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Simple Harmonic
Motion (SHM)
Harmonic motion is a type of motion in which an object
moves along a repeating path over and over again.
Examples: pendulum motion of a child on a swing; the
motion of the Earth around the Sun.
Force and energy are common to all forms of harmonic
motion.
The figure shows a person hanging from a spring acting
like a bungee cord.
The spring is the dominant force acting on the person;
gravity also acts on the person.
The figure below shows snapshots of the person as he
oscillates up and down due to the force of the spring.
The position of the person on the spring can be
measured as a location on the vertical y-axis.
The figure below shows a plot of y as a function of time
where each point on the y-axis matches the physical
location of the person on the spring.
The person’s position varies in a repeating manner as he
moves up and down.
This is oscillatory motion.
In the absence of friction and air resistance, the
repeating motion would continue forever.
Period T is the repeat time for the oscillatory motion; the
time it takes for the man on the spring to travel through
one complete cycle of the motion.
Period can be measured from equilibrium point (1) to
equilibrium point (5).
Period can be measured from highest point (2) to highest point
(6).
Period can be measured between any two equal points on the
y vs. t curve.
Frequency f is the number of oscillations that occur in
one unit of time.
Frequency f is the number of cycles completed in one second.
Since period is the number of seconds to perform one
cycle, frequency and period are reciprocals of one
another:
1
1
f 
and T 
T
f
Units: frequency f is in cycles/second, otherwise known
as a hertz (Hz).
Simple Harmonic Motion
The key property of an oscillator is that its motion is
periodic or repeating.
Oscillatory motion is described as a sine wave:
y = A·sin(2·π·f·t)
For the man on the spring, y is the location along the
vertical axis and the equation y = A·sin(2·π·f·t) tells us
how y changes with time.
Systems that oscillate in a sinusoidal manner are called
simple harmonic oscillators.
The frequency f tells us the rate at which the oscillator
repeats its motion.
The range of the oscillating motion depends on the
quantity A, the amplitude of the motion.
The man on the spring moves back and forth between
the maximum and minimum values of y = ±A.
The velocity of the man on the spring also changes with
time and oscillates between positive and negative values
with the same frequency f as the position y.
The maximum values of the velocity do not happen when
y has its maximum values:
the largest values of v occur when y =0;
The velocity v = 0 when y = ±A
For the figure:
One cycle (a period) is
from (1) to (5) or from
(2) to (6).
The maximum positive
value of y (maximum
amplitude) occurs at (2)
and (6); the minimum
value of y (also the
maximum amplitude)
occurs at (4).
The velocity v = 0 at
these points.
The maximum velocity v
occurs at (1) and (3) and
(5).
The value of y = 0 at
these points.
Connection Between SHM and Circular Motion
A CD spinning at a constant rate is an example of periodic
motion.
There is a close relationship between uniform circular
motion and SHM if you follow a point on the edge of the
CD.
A particle at the edge of the
CD moves at a constant speed
vc as it moves around a circle
of radius A.
The particle’s position is
marked by a reference line
drawn on the CD.
If  is the angle between the reference line and the xaxis, then  increases with time according to  = ω·t,
where ω is the angular velocity of the rotational motion.
As the particle completes one full trip around the circle,
varies from 0 rad to 2·π rad; this is one full period for the
harmonic motion.
  2     t   T
Solving for T:
T 
2

Harmonic motion can be
described in terms of angular
velocity ω or a frequency f:

f 
2
The frequency f is the number of cycles a system
completes in 1 second.
For a rotating object, the system completes an angular
displacement of ω each second, so ω is called the
angular frequency.
SHM: Position as a Function of Time
Let’s examine the circular
motion and SHM in terms of
the particles coordinates in
the x-y plane.
The y-component of the
particle’s position is y = A·sin 
Since  = ω·t:
y  A  sin  A  sin   t 
Using ω = 2·π· f:
y  A  sin 2    f  t 
The x-component of the particle:
x  A  cos   A  cos   t 
x  A  cos  2    f  t 
Both equations describe SHM
with the same period T  1
f
The only difference between the two functions is a
“shift” along the time axis.
The sine function begins at y = 0 when t = 0.
The cosine function begins at y = +A when t = 0.
SHM: Velocity as a Function of Time
While vc is constant, the y-component of the velocity vy is
not constant.
v y  vc  cos   vc  cos   t 
v y  vc  cos  2    f  t 
The particle travels
once around the circle a
distance of 2·π·A in a
time period equal to
one period, so:
2  A
vc 
 2  f  A
T
Substituting this into
the vy equation:
vy  2    f  A  cos 2    f  t 
Vy is the velocity of the
simple harmonic
oscillator and is just
called v.
The position and velocity of a simple harmonic oscillator
is:
y  A  sin  2    f  t  v  2    f  A  cos 2    f  t 
We can also use the x-component:
x  A  cos  2    f  t 
vc  2    f  A
v x  vc  sin 
v x  2    f  A  sin  2    f  t 
The only difference between the x and y components of
the simple harmonic oscillator is in how the oscillator is
initially set into motion.
Conditions for SHM
Any system for which the acceleration varies with the negative of
the displacement will exhibit SHM (College Board loves this!). The
coefficient between a and x defines the square of the angular
frequency 𝜔 2.
a( x)    x
2
x(t )  xm  cos(  t   )
Descriptive features of SHM
Although the causes of SHM will vary from one system to another, the
sinusoidal variation is a common element. All solutions are directly
characterized by three features:
xm : maximum displacement amplitude (or amplitude)
 : angular frequency
: phase constant (or phase angle)
ϕ and ω can alternately be specified by either of the following:
f : frequency, f =  /2
measured in Hertz (1 Hz = 1 s-1)
T : period,
T = 1/f = 2/
Example of SHM: Mass on a Spring
A mass m is attached to a spring with the opposite end of
the spring attached to the wall; the block is free to move
horizontally on a frictionless surface.
Apply Newton’s second law to the mass.
In the vertical direction, the forces acting on the block are
gravity and the normal force exerted by the table.
These two forces are equal in magnitude and opposite in
direction and cancel (and are not shown in the figure).
When stressed, there is a horizontal force exerted by the
spring given by Hooke’s law: Fspring =  k·x
where x is the amount the spring is stretched or compressed
from its relaxed (equilibrium) position.
The spring constant k is a measure of the spring’s strength.
The equilibrium position of the mass in the figure is at
x = 0; when the mass is at that position, the spring is in
its relaxed state and the force exerted by the spring on
the mass is zero.
When the mass is displaced to the right so that x > 0, the
force exerted by the spring on the mass will be in the
negative direction.
If the mass is displaced to the left so x < 0, the spring
force will be to the right.
Fspring is called the restoring force because it always
opposes the displacement away from the equilibrium
position.
Whenever there is a restoring force described by
Fspring =  k·x, the system will exhibit simple harmonic
motion. College Board definition of SHM.
Applying F = m·a, where F is the total force in the
horizontal direction parallel to x: F = Fspring =  k·x = m·a
The frequency f for the mass1
k
spring system is:
f 
2

m
The frequency f increases as the spring constant k is
increased and gets smaller as the mass m is increased.
Frequency f does not depend on the amplitude A.
Linear Restoring Forces and SHM Summary
A linear restoring force tends to push a system back toward a
point of stable equilibrium, with a magnitude that varies linearly
with the displacement away from equilibrium. An example is
Hooke’s law for an ideal spring
F  kx
Applying Newton’s second law gives a second-order ordinary
differential equation
2
d x
k
 a   x
2
dt
m

the solution of which is a sinusoidal variation of position in time
x(t )  xm  cos(  t   )
k
( 
)
m
Any system with displacement following this form is said to be
undergoing simple harmonic motion (SHM).

x(t )  A  cos(  t  )
3
k 2 


m
T
v = dx/dt
v=0
at x = A
|v| = max at x = 0
a = dv/dt
|a| = max at x = A
a =0
at x = 0
SHM Example: Mass on a Vertical Spring
For the person suspended vertically from the spring in
the figure, two forces act on the mass:
The force from the spring;
The force of gravity.
The vertical
oscillation
frequency f is the
same as the
horizontal
oscillation
frequency:
1
k
f 

2 m
SHM Example: The Pendulum
Pendulums come in many forms: a kid on a playground
swing; a mass tied to the end of a string; the pendulum
in an old clock.
The pendulum mass is also called the “bob”.
Applying Newton’s 2nd law, the two
forces acting on the mass are the
upward tension in the string and
the downward force of gravity.
Air resistance is ignored.
It is convenient to measure the
bob’s displacement in terms of y,
the displacement along the
circular arc.
When the pendulum is displaced from its equilibrium
position, the resultant force Fparallel is parallel to the
circular arc and is responsible for the simple harmonic
motion.
Fparallel is the restoring force in the pendulum.
The bottom point on the trajectory (y = 0) is the
equilibrium point of the pendulum; Fparallel = 0 at this
point.
If the bob is placed at this point and given no initial
velocity, it will remain there motionless.
As the pendulum oscillates, the bob moves back and
forth past the equilibrium point and the magnitude and
direction of Fparallel changes with time.
When the bob is to the right where y > 0, Fparallel is
directed to the left (in the negative direction).
When the bob is to the left where y < 0, Fparallel is directed
to the right (in the positive direction).
Fparallel always opposes the bob’s displacement away from
the equilibrium position (the restoring force).
The restoring force is due to the
component of the gravitational
force directed along the bob’s
path.
For a pendulum of length L, with
a mass m, the restoring force is:
Frestoring  Fparallel  m  g  sin 
If the angle  is small, the function
sin  is approximately equal to 
when the angle is measured in
radians; sin  = .
The angle  is also related to the displacement by
y

L
For a small angle displacement (generally less than 15º):
Frestoring  Fparallel
Applying Newton’s
2nd
m  g  y
 m  g   
L
law: Fparallel
m  g  y

 m a
L
To determine the frequency f, we can use the massspring equation by replacing the spring constant k with:
m g
L
1
k
f 

2
m
1
m g
f 

2
m L
replacing k
1
g
f 

2
L
The frequency of the pendulum is independent of the
mass of the bob.
You might have expected that a heavier bob would move more
slowly and have a lower frequency than a lighter bob.
The restoring force is due to gravity and is proportional to the
bob’s mass; m·g·sin  = m·a; the mass cancels; g·sin  = a.
The frequency is also independent of the amplitude of
the motion.
You might have expected the frequency to depend on the
amplitude since if the amplitude is increased (by starting the
pendulum farther from y = 0), the pendulum bob must travel
over a greater distance.
If the amplitude A is increased, the bob’s velocity at all points
along the path is increased by the same amount because A
enters into the results for y (displacement) and v (velocity) in
the same way.
Features Common to All Simple Harmonic Oscillators
The motion of a simple harmonic oscillator exhibits a
sinusoidal time dependence for position or velocity.
y  A  sin  2    f  t 
v  2    f  A  cos 2    f  t 
x  A  cos  2    f  t 
v x  2    f  A  sin 2    f  t 
All simple harmonic oscillators involve
a restoring force.
For a horizontal mass on a spring, F is
the force exerted by the spring on the
mass.
For a pendulum, f is the component of
the gravitational force along the path
of the pendulum bob.
Both displacement and restoring force vary sinusoidally
with time, and F always has a sign opposite that of the
displacement.
The magnitude of the restoring force is proportional to
the displacement.
The frequency of a simple harmonic
oscillator is independent of the
amplitude A and depends on either
the length of a pendulum string or the
mass and spring constant of the
oscillating system.
The amplitude is determined entirely
by how the oscillator is set into
motion.
Harmonic Motion and Energy
• For the mass-spring system,
the mechanical energy of the
system is the sum of the kinetic
energy K (½·m·v²) and elastic
potential energy Ue (½·k·x²).
• For figure A and C: mass is at
x = 0, so Ue = 0 because the spring
is relaxed; the speed is maximum,
so K is equal to the total energy.
• For figure B and D: when x = ±A,
the speed = 0, so K = 0; the mass
has its maximum displacement, so
Ue is equal to the total energy.
• Energy oscillates between elastic
potential and kinetic energy.
• The energy oscillates back
and forth between kinetic
potential energies as shown.
• An ideal oscillator has no
friction so the total
mechanical energy is
conserved.
• The sum of the kinetic and potential energy is constant;
the maximum kinetic energy is equal to the maximum
potential energy.
1
1
2
2
 m  vmax  K max  U max   k  A
2
2
Derivation of the Frequency for a Mass-Spring System Using
Circular Motion Concepts
X-components of the mass’s position and velocity:
x  A  cos  2    f  t 
vc  2    f  A
v x  vc  sin 
v x  2    f  A  sin  2    f  t 
The centripetal acceleration of
the particle is directed toward
the center of the circle is:
vc 2
ac 
A
The simple harmonic motion
is described by the particle’s
position on the x-axis.
The x-component of the acceleration is:
vc 2
a  ac  cos   
 cos 
A
Substituting into the  k·x = m·a equation:
 vc 2

k   A  cos    m   
cos  
 A

m  vc 2
k A 
A
m  2    f  A 
m  vc
k A 

A
A
2
2
2
m  4  f  A
k A 
A
2
2
k  4   f m
2
Substituting
vc = 2·π·f·A:
2
Solving for f:
f
2

k
4  m
2
1
k
f 

2
m
The frequency f increases as the spring constant k is
increased and gets smaller as the mass m is increased.
Frequency f does not depend on the amplitude A.
For situations with two springs as shown:
1
2k
f 

2
m
If there is no friction, an oscillator will oscillate forever.
Most oscillators eventually come to rest as their
vibrations decay with time.
The friction in an oscillating system is referred to as
damping.
When a damped oscillator is
given a nonzero displacement at
t = 0 and released, it can exhibit
three different types of behavior:
(1) overdamped
(2) critically damped
(3) underdamped
For a damped harmonic oscillator, such as a pendulum
swinging in air, the displacement varies with time when
the damping is weak due to a small amount of friction.
The system still oscillates because the displacement alternates
between positive and negative values, but the amplitude of the
oscillation gradually decreases over time.
After undergoing many oscillations, the damping eventually
reduces the oscillations to zero.
This is an example of an underdamped oscillation.
An oscillator is said to be overdamped when friction is
very large.
The resulting displacement as a function
of time is represented by curve 1 on the
right.
• Think of a pendulum mass swinging
through a thick fluid rather than air; the
mass swings very slowly back to the bottom.
How to distinguish underdamped and overdamped
oscillations?
Underdamped – displaced mass always passes through zero at
least once, usually many times, before the system comes to
rest.
Overdamped – displaced mass does not swing past the
bottom, but moves to the equilibrium position without going
beyond it.
In critically damped oscillations, the displaced mass falls
to zero as rapidly as possible without moving past the
equilibrium position.
Shock absorbers on cars:
Essentially springs that support the car’s
body and allow the tires to move up and
down over bumps in the road without
directly passing these vibrations to the
car and passengers.
When car hits a bump, shock absorber springs are
compressed to the initial value of the
displacement.
Shocks are designed to be critically damped,
which allows the body of the car to return to its
original height as quickly as
possible.
Worn out (underdamped) shocks
allow the car to oscillate up and
down.
• Overdamped shocks give the car a
soft ride with poor steering and
handling.
A pendulum can be set into motion by pulling the
pendulum bob to one side and letting go or by giving the
pendulum bob an initial impulse.
In many applications, an oscillator is given regular or
continuous impulses. For example, pushing a kid on a
swing.
The oscillator is “driven” by the applied
force. For example, the pendulum has an
oscillation frequency of:
f natural
1
g


2  L
• If the driving force is not applied at the natural frequency,
but with a period called Tdrive, we can define the driving
frequency as:
1
f drive 
Tdrive
• The amplitude of the oscillations
depends on the driving frequency
as
shown.
• If fdrive is not close to fnatural, the
amplitude of the oscillation is small
no matter how large a driving force
is applied.
• The amplitude is largest when the driving frequency is
close to the natural frequency of the system, which is
called resonance.
• When the frequency of the driving force matches the
natural frequency of an oscillator, the amplitude of the
oscillation will be large.
• Resonance curve is narrowest
and tallest when damping is
weakest.
• Resonance curve is lower and
wider as the damping is
increased.
• The peak of the resonance curve
occurs at the resonant frequency.
x
a
v
m
x = -A
x=0
F  m  a  k  x
x = +A
k  x
a
m
Conservation of Energy:
½·m·vA2 + ½·k·xA 2 = ½·m·vB2 + ½·k·xB 2
1
2
mv  k  x  k  A
2
1
2
k
v
 A2  x 2
m
x  A  cos(2    f  t )
2
1
2
v0 
2
k
A
m
a  4    f  x
2
2
v  2    f  A  sin(2    f  t )
x
a
v
m
x = -A
1
f 
2 
x=0
a
x
1
k
f 

2  m
x = +A
x
T  2  
a
m
T  2  
k
Condition for SHM
Natural frequency
[rad/s]
Period
Mass on a
Spring
Pendulum
Small oscillations
(spring obeys
Hooke’s Law)
Small angles
(sinθ ≈ θ)
k

m
m
T  2  
k

g
L
T  2  
L
g