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Ch 10. Harmonic Motion & Elasticity
Example 1 A Tire Pressure Gauge
Ideal Spring & Simple Harmonic Motion
The spring constant is 320 N/m and the bar
indicator extends 2.0 cm. What force does the
air in the tire apply on the spring?
HOOKE’S LAW:
Fx  k x
The restoring force on an ideal spring
k = spring constant
Units: N/m
FxApplied  k x
 320 N m 0.020 m   6.4 N
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10.2 Simple Harmonic Motion and Reference Circle
x  A cos  A cos t
amplitude A: maximum displacement
period T: time to complete one cycle
frequency f: number of cycles per second (Hz)
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f 
T
2
  2 f 
T
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SHM & the Reference Circle
VELOCITY
Ex. 3 Maximum Speed of Loudspeaker Diaphragm
Frequency of motion is 1.0 KHz, amplitude is 0.20 mm.
(a) What is the maximum speed of the diaphragm?
(b) Where in the motion does this max speed occur?
v x  vT sin    
A sin t
vmax
(a)
vmax  A  A2 f   0.20 103 m 2 1.0 103 Hz 
 1.3 m s
v x  vT sin    
A sin t
vmax
(b) The maximum speed occurs midway between the
ends of its motion.
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Simple Harmonic Motion
ACCELERATION FREQUENCY OF VIBRATION
x  A cos  t
a x   A 2 cos  t
 F  kx  ma
x
 kA  mA 2

k
m
ax  ac cos   
A 2 cos t
amax
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Simple Harmonic Motion
Example 6 A Body Mass Measurement
Device
k

mtotal
The device below consists of a springmounted 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.
mtotal  k  2
2
  2 f 
T
mtotal 
k
 mchair  mastro
2
2 T 
k
mastro 
 mchair
2
2 T 
2

606 N m 2.41 s 

 12.0 kg  77.2 kg
4 2
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10.3 Energy and Simple Harmonic Motion
A compressed spring can do work.
Welastic  F cos s  12 kxo  kxf cos 0 xo  x f 
Welastic  12 kxo2  12 kx2f
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
PEelastic  12 kx2
Units: joule (J)
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Energy and Simple Harmonic Motion
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?
E f  Eo
1
2
1
2
mv2f  12 I 2f  mghf  12 ky2f  12 mvo2  12 Io2  mgho  12 kyo2
kho2  mgho
2mg
ho 
k
20.20 kg  9.8 m s 2

 0.14 m
28 N m


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10.4 The Pendulum
A simple pendulum consists
of a particle attached to a
frictionless pivot by a cable of
negligible mass.

g
L

mgL
I
(small angles only)
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 f 

T
g
L
T 2g
L
4 2
(small angles only)


T 2 g 1.00 s  9.80 m s 2
L

 0.248 m
2
2
4
4
2
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Damped Harmonic Motion
In simple harmonic motion, an object
oscillated with a constant amplitude.
In reality, friction is always present
and amplitude decreases with time.
This is damped harmonic motion.
1) simple harmonic motion
2&3) underdamped
4) critically damped
5) overdamped
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Driven Harmonic Motion & 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.
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.
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Elastic Deformation
Because of these atomic-level “springs”,
a material tends to return to its initial
shape once forces have been removed.
STRETCHING, COMPRESSION, &
YOUNG’S MODULUS
 L 
 A
F  Y 
 Lo 
Young’s modulus has units of pressure: N/m2
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Elastic Deformation
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.
 L 
 A
F  Y 
 Lo 
FLo
L 

YA

540 N 0.55 m 
9.4 10
9

N m 2 7.7 10  4 m 2

 4.1105 m
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Elastic Deformation
SHEAR DEFORMATION & SHEAR MODULUS
 x 
F  S   A
 Lo 
The shear modulus has the units of
pressure: N/m2
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Elastic Deformation
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?
 x 
F  S   A
 Lo 
FLo
S
Ax

0.45 N 0.030 m 
S
 460 N
2
3
0.070 m  6.0 10 m 
m2
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Elastic Deformation
VOLUME DEFORMATION & BULK MODULUS
 V 

P   B
 Vo 
The Bulk modulus has the units of
pressure: N/m2
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10.8 Stress, Strain, and Hooke’s Law
In general the quantity F/A is called the
stress.
The change in the quantity divided by that
quantity is called the
strain:
V Vo L Lo x Lo
HOOKE’S LAW FOR STRESS AND
STRAIN
Stress is directly proportional to
strain.
Strain is a unitless quantitiy.
SI Unit of Stress: N/m2
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