Simple harmonic motion and elasticity

Chapter 10
Simple Harmonic
Motion and Elasticity

Simple Harmonic Motion
Back and forth motion that is caused by a force that is directly
proportional to the displacement. The displacement centers
around an equilibrium position.
xFs

Springs – Hooke’s Law
One of the simplest type
of simple harmonic
motion is called
Hooke's Law. This is
primarily in reference to
SPRINGS.
kxorkxF
k
k
xF
s
s



N/m):nitConstant(USpring
alityProportionofConstant

The negative sign only
tells us that “F” is what is
called a RESTORING
FORCE, in that it works in
the OPPOSITE direction
of the displacement.

10.1 The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
The restoring force on an ideal spring is xkFx 

10.2 Simple Harmonic Motion and the Reference Circle
tAAx  coscos 
DISPLACEMENT

tAAx  coscos 

period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
T
f
1

T
f


2
2 
amplitude A: the maximum displacement

VELOCITY
 tAvv
v
Tx  sinsin
max


Example 3 The Maximum Speed of a Loudspeaker Diaphragm
The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm.
(a)What is the maximum speed of the diaphragm?
(b)Where in the motion does this maximum speed occur?

 tAvv
v
Tx  sinsin
max

(a)      
sm3.1
Hz100.12m1020.02 33
max

 
 fAAv
(b)The maximum speed
occurs midway between
the ends of its motion.

ACCELERATION
 tAaa
a
cx  coscos
max
2


FREQUENCY OF VIBRATION
m
k

tAax  cos2
tAx cos
xmakxF 
2
mAkA 

Example 6 A Body Mass Measurement Device
The device consists of a spring-mounted chair in which the astronaut
sits. The spring has a spring constant of 606 N/m and the mass of
the chair is 12.0 kg. The measured
period is 2.41 s. Find the mass of the
astronaut.

totalm
k
 2
total km 
T
f


2
2 
  astrochair2total
2
mm
T
k
m 

 
   kg77.2kg0.12
4
s41.2mN606
2
2
2
chair2astro




m
T
k
m

10.3 Energy and Simple Harmonic Motion
A compressed spring can do work.

     fofo xxkxkxsFW  
0coscos 2
1
elastic 
2
2
12
2
1
elastic fo kxkxW 

DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy is the energy that a spring
has by virtue of being stretched or compressed. For an
ideal spring, the elastic potential energy is
2
2
1
elasticPE kx
SI Unit of Elastic Potential Energy: joule (J)

Conceptual Example 8 Changing the Mass of a Simple
Harmonic Oscilator
The box rests on a horizontal, frictionless
surface. The spring is stretched to x=A
and released. When the box is passing
through x=0, a second box of the same
mass is attached to it. Discuss what
happens to the (a) maximum speed
(b) amplitude (c) angular frequency.

Example 8 Changing the Mass of a Simple Harmonic Oscilator
A 0.20-kg ball is attached to a vertical spring. The spring constant
is 28 N/m. When released from rest, how far does the ball fall
before being brought to a momentary stop by the spring?

of EE 
2
2
12
2
12
2
12
2
12
2
12
2
1
ooooffff kymghImvkymghImv  
oo mghkh 2
2
1
   m14.0
mN28
sm8.9kg20.02
2
2


k
mg
ho

10.4 The Pendulum
A simple pendulum consists of a particle attached to a frictionless
pivot by a cable of negligible mass.
only)angles(small
L
g

only)angles(small
I
mgL


A simple pendulum consists of a particle attached
to a frictionless pivot by a cable of negligible mass
When the particle is pulled away from its
equilibrium position by an angle and
released, it swings back and forth
Gravity causes the back-and-forth rotation about the axis. The rotation speeds
up as the particle approaches the lowest point and slows down on the upward
part of the swing.
We denote the position of the pendulum along the circle of arc by, s, where we
choose s=0 to correspond to the equilibrium. In terms of the angle, which the
cord of the pendulum makes with the vertical direction we have
There is no acceleration along the direction of the cord so we have a tangential net
force:
But for small angles
From Hook’s law, the restoring force is
Equating (4) and (6) substituting (5)
But the a frequency which gives
Previously we found that
𝜔 = 2𝜋𝑓 =
2𝜋
𝑇
then the period (T) of the pendulum given by
𝑇 =
2𝜋
𝜔
= 2𝜋
𝑙
𝑔

13. REASONING AND SOLUTION From the drawing given with the problem statement, we see that
the kinetic frictional force on the bottom block (#1) is given by
fk1 = µk(m1 + m2)g (1)
and the maximum static frictional force on the top block (#2) is
MAX
s2 s 2f m g
(2)
Newton’s second law horizontally applied to the bottom block gives
F – fk1 – kx = 0 (3)
Newton’s second law applied to the top block gives MAX
s2 0f kx  (4)
a. To find the compression x, we have from Equation (4) that
x = MAX
s2f /k = µsm2g/k = (0.900)(15.0 kg)(9.80 m/s
2
)/(325 N/m) = 0.407 m
b. Solving Equation (3) for F and then using Equation (1) to substitute for fk1, we find
that
F = kx + fk1 = kx + µk(m1 + m2)g
F = (325 N/m)(0.407 m) + (0.600)(45.0 kg)(9.80 m/s
2
) = 397 N

26. REASONING The work done in stretching or compressing a spring is given
directly by Equation 10.12 as  2 21
0 f2
W k x x  , where k is the spring constant and x0
and xf are, respectively, the initial and final displacements of the spring from its
equilibrium position. The work is positive if the restoring force and the displacement
have the same direction and negative if they have opposite directions.
SOLUTION
a. The work done in stretching the spring from +1.00 to +3.00 m is
       2 22 2 21 1
0 f2 2
46.0 N/m 1.00 m 3.00 m 1.84 10 JW k x x          
b. The work done in stretching the spring from –3.00 m to +1.00 m is
       2 22 2 21 1
0 f2 2
46.0 N/m 3.00 m 1.00 m 1.84 10 JW k x x           
c. The work done in stretching the spring from –3.00 to +3.00 m is
       2 22 21 1
0 f2 2
46.0 N/m 3.00 m 3.00 m 0 JW k x x         

27. REASONING As the block falls, only two forces act on it: its weight and the
elastic force of the spring. Both of these forces are conservative forces, so the falling
block obeys the principle of conservation of mechanical energy. We will use this
conservation principle to determine the spring constant of the spring. Once the spring
constant is known, Equation 10.11, /k m  , may be used to find the angular
frequency of the block’s vibrations.
SOLUTION
a. The conservation of mechanical energy states that the final total
mechanical energy Ef is equal to the initial total mechanical energy E0, or Ef = E0
(Equation 6.9a). The expression for the total mechanical energy of an object
oscillating on a spring is given by Equation 10.14. Thus, the conservation of total
mechanical energy can be written as
2 2 2 2 2 21 1 1 1 1 1
f f f f 0 0 0 02 2 2 2 2 2
f 0
mv I m g h k y mv I m g h k y
E E
       
Before going any further, let’s simplify this equation by noting which variables are
zero. Since the block starts and ends at rest, vf = v0 = 0 m/s. The block does not rotate,
so its angular speed is zero, f = 0 = 0 rad/s. Initially, the spring is unstretched, so
that y0 = 0 m. Setting these terms equal to zero in the equation above gives 21
f f 02
m g h k y m g h 
Solving this equation for the spring constant k, we have that      
 
2
0 f
2 21 1
f2 2
0.510 kg 9.80 m/s 0.120 m
83.3 N/m
0.120 m
mg h h
k
y

  
b. The angular frequency  of the block’s vibrations depends on the spring
constant k and the mass m of the block:

43. REASONING As the ball swings down, it reaches it greatest speed at the lowest
point in the motion. One complete cycle of the pendulum has four parts: the
downward motion in which the ball attains its greatest speed at the lowest point, the
subsequent upward motion in which the ball slows down and then momentarily comes
to rest. The ball then retraces its motion, finally ending up where it originally began.
The time it takes to reach the lowest point is one-quarter of the period of the
pendulum, or t = (1/4)T. The period is related to the angular frequency  of the
pendulum by Equation 10.4, T = 2/. Thus, the time for the ball to reach its lowest
point is
1
4
1 2
4
t T


 
   
 
The angular frequency  of the pendulum depends on its length L and the acceleration
g due to gravity through the relation /g L  (Equation 10.16). Thus, the time is
1 2 1 2
4 4 2
L
t
gg
L
  

   
      
  
 
2
0.85 m
0.46 s
2 2 9.80 m/s
L
t
g
 
  

45. REASONING
a. The angular frequency  of a simple pendulum can be found directly from
Equation 10.16 as /g L  , where g is the magnitude of the acceleration due
to gravity and L is the length of the pendulum.
b. The total mechanical energy of the pendulum as it swings back and forth is the
gravitational potential energy it has just before it is released, since the pendulum
is released from rest and has no initial kinetic energy. The reason is that friction
is being neglected, and the tension in the cable is always perpendicular to the
motion of the bob, so the tension does no work. Thus, the work done by
nonconservative forces, such as friction and tension, is zero. This means that the
total mechanical energy is conserved (see Equation 6.9b) and is the same at all
points along the motion, including the initial point where the bob is released.
c. To find the speed of the bob as it passes through the lowest point of the swing, we
will use the conservation of energy, which relates the total mechanical energy at
the lowest point to that at the highest point.
a. The angular frequency of the pendulum is
2
9.80 m/s
3.5 rad/s
0.79 m
g
L
    (10.16)
b. At the moment the pendulum is released, the only type of energy it has is its
gravitational potential energy. Thus, its potential energy PE is equal to its initial
total mechanical energy E0, so PE = E0. According to Equation 6.5, the potential
energy of the pendulum is PE = mgh, where m is the mass of the bob and h is its
height above its equilibrium position (i.e., its position when the pendulum hangs
straight down). The drawing shows that this height is related to the length L of
the pendulum by  1 cos5.50h L   . Thus, the total mechanical energy of the
pendulum is
 
    
0
2
1 cos5.50
0.24 kg 9.80 m/s 0.79 m 1 cos5.50 10 J
E mgh mgL

   
    
c. As the bob passes through the lowest point of the swing, it has only kinetic
energy, so its total mechanical energy is 21
f f2
E mv . Since the total mechanical
energy is conserved  f 0 ,E E we have that
21
f 02
mv E
Solving for the final speed gives
 0
f
2 2 10 J
0.27 m/s
0.24 kg
E
v
m


  

10.4 The Pendulum
Example 10 Keeping Time
Determine the length of a simple pendulum that will
swing back and forth in simple harmonic motion with
a period of 1.00 s.
2
2
L
g
T
f 


    m248.0
4
sm80.9s00.1
4 2
22
2
2


gT
L
2
2
4
gT
L 

10.5 Damped Harmonic Motion
In simple harmonic motion, an object oscillated
with a constant amplitude.
In reality, friction or some other energy
dissipating mechanism is always present
and the amplitude decreases as time
passes.
This is referred to as damped harmonic
motion.

10.5 Damped Harmonic Motion
1) simple harmonic motion
2&3) underdamped
4) critically damped
5) overdamped

10.6 Driven Harmonic Motion and Resonance
When a force is applied to an oscillating system at all times,
the result is driven harmonic motion.
Here, the driving force has the same frequency as the
spring system and always points in the direction of the
object’s velocity.

10.6 Driven Harmonic Motion and Resonance
RESONANCE
Resonance is the condition in which a time-dependent force can transmit
large amounts of energy to an oscillating object, leading to a large amplitude
motion.
Resonance occurs when the frequency of the force matches a natural
frequency at which the object will oscillate.

10.7 Elastic Deformation
Because of these atomic-level “springs”, a material tends to
return to its initial shape once forces have been removed.
ATOMS
FORCES

STRETCHING, COMPRESSION, AND YOUNG’S MODULUS
A
L
L
YF
o





 

Young’s modulus has the units of pressure: N/m2

Example 12 Bone Compression
In a circus act, a performer supports the combined weight (1080 N) of
a number of colleagues. Each thighbone of this performer has a length
of 0.55 m and an effective cross sectional area of 7.7×10-4 m2. Determine
the amount that each thighbone compresses under the extra weight.

A
L
L
YF
o





 

  
   m101.4
m107.7mN104.9
m55.0N540 5
2429





YA
FL
L o

SHEAR DEFORMATION AND THE SHEAR MODULUS
A
L
x
SF
o





 

The shear modulus has the units of pressure: N/m2

Example 14 J-E-L-L-O
You push tangentially across the top
surface with a force of 0.45 N. The
top surface moves a distance of 6.0 mm
relative to the bottom surface. What is
the shear modulus of Jell-O?
A
L
x
SF
o





 

xA
FL
S o


  
   
2
32
mN460
m100.6m070.0
m030.0N45.0


 
S

VOLUME DEFORMATION AND THE BULK MODULUS





 

oV
V
BP
The Bulk modulus has the units of pressure: N/m2

10.8 Stress, Strain, and Hooke’s Law
HOOKE’S LAW FOR STRESS AND STRAIN
Stress is directly proportional to strain.
Strain is a unit less quantity.
SI Unit of Stress: N/m2
In general the quantity F/A is called the stress.
The change in the quantity divided by that quantity is called the
strain:
ooo LxLLVV 

10.8 Stress, Strain, and Hooke’s Law

HW
Page 307 no 58, 59, 60, 61, 63 and 81
Page 303 no 2

The change in length of the wire is, L  FL0 /YA
, where the force F is equal to the tension T in the wire. The tension in the wire can be found by
applying Newton's second law to the two crates.
SOLUTION The drawing shows the free-body diagrams for
the two crates. Taking up as the positive direction, Newton's
second law for each of the two crates gives
(1)
(2)
T  m1g  m1a
T  m2g  –m2a
TT
m1g
m2g
Solving Equation (2) for a, we find a  
T  m2g
m2


 

. Substituting into Equation (1) gives
T  2m1g 
m1
m2
T  0
Solving for T we find T 
2m1m2g
m1  m2

2(3.0 kg)(5.0 kg)(9.80 m/s2
)
3.0 kg + 5.0 kg
 37 N
Using the value given in Table 10.1 for Young’s modulus Y of steel, we find,
therefore, that the change in length of the wire is given by Equation 10.17 as
L 
(37 N)(1.5 m)
(2.0 1011
N/m2
)(1.3 10–5
m2
)
 2.1 10–5
m

sin12F f mg m a   
     
       
11 2 5 2 4
0 2 2
2.0 10 N/m 7.8 10 m 2.0 10 m
m
sin12 61 kg 1.1 m/s 68 N 61 kg 9.80 m/s sin12
Y A L
L
m a f mg
 
   
   
     
 
0
Y A L
L
F


To determine F, we examine the following free-body diagram of the skier. For
convenience, the +x direction is taken to be parallel to the slope and to point
upward (see the drawing).
Three forces act on the skier in the x direction: (1) the towing force (magnitude = F),
(2) the frictional force (magnitude = f ) exerted on the skis by the snow, and (3) the
component of the skier’s weight that is parallel to the x axis (magnitude =
W sin12 = mg sin 12). This component is shown to the right of the free-body
The net force acting on the skier has a magnitude of sin12F f mg  
. According to Newton’s second law (see Section 4.3), this net force is equal to the skier’s mass times the
magnitude of her acceleration, or

58. REASONING AND SOLUTION F = S(X/L0)A for the shearing force. The shear modulus S for copper is given in Table 10.2.
From the figure we also see that tan  = (X/L0) so that
  
6
1 1
10 2 2
6.0 10 N
tan tan 0.091
4.2 10 N/m 0.090 m
F
SA
  
 
           
 
59. REASONING AND SOLUTION The shearing stress is equal to the force per
unit area applied to the rivet. Thus, when a shearing stress of 5.0 10
8
Pa is applied
to each rivet, the force experienced by each rivet is
  Nm)(5.0Pa) 42–38
109.310105.0())(Stress()Stress( 2
 rAF
Therefore, the maximum tension T that can be applied to each beam, assuming
that each rivet carries one-fourth of the total load, is 5
4 1.6 10 NF   .
60. REASONING Both cylinders experience the same force F. The magnitude of
this force is related to the change in length of each cylinder according to Equation
10.17: F Y(L/ L0)A. See Table 10.1 for values of Young’s modulus Y. Each
cylinder decreases in length; the total decrease being the sum of the decreases for each
cylinder.
SOLUTION The length of the copper cylinder decreases by
Lcopper
FL0
YA

FL0
Y(r
2
)

(6500 N)(3.0 10–2
m)
(1.110
11
N/m
2
)(0.25 10
–2
m)
2  9.0 10
–5
m
Similarly, the length of the brass decreases by
Lbrass 
(6500 N)(5.0 10–2
m)
(9.01010
N/m2
) (0.2510–2
m)2
1.810–4
m
Therefore, the amount by which the length of the stack decreases is 2.7 10–4
m

61. REASONING AND SOLUTION Equation 10.20 gives the desired result.
Solving for V/V0 and taking the value for the bulk modulus B of aluminum from
Table 10.3, we obtain
V
V0
 
P
B
 
 1.01105
Pa
7.1 10
10
N/m
2  1.4 10
–6
63. REASONING AND SOLUTION From the drawing we have x = 3.0 × 10−3
m
and
A = 2 rx = 2(1.00  10
–2
m)(3.0  10
–3
m)
We now have Stress = F/A. Therefore,
F = (Stress)A = (3.5  10
8
Pa)[21.00  10
–2
m)(3.0  10
–3
m)] = 6.6 104
N

81. REASONING Since the
surface is frictionless, we
can apply the principle of
conservation of mechanical
energy, which indicates
that the total mechanical
energy of the spring/mass
system is the same at the
instant the block contacts the bottle (the final state of the system) and at
the instant shown in the drawing (the initial state). Kinetic energy 21
2
mv
is one part of the total mechanical energy, and depends on the mass m and
the speed v of the block. The dependence of the kinetic energy on speed
is critical to our solution. In order for the block to knock over the bottle,
it must at least reach the bottle. When launched with the minimum speed
v0
shown in the drawing, the block will reach the bottle with a final speed
of vf
= 0 m/s. We will obtain the desired initial speed v0
by solving the
energy-conservation equation for this variable.
0.080 m
v0
x = 0 m
0.050 m

SOLUTION The conservation of mechanical energy states that the final
total mechanical energy Ef
is equal to the initial total mechanical energy
E0
. The expression for the total mechanical energy for a spring/mass
system is given by Equation 10.14, so that we have
2 2 2 2 2 21 1 1 1 1 1
2 2 2 2 2 2f f f f 0 0 0 0
f 0
mv I mgh kx mv I mgh kx
E E
       
Since the block does not rotate, the angular speeds f
and 0
are zero.
Moreover, the block reaches the bottle with a final speed of vf
= 0 m/s
when the block is launched with the minimum initial speed v0
. In
addition, the surface is horizontal, so that the final and initial heights, hf
and h0
, are the same. Thus, the above expression can be simplified as
follows: 2 2 21 1 1
2 2 2f 0 0
kx kx mv 
In this result, we are given no values for the spring constant k and the mass m.
However, we are given a value for the angular frequency . This frequency is given
by Equation 10.11
k
m

 
  
 
, which involves only the ratio k/m. Therefore, in
solving the simplified energy-conservation expression for the speed v0
, we will divide
both sides by m, so that the ratio k/m can be expressed using Equation 10.11.
 
2 22 1 11
2 22 2 20 0f
0 f 0
or
kx mvkx k
v x x
m m m
  
   
 
Substituting
k
m
  from Equation 10.11, we find
       
2 22 2
0 f 0
7.0 rad/s 0.080 m 0.050 m 0.44 m/sv x x    

Can we apply work energy theorem?
Work done by non-conservative forces
is equal to the change in the
energy of the system
𝑊𝑁𝐶 = 𝐸𝑓 − 𝐸𝑖
𝑊ℎ𝑎𝑡 𝑖𝑠 𝑜𝑢𝑟 𝐸?

Simple harmonic motion and elasticity

Simple harmonic motion and elasticity

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