graph of x. When the displacement is
maximum (x = ±A), the velocity is zero. At the equilibrium point
(x = 0), the speed is maximum (vx = ±vm). The velocity graph is
one-quarter of a cycle ahead of the position graph. That is, vx(t)
reaches a maximum one-quarter period before x(t) reaches its
maximum.

Graphical Analysis of SHM Continued Appendix
Graphs of acceleration, kinetic and potential energy for SHM. At
any time, the value of ax is the slope of the graph of vx. The
acceleration is always proportional to the displacement; its
direction is always toward the equilibrium point. Kinetic and
potential energy oscillate out of phase with each other. When
the kinetic energy is maximum, the potential energy is zero, and
vice versa. The total mechanical energy E = K + U is constant.

Simple Pendulum Appendix
A simple pendulum of length L, displaced so the string makes an
angle  with the vertical. The bob travels in a circular path. The
arc length from the bottom of the swing to the position of the
bob is s. The forces acting on the bob are the tension in the
string and the weight.

PHY300 Chapter 10 physics 5e

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to PHY300 Chapter 10 physics 5e

Similar to PHY300 Chapter 10 physics 5e (20)

More from BealCollegeOnline

More from BealCollegeOnline (20)

Recently uploaded

Recently uploaded (20)

PHY300 Chapter 10 physics 5e