Simple Harmonic Motion
Holt Physics
Pages 438 - 451
Distinguish simple harmonic motion
from other forms of periodic
motion.
Periodic motion is motion in which a
body moves repeatedly over the same
path in equal time intervals.
 Examples: uniform circular motion and
simple harmonic motion.
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Cont’d
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Simple Harmonic Motion (SHM) is a special type of
periodic motion in which an object moves back and
forth, along a straight line or arc.
Examples: pendulum, swings, vibrating spring,
piston in an engine.
In SHM, we ignore the effects of friction.
Friction damps or slows down the motion of the
particles. If we included the affect of friction then
it’s called damped harmonic motion.
Cont’d
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For instance a person oscillating on a
bungee cord would experience damped
harmonic motion. Over time the amplitude of
the oscillation changes due to the energy lost
to friction.
http://departments.weber.edu/physics/amiri/di
rector/DCRfiles/Energy/bungee4s.dcr
State the conditions necessary for
simple harmonic motion.
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A spring wants to stay at its equilibrium or resting
position.
However, if a distorting force pulls down on the
spring (when hanging an object from the spring,
the distorting force is the weight of the object), the
spring stretches to a point below the equilibrium
position.
The spring then creates a restoring force, which
tries to bring the spring back to the equilibrium
position.
Cont’d
The distorting force and the restoring
force are equal in magnitude and
opposite in direction.
 FNET and the acceleration are always
directed toward the equilibrium position.

Cont’d

Applet showing the forces, displacement,
and velocity of an object oscillating on a
spring.

https://ngsir.netfirms.com/englishhtm/SpringS
HM.htm
Displacement
Velocity
Acceleration
Cont’d
 at
equilibrium:
speed or velocity is at a maximum
– displacement (x) is zero
– acceleration is zero
– FNET is zero (magnitude of Restoring
Force = magnitude of Distorting Force
which is the weight)
– object continues to move due to inertia
–
Cont’d

at endpoints:
– speed or velocity is zero
– displacement (x) is at a maximum
equal to the amplitude
– acceleration is at a maximum
– restoring force is at a maximum
 F=
–
-kx (hook’s law)
FNET is at a maximum
State Hooke’s law and apply it to the
solution of problems.

Hooke’s Law relates the distorting force
and the restoring force of a spring to
the displacement from equilibrium.
Cont’d


F  kx


F –restoring force in Newtons
k – spring constant or force
constant (stiffness of a
spring) in Newtons per meter
(N/m)
x –displacement from
equilibrium in meters
The distorting force is equal
in magnitude but opposite in
direction to the restoring
force.
The spring shown to the right
has an unstretched length of
3 cm. When a 2 kg object is
hung from the spring, it
comes to rest at the 7 cm
mark. What is the spring
constant of this spring?
490 N/m
What direction is the
restoring force?
upward
Calculate the frequency and period of
any simple harmonic motion.
T – period (time
required for a complete
vibration) in seconds
 f – frequency in
vibrations / second or
Hertz

1
T
f
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A particle is moving in simple harmonic
motion with a frequency of 10 Hz.
What is its period?
0.1 sec
How many complete oscillations does it
make in one minute?
600 oscilations
Relate uniform circular motion to
simple harmonic motion.
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The reference circle relates uniform circular motion to
SHM.
The shadow of an object moving in uniform circular
motion acts like SHM.
The speed of an object moving in uniform circular
motion may be constant but the shadow won’t move at
a constant speed.
The speed at the endpoints is zero and a maximum in
the middle.
The shadow only shows one component of the motion.
Cont’d

Applet showing the forces, displacement,
and velocity of an object oscillating on a
spring and an object in uniform circular
motion.

http://www.physics.uoguelph.ca/tutorials/shm
/phase0.html
Identify the positions of and calculate
the maximum velocity and
maximum accelerations of a
particle in simple harmonic motion.
The acceleration is a maximum at the
endpoints and zero at the midpoint.
 The acceleration is directly proportional
to the displacement, x.
 The radius of the reference circle is
equal to the amplitude.
 The force and acceleration are always
directed toward the midpoint.

The mass on the end of a spring
(which stretches linearly) is in
equilibrium as shown. It is pulled
down so that the pointer is opposite
the 11 cm mark and then released.
What is the amplitude of the vibration?
4 cm
What two places will the restoring
force be greatest?
11 cm and 3 cm
Where will the restoring force be least?
7 cm
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Where is the speed greatest?
7 cm
What two places is the speed
least?
3 cm and 11 cm
Where is the magnitude of the
displacement greatest?
3 cm and 11 cm
Where is the displacement least?
7 cm
Where is the magnitude of the
acceleration greatest?
3 cm and 11 cm
Where is the acceleration least?
7 cm
Where is the elastic potential
energy the greatest?
Where is the kinetic energy the
greatest?
Cont’d
Remember that in uniform
circular motion, the
velocity is calculated
using:
2r
v
T
In SHM, the maximum velocity
would be equal to the velocity
of the object in uniform circular
motion. The radius of the circle
correlates to the Amplitude (A)
in SHM.
v max
2A

T
Cont’d
Remember that in uniform
circular motion, the
centripetal acceleration
is calculated using:
2
v
ac 
r
In SHM, the maximum
acceleration would be equal to
the acceleration of the object in
uniform circular motion. The
radius of the circle correlates to
the Amplitude (A) in SHM.
amax 
v
2
max
A
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A mass hanging on a spring oscillates with an
amplitude of 10 cm and a period of 2 seconds. What
is the maximum speed of the object and where does
it occur?
0.314 m/s at equilibrium
What is the minimum speed of the object and where
does it occur?
0 m/s at the end points.
What is its maximum acceleration?
0.987 m/s2 at the end points

An object moving is
simple harmonic motion
can be located using:
–
–
–
–
A is amplitude
f is frequency
x is displacement from
equilibrium
ω is angular velocity
x  A cos( 2ft )
or
x  A cos(t )
2

 2f
T

The mass on the end of a
spring (which stretches
linearly) is in equilibrium as
shown.

It is pulled down so that the
pointer is opposite the 11 cm
mark and then released. A
spring vibrates in SHM
according to the equation
x = 4 cosπt.
How many complete
vibrations does it make in 10
seconds?
5 vibrations
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The elastic potential energy content of the system is
2
1
s
2
U  kx
So the maximum elastic potential energy is stored at the end points of the
oscillations where the displacement is equal to the amplitude of the vibration
U s  kA
2
1
2
At the end point, the object is not moving so there is no kinetic energy.
Therefore the total energy content of the system is equal to
Eo  kA
1
2
2
A mass on a spring oscillates horizontally on a
frictionless table with an amplitude of A. In terms of
Eo (total mechanical energy of the system) when the
mass is at A, Us = ______ and K = _________.
Us = Eo and K = 0
When the mass is at 0.5 A, then Us = __________ and
K = _________.
Us = 0.25 Eo and K = 0.75Eo
When the mass is at the equilibrium position, then Us =
_________ and K = ________
Us = 0 and K = Eo

A 2 kg object is attached to a spring of force constant
k = 500 N / m. The spring is then stretched 3 cm
from the equilibrium position and released. What is
the maximum kinetic energy of this system?

0.225 J
What is the maximum velocity it will attain?
0.47 m/s
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Cont’d
m
T  2
k
T = period (s)
m = mass (kg)
k = spring
constant (N/m)
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You want a mass that, when hung on the end
of the spring, oscillates with a period of 3
seconds. If the spring constant is 5 N/m, the
mass should be _______.
1.14 kg
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The period for a mass vibrating on very stiff springs
(large values of k) will be (larger / smaller) compared
to the same mass vibrating on a less stiff spring.
Smaller
If the value of k halves, the period will be ______
times as long.
2
Relate the motion of a simple
pendulum to simple harmonic
motion.
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A pendulum is a type of SHM.
A simple pendulum is a small, dense mass
suspended by a cord of negligible mass.
The period of the pendulum is directly
proportional to the square root of the length
and inversely proportional to the square root
of the acceleration due to gravity.
Cont’d
l
T  2
g
T = period (s)
l = length (m)
g = acceleration due
to gravity (m/s2)
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A pendulum has a period of 2 seconds here
on the surface of the earth. That pendulum
is taken to the moon where the acceleration
due to gravity is 1/6 as much. What is the
period of the pendulum on the moon?
Squareroot of 6 times as much or 4.9
seconds.