Physics 102
Waves
Lecture 1
The Harmonic Oscillator
Moza m. Al-Rabban
Associate Professor of Physics
mmr@qu.edu.qa
Useful links
http://www.physics.uoguelph.ca/tutorials/shm/Q.shm.html
Physics 102 - Lecture 1
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Periodic Motion
A motion that repeat itself over and over is referred to as periodic motion.
Examples:
• The beating of your heart,
•The ticking of a clock,
•The movement of a child on a swing….
The time required for the completion of one cycle of a periodic
system’s repetitive motion called the period, T, of that system.
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Period, T
• Definition of Period:
T = time required for one cycle of a periodic motion.
SI unit: second/cycle = s
Note that a cycle (that is, an oscillation) is dimensionless.
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Frequency, f
• The frequency of an oscillation is the number of
oscillations per unit time.
• It tells us how frequently, or rapidly, an oscillation
takes place _ the higher frequency, the more rapid
the oscillations.
• Definition of Frequency, f
1
f 
T
• SI unit: cycle/second = 1/s = s 1
• 1 Hz = 1 cycle/second
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Example 1: If the processing speed of a personal computer
is 1.80 GHz, how much time is required for one processing
cycle?
T
1
1
1
10



5
.
56

10
s
9
f 1.80GHz 1.8010 cycle / s
The high frequency of the computer corresponds to a very small
period to complete one operation.
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Example 2: A tennis ball is hit back and forth between two
players warming up for a match. If it takes 2.31 s for the
ball to go from one player to the other, what are the period
and frequency of the ball’s motion?
The period of this motion is the time for the ball to complete one
round trip. Therefore,
T  22.31s   4.62s
1
1
f  
 0.216 Hz
T 4.62s
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Example: Frequency and Period
of a Radio Station
What is the oscillation period for the broadcast of a 100 MHz FM
radio station?
1
1
-8
T 

1.0

10
s  10 ns
6
f 1.0 10 Hz
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Simple Harmonic Motion
An oscillator is an object or system of objects that undergoes periodic
oscillatory motion or behavior. Example: a mass and spring system.
Characteristics:
1. Oscillatory motion is about some (energy=0) equilibrium position;
2. Oscillatory motion is periodic, with a definite period or cycle time.
f 
1
T
or
T
1
f
1 hertz  1 Hz  1 cycle per second  1 s1
Physics 102 - Lecture 1
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SHM Prototype Experiment
Consider Fig. (a). An
air-track glider attached
to a spring. The glider is
pulled a distance A from its
rest position and released.
Fig. (b) shows a graph of
the motion of the glider, as
measured each 1/20 of a
second.
The object’s maximum
displacement from
equilibrium is called the
amplitude A of the motion.
The object’s position
oscillates between x=-A and
x=+A.
Physics 102 - Lecture 1
The graph shows the
position of the glider from
the same measurements.
We see that A=0.17 m and
T=1.60 s. Therefore the
oscillation frequency of
the system is f = 0.625 Hz
The Figure down is a velocity-versus-time graph.
The velocity graph is also sinusoidal, oscillating
between –vmax ( maximum speed to the left) and +vmax
(maximum speed to the right). As the figure shows,
 The instantaneous velocity is zero at the points
where x = ±A. These are the turning points in the
motion.
 The maximum speed vmax is reached as the object
passes through the equilibrium position at x = 0 m.
The velocity is positive as the object moves to the
right but negative as it moves to the left.
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We can ask three important questions about
this oscillating system:
1. How is the maximum speed vmax related to the
amplitude A?
2. How are the period and frequency related to
the object’s mass m, the spring constant k,
and the amplitude A?
3. Is the sinusoidal oscillation a consequence of
Newton’s laws?
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Kinematics of SHM
x(t )  A cos
f 
2 t
 A cos 2 ft  A cos t
T
1 

(frequency units: cycles/s  Hz)
T 2
  2 f 
v(t ) 
2
(angular frequency units: rad/s  s-1 )
T
dx
2 A
2 t

sin
 2 fA sin 2 ft   A sin t
dt
T
T
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v(t ) 
dx
2 A
2 t

sin
 2 fA sin 2 ft   A sin t
dt
T
T
2A
v max 
 2fA  A
It’s
T especially important to
remember to set your calculator to
radian mode before working
oscillation problems and use these
equations.
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Example: A System in SHM
An air-track glider is attached to a spring,
pulled 20 cm to the right, and released
at t=0. It makes 15 oscillations in 10 s.
(a) What is the period T of oscillation?
(b) What is the object’s maximum speed?
(c) What is the position and velocity at t=80 s?
1
(15 oscillations)
T

 0.667 s
f 
 1.5 Hz
f
(10 s)
2 A 2 (0.20 m)
vmax 

 1.88 m
T
(0.667 s)
2 t
2 (0.80 s)
 (0.20 m) cos
 0.062 m
T
(0.667 s)
2 t
2 (0.80 s)
v(t )  vmax sin
 (1.88 m)sin
 1.79 m/s
T
(0.667 s)
x(t )  A cos
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Example: Finding the Time
A mass on a spring oscillating in simple
harmonic motion (SHM) starts at x=A at t=0
and has period T.
At what time, as a fraction of T, does the
object first pass through x = ½A?
x
A
2 t
 A cos
2
T
T
T  T
t
cos 1  12  

2
2 3 6
2
√3
600 =/3
1
Physics 102 - Lecture 1
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Clicker Question 1
An object moves with simple harmonic motion.
If the amplitude and period of the oscillation are both
doubled, the object’s maximum speed is:
(a)
(b)
(c)
(d)
(e)
quadrupled;
doubled;
unchanged
halved;
quartered.
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SHM and Circular Motion
Uniform circular
motion projected into
one dimension is simple
harmonic motion (SHM).
Consider a particle
rotating ccw, with the
angle f increasing
linearly with time:
x  A cos f

df
, so f  t if f  0 at t  0.
dt
x(t )  A cos t
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Note » when used to describe oscillatory motion,  is called
the angular frequency rather than the angular velocity.
Note » Hertz is specifically “ cycle per second” or “ oscillations
per second” . It is used for ƒ but not for  .
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The Phase Constant
But what if f is not zero at t = 0 ?
f  t  f0
x(t )  A cos t  f0 
v(t )   A sin t  f0 
 vmax sin t  f0 
Set t  0:
x0  x(0)  A cos f0
v0 x  v(0)   A sin f0
The position xo and velocity vox at t = 0 are the
initial conditions. Different values of the phase
constant correspond to different starting
points on circle and thus different initial
conditions.
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Phases and Oscillations
Here are three examples of
differing initial conditions:
1
2
3
1. f0  /3, implying x0 = A/2
and moving to the left
(v<0);
2. f0  /3, implying x0 = A/2
and moving to the right
(v>0);
3. f0  , implying x0 = A and
momentarily at rest (v=0).
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Example: Using Initial
Conditions
A mass on a spring oscillates with a period of
0.80 s and an amplitude of 10 cm. At t=0, it is 5.0
cm to the left of equilibrium and moving to the left.
What is the position and direction of motion at
t=2.0 s?
2
T  0.80 s, so  
 7.854 rad/s
A  10cm, and x  5cm
T
2
x 
x0  A cos f0 so f0  cos 1  0   cos 1 -12   
rad
A
3
 
v0 x  0, so
f0 
2
rad
3
x(2.0 s)  A cos t  f0 
2


 (10 cm) cos (7.854 rad/s)(2.0 s)  (
rad) 
3


 (10 cm) cos(17.80 rad)  5.00 cm
v(2.0 s)   A sin t  f0 
Physics 102 - Lecture 1
 (7.854 rad/s)(10 cm)sin(17.80 rad)
 68.1 cm/s (i.e., moving to the right)
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Clicker Question 2
These figures show four
oscillator systems at t=0.
Which one has the phase
constant f0  /4 rad?
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The Energy of SHM
U  12 kx2
K  12 mv2
E  K U
 12 mv2  12 kx2
At x   A, E  U  12 kA2
E  constant, so
vmax  A

k
;
m
1
2
At x  0, E  K  12 mvmax 2
kA2  12 mvmax 2
k
k
  A, so  
m
m
f 
Physics 102 - Lecture 1
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2
k
m
; T  2
m
k
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Velocity and Amplitude of SHM
E  12 mv 2  12 kx 2
 12 kA2  12 mvmax 2
v
k 2
A  x 2    A2  x2

m
mv02
v 
2
A  x0 
 x02   0 
k
 
Physics 102 - Lecture 1
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Example:
Using Conservation of Energy
A 500 g block on a spring is pulled a distance of
20 cm and released. The subsequent oscillation is
measured to have a period of 0.80 s.
At what position(s) is the block’s speed 1.0 m/s?
T  0.80 s so   2 / T  7.85 rad/s
v  A  x
2
2
x  A2   v /  
2
 (0.20 m)   (1.0 m/s) /(7.85 rad/s) 
2
2
 0.154 m
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Clicker Question 3
These figures show four
springs that have been
compressed from their
equilibrium positions at x=0.
When released, they will
begin to oscillate.
Which one will have the
largest maximum speed of
oscillation?
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The Dynamics of SHM
x(t )  A cos t  f0 
d2
ax (t )  2  A cos t  f0  
dt
  2 A cos t  f0 
  x(t )
2
The acceleration is proportional
to the negative of the displacement
at any time.
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The Equation of Motion in SHM
( Fsp ) x  k x
(Fnet ) x  ( Fsp ) x  kx  max
k
ax   x
m
d 2x k
 x0
2
dt
m
This is the equation of motion for the system. It is a homogeneous
linear 2nd order differential equation.
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Solving the Equation of Motion
d 2x k
 x0
2
dt
m
Q: How do you solve this differential equation?
A: Our method - guess the solution and see if it works.
x  A cos t  f0 
 2 A cos t  f0  
d
d
x   A cos t  f0     A sin t  f0 
dt
dt
d2
d
2
x



A
sin

t

f



A cos t  f0 




0
2


dt
dt
Physics 102 - Lecture 1
2 
k
A cos t  f0   0
m
k
k
or  
m
m
It works!
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Example:
Analyzing an Oscillator
At t=0, a 500 g block oscillating on a
spring is observed to be moving to the right
at x=15 cm. At t=0.30 s, it reaches its
maximum displacement of 25 cm.
(a) Draw a graph of the motion for one cycle.
(b) At what time in the first cycle is x=20 cm?
x0  15 cm, A  25 cm
f0  cos1  x0 / A  cos1  0.60  0.927 rad
At t  0.30 s, cos t  f0   1
Therefore,   f0 / t  (0.927 rad)/0.30 s  3.09 rad/s
T  2 /   2.03 s
t  cos 1 ( x / A)  f0  /   0.508 s
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Clicker Question 4
The position vs. time of a
mass-and-spring oscillator is
shown.
What can you say about the
velocity and force at the time
indicated by the dashed line?
(a) Velocity is positive; force is to the right.
(b) Velocity is zero; force is to the right.
(c) Velocity is negative; force is to the right.
(d) Velocity is positive; force is to the left.
(e) Velocity is zero; force is to the left.
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End of Lecture 1
 Before the next lecture, read Knight,
Chapters 14.6 through 14.8