Chapter 10
Simple Harmonic Motion and Elasticity
Goals for Chapter 10
• To follow periodic motion to a study of simple
harmonic motion.
• To solve equations of simple harmonic motion.
• To use the pendulum as a prototypical system
undergoing simple harmonic motion.
• To study how oscillations may be damped or driven.
• To study stress, strain and elastic deformation.
• To define elasticity and plasticity.
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The Ideal Spring and Simple Harmonic Motion
Periodic Motion
Periodic motion or oscillations: An object repeats the same
motion over and over; e.g. rising of the sun, rotation of a bicycle
wheel, mass on a spring, ...
FxApplied = k x
spring constant
Units: N/m
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Example: A Tire Pressure Gauge
The spring constant of the spring
is 320 N/m and the bar indicator
extends 2.0 cm. What force does the
air in the tire apply to the spring?
Hooke’s Law: Restoring Force of an Ideal Spring
The restoring force of an ideal spring is
Fx = −k x
FxApplied = k x
= (320 N m )(0.020 m )
= 6 .4 N
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Simple Harmonic Motion and Uniform Circular Motion
Connection between Rotational and Oscillatory 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
y = R sin φ
φ
t
Y motion
X motion
x = R cos φ
t
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Simple Harmonic Motion and the Reference Circle
Simple Harmonic Motion and the Reference Circle
DISPLACEMENT
x(t ) = A cos θ = A cos ωt
θ = ωt
amplitude
li d A:
A the
th maximum
i
di
displacement
l
t
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
f=
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1
T
ω = 2πf =
2π
T
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Amplitude
Period and Frequency
• Amplitude, A
• The period, T, is the time that it takes for the
object to complete one complete cycle of motion
– The amplitude is the maximum position of the object
relative to the equilibrium position
– In the absence of friction, an object in simple
harmonic motion will oscillate between ±A on each
side of the equilibrium position
– From x = A to x = - A and back to x = A
• The frequency, ƒ, is the number of complete
cycles or vibrations per unit time
f=
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1
T
Velocity
Example
vT
θ
Vx
(A = 0.4mm)
(b) period,
(T = 2 ms)
(c) frequency
(f = 1/T = 1/(2ms) = 500 Hz)
v x = − v T sin θ
= −A ω sin ωt
θ = ωt
= −A
ω sin ωt
{
The graph shown in the figure closely approximates the displacement x of a
tuning fork as a function of time t as it is playing a single note. What are
(a) the amplitude,
v T = ωR
= ωA = 2πfA
v max
v x , max = ω A
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Example: Maximum Speed of a Loudspeaker Diaphragm
Acceleration
The frequency of motion is 1.0
KHz and the amplitude is 0.20
mm.
• What is the maximum speed
of the diaphragm?
• Where in the motion does
this maximum speed occur?
θ
ac
θ = ωt
v x = − v T sin θ = −A
ω sin ωt
{
)
(
v2
cos ωt
R
= −Rω2 cos ωt
=−
a x = −ω 2 A cos ωt
= −ω 2 ⋅ x
v max
v max = Aω = A(2πf )
(
a x = −a c cos θ
ax
)
= 0.20 ×10 −3 m (2π) 1.0 ×103 Hz = 1.3 m s
a x ,max = ω2 A
The maximum speed occurs midway between the ends of its motion.
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Example: Maximum Acceleration of a Loudspeaker Diaphragm
The frequency of motion is 1.0
KHz and the amplitude is 0.20
mm.
• What is the maximum
acceleration of the diaphragm?
a x = −a c cos θ = − A
ω2 cos ωt
{
a max
a max = Aω = A(2πf )
2
2
(
)
(
= 0.20 × 10 −3 m (2π ) 1.0 × 103 Hz
= 7.9 ×103 m s 2
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2
huge !
)
2
Simple Harmonic Motion (SHM)
For SHM: Restoring force is directly proportional to the
displacement from equilibrium
Hooke’s law force
• The force always acts toward the
equilibrium position
– It is called the restoring force
• The direction of the restoring
force is such that the object is
being either pushed or pulled
toward the equilibrium position
Spring - mass system
example for Hooke’s law
and simple harmonic motion
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Frequency of Vibration
x = A cos ωt
Oscillation frequency of a mass spring system
a x = −Aω cos ωt
2
∑ F = −kx = ma
• The force is given by Hooke’s Law
• F = - k x = m ax
ax = - k x / m
• acceleration for simple harmonic motion a x = − ω x
k
k angular
⇒ − x = − ω2 x ⇒ ω = 2π
2π ƒ =
m
m frequency
2
x
− kA = −mAω2
• Frequency
– Units are cycles/second or Hertz, Hz
ω=
• Period
– This gives the time required for an object
of mass m attached to a spring of constant
k to complete one cycle of its motion
k
m
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ƒ=
1
1 k
=
T 2π m
T = 2π
m
k
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Example: A Body Mass Measurement Device
Given k and T, find mastro
ω=
k
m total
m total =
m total = k ω2
k
= m chair + m astro
(2π T )2
m astro =
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.
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=
ω = 2 πf =
2π
T
k
−m
(2π T )2 chair
(606 N m )(2.41 s )2 − 12.0 kg = 77.2 kg
4π 2
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Summary: Simple Harmonic Oscillator (t)
Initial Position and Phase Shift:
x = A cos ωt
v x = − A ω sin ωt
The most general solution is x = A cos(ωt + φ)
p
where A = amplitude
ω = frequency
φ = phase
a x = −A ω2 cos ωt
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Energy and Simple Harmonic Motion
Example
A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm.
At t = 0 its speed is maximum, and is vmax = +2 m/s.
What is the angular frequency of oscillation ω?
What is the spring constant k ?
k
A compressed spring can do work.
Welastic = (F cos θ)s
=
m
1
2
(kx o + kx f ) cos 0o (x o − x f )
x
vMAX = ωA
Also: ω =
ω=
vMAX 2 m s
=
= 20s −1
A 10cm
k
m
Welastic = 12 kx o2 − 12 kx f2
k = mω2
So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m
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Energy and Simple Harmonic Motion
Example: Landing a jet on an aircraft carrier
A Navy jet of mass 10,000 kg lands on
an aircraft carrier and snags a cable
to slow it down. The cable is attached
to a spring with spring constant
40,000 N/m If the spring stretches 25
m to stop the plane, what was the
landing speed of the plane?
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
Ki + U i = K f + U f
PE elastic = 12 kx 2
assume no change in
gravitational
potential energy
SI Unit of Elastic Potential Energy: joule (J)
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1
1
k
2
2
mvi = kx 2 , vi = x 2
2
2
m
40
,
000
k
v=
x=
⋅ 25
m
10,000
= 50 m / s.
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Conceptual Example 2
Are Shorter Spring stiffer Spring?
A 10-coil spring has a spring constant k. If the spring is
cut in half, so there are two 5-coil springs, what is the
spring
constant of each of the smaller springs?
Final covers upp to page
p g 296.
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