Simple Harmonic Motion
• Simple harmonic motion (SHM) refers to a
certain kind of oscillatory, or wave-like motion
that describes the behavior of many physical
phenomena:
–
–
–
–
–
–
–
a pendulum
a bob attached to a spring
low amplitude waves in air (sound), water, the ground
the electromagnetic field of laser light
vibration of a plucked guitar string
the electric current of most AC power supplies
…
SHM Position, Velocity, and Acceleration
Simple Harmonic Motion
Periodic Motion: any motion of system
which repeats itself at regular, equal
intervals of time.
Simple Harmonic Motion
Simple Harmonic Motion
• Equilibrium: the position at which no net force acts on
the particle.
• Displacement: The distance of the particle from its
equilibrium position. Usually denoted as x(t) with x=0 as
the equilibrium position.
• Amplitude: the maximum value of the displacement with
out regard to sign. Denoted as xmax or A.
The period and frequency of a
wave
•
•
the period T of a wave is the amount of time it takes to
go through 1 cycle
the frequency f is the number of cycles per second
– the unit of a cycle-per-second is commonly referred to as
a hertz (Hz), after Heinrich Hertz (1847-1894), who
discovered radio waves.
•
frequency and period are related as follows:
1
f 
T
•
t
Since a cycle is 2p radians, the relationship between
frequency and angular frequency is:
  2pf
T
• http://www.physics.uoguelph.ca/tutorials/s
hm/Q.shm.html
Here is a ball moving back and forth with simple
harmonic motion (SHM):
Its position x as a function of time t is:
where A is the amplitude of motion : the distance from the
centre of motion to either extreme
T is the period of motion: the time for one complete cycle
of the motion.
Restoring Force
How does the restoring force act with respect to the displacement from the
equilibrium position?
F is proportional to -x
Simple harmonic motion is the motion executed by a particle of mass m subject to
a force that is proportional to the displacement of the particle but opposite in
sign.
Springs and SHM
•
Attach an object of mass m to the end of a spring, pull it out to a
distance A, and let it go from rest. The object will then undergo simple
harmonic motion:
x(t )  A cos(t )
v(t )   A sin(t )
a(t )   A2 cos(t )
•
What is the angular frequency in this case?
– Use Newton’s 2nd law, together with Hooke’s law, and the
above description of the acceleration to find:
k

m
Springs and Simple Harmonic Motion
Equations of Motion
Conservation of Energy allows a calculation of the velocity
of the object at any position in its motion…
Conservation of Energy For A Spring in Horizontal Motion
E =
E =
Kinetic + Elastic Potential
½ mv2 +
½ kx2
=
Constant
• At maximum displacement, velocity is zero and all energy is
elastic potential,
so total energy is equal to
½ kA2
• To find the velocity @ any displacement… do
conservation of Energy… @ some point at max
displacement
• ½ mv2 + ½ kx2 = 1/2kA2
• Solving for v
k
2
2
v 
A x
m


• This is wonderful but what about time??!
v=dx/dt and separate variables!
dx

dt
k
A2  x 2 
m
dx
 A sin 
dt
k
t c
m

k
2
2
v 
A x
m


SHM Solution...
• Drawing of A cos(t )
• A = amplitude of oscillation
T = 2p/
A
2p
p
A
p
2p

SHM Solution...
• Drawing of A cos(t + )

2p
p
p
2p

SHM Solution...
• Drawing of A cos(t - p/2)
=
p/2
A
2p
p
p
= A sin(t)!
2p

1989.M3 3.0 m/s, 0.63 s,
0.098 m, 0.41 m, 0.31 m
1989M3. A 2-kilogram block is dropped from a height of 0.45 m above an
uncompressed spring, as shown above. The spring has an elastic
constant of 200 N per meter and negligible mass. The block strikes the
end of the spring and sticks to it.
a. Determine the speed of the block at the instant it hits the end of the
spring. 3 m/s
b. Determine the period of the simple harmonic motion that ensues. 0.63s
c. Determine the distance that the spring is compressed at the instant the
speed of the block is maximum. 0.098 m
d. Determine the maximum compression of the spring. 0.41 m
e. Determine the amplitude of the simple harmonic motion. 0.31 m
SHM So Far
• The most general solution is x = A cos(t + )
where A = amplitude
 = angular frequency
 = phase
• For a mass on a spring
k

m
– The frequency does not depend on the amplitude!!!
– We will see that this is true of all simple harmonic
motion!
Pendulums
• When we were discussing the energy in a
simple harmonic system, we talked about the
‘springiness’ of the system as storing the
potential energy
• But when we talk about a regular pendulum
there is nothing ‘springy’ – so where is the
potential energy stored?
The Simple Pendulum
• As we have already seen,
the potential energy in a
simple pendulum is stored in
raising the bob up against
the gravitational force
• The pendulum bob is clearly
oscillating as it moves back
and forth – but is it exhibiting
SHM?
• We can see that the
restoring force is:
F = -mg sinθ
If we assume that the
angle θ is small, then
sinθ ~ θ
and our equation
becomes
F ~ -mgθ = - mgs/L
= - (mg/L)s
F ~ -mgθ = - mgs/L
= - (mg/L)s
dividing both sides by m,
a = -(g/L)s.
Equate to a = -ω 2 x, and
ω = (g/L)1/2
The Physical Pendulum
• Now suppose that the
mass is not all
concentrated in the
bob?
• In this case the
equations are exactly
the same, but the
restoring force acts
through the center of
mass of the body (C in
the diagram) which is
a distance h from the
pivot point
• Going back to our
definition of torque,
we can see that
the restoring force
is producing a
torque around the
pivot point of:
   LFg sin 
•where L is the moment
arm of the applied
force
The Simple Pendulum
If we substitute τ = Iα, we get:
 LFg sin   I
• This doesn’t appear too promising until we make the
following assumption –
•
that θ is small…
• If θ is small we can use the approximation that
sin θ  θ
• (as long as we remember to express θ in radians)
The Physical Pendulum
• Making the substitution we then get:
 LFg   I
which we can then rearrange to get:
mgL
 

I
which is the angular equivalent to
a = -ω 2 x
•So, we can reasonably say that the motion of a
pendulum is approximately SHM if the maximum
angular amplitude is small
The Physical Pendulum
• Making the substitution we then get:
 LFg   I
which we can then rearrange to get:
mgL
 

I
which is the angular equivalent to
a = -ω 2 x
•So, we can reasonably say that the motion of a
pendulum is approximately SHM if the maximum
angular amplitude is small
The Physical Pendulum
• So we go back to
our previous
equation for the
period and
replace L with h to
get:
I
T  2p
mgh
The Simple Pendulum
The period of a pendulum is given by:
T  2p
I
mgL
where I is the moment of inertia of the pendulum
•If all of the mass of the pendulum is concentrated in the
bob, then I = mL2 and we get:
T  2p
L
g
An Angular
Simple Harmonic Oscillator
• The figure shows an
angular version of a
simple harmonic oscillator
• In this case the mass
rotates around it’s center
point and twists the
suspending wire
• This is called a torsional
pendulum with torsion
referring to the twisting
motion
Torsional Oscillator
• If the disk is rotated through an angle (in either
direction) of θ, the restoring torque is given by the
equation:
  
Comparing with
F  kx
Which gives T  2p
I

Where  has replaced k and I has replaced m
Sample Problem
A uniform bar with mass m lies symmetrically across two
rapidly rotating, fixed rollers, A and B, with distance L = 2.0
cm between the bar’s center of mass and each roller. The
rollers slip against the bar with coefficient of friction µk = 0.40.
Suppose the bar is displaced horizontally by a distance x and
then released. What is the angular frequency of the resulting
horizontal motion of the bar?
ω = 14 rad/s
Sample Problem 1
• A 1 meter stick swings about a
pivot point at one end at a
distance h from its center of
mass
• What is the period of oscillation?
A penguin dives from a uniform board that is hinged
at the left and attached to a spring at the right. The
board has length L = 2.0 m and mass m = 12 kg; the
spring constant k = 1300 N/m. When the penguin
dives, it leaves the board and spring oscillating with a
small amplitude.
Assume that the board is stiff enough not to bend,
and find the period T of the oscillations.
T = 0.35 s