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Simple Harmonic Motion
A restoring force is one that moves a system back to an
equilibrium position.
Example: mass on
frictionless table,
attached to spring.
Example: gravity
acting on a mass
hanging from a
string.
*
Example: gravity
acting on a mass
hanging from a
spring.
*Lots of material borrowed from:
http://hyperphysics.phy-astr.gsu.edu/hbase/permot.html#permot
When the restoring force is linearly proportional to the
amount of the displacement from equilibrium, the force
is said to be a Hooke’s Law force.
Physicists are always sloppy with their signs when doing
Hooke’s law calculations.
We have studied Hooke’s Law already. Hooke’s Law is
valid for forces that take the form (in one dimension) Fx
= - kx, where x is the displacement from equilibrium.
OSE: F= kx, restoring
It is easy to show with calculus that, in the limit of small
displacement, all restoring forces are Hooke’s Law forces.
Systems consisting of Hooke’s Law forces undergo simple
harmonic motion when displaced away from equilibrium.
Such systems are often called simple harmonic
oscillators.
Examples of simple harmonic motion: mass
bouncing on spring, swinging pendulum.
Here’s that calculus I mentioned. Perhaps you recall
from math how functions can be written in the form of
a Maclaurin’s series (a Taylor series about the origin):
x 2  d2F 
x 3  d 3F 
 dF 
F(x) = F(0) + x   +
 2 +
 3 +...
 dx 0 2!  dx 0 3!  dx 0
If F represents a restoring force (a force that “pulls the
system back to the origin”) then F(0) = 0.
For small displacements x, all the higher order terms
(involving x2, x3, etc.) are small, so
 dF 
F(x)  x   = - k x .
 dx 0
The – sign enters because F is a restoring force, so the derivative is negative.
The displacement (x, or in this case, y) is measured
away the equilibrium position.
The maximum displacement is called the amplitude A.
A complete back-and-forth (or whatever) motion is called
a cycle.
The time for one cycle is the period, T.
The number of cycles per second is the frequency.
We’ll learn about angular frequency shortly.
Energy in the Simple Harmonic Oscillator
We get our OSE for the total mechanical energy of a
simple harmonic oscillator by considering a mass
attached to a spring on a horizontal, frictionless surface:
E = mv2/2 + kx2/2
(here I am using x instead of s).
At the equilibrium position (x=0), the energy is all
kinetic. At x=A the energy is all potential. Thus
mv2/2 + kx2/2 = kA2/2 = mv02/2
where v0 is the speed at x=0 where the energy is all
kinetic. Your text solves this for v(t) and makes it look
like a brand new equation, when it is really just old stuff.
Example 11-3. A spring stretches 0.15 m when a 0.3
kg mass is hung from it. The spring is then stretched an
additional 0.1 m from this equilibrium point and
released. Determine (a) the spring constant k, (b) the
amplitude of oscillation A, (c) the maximum velocity v0,
(d) the magnitude of the velocity when the mass is 0.04
m from equilibrium, and (e) the magnitude of the
maximum acceleration of the mass.
Wait a minute! Where did the gravitational potential
energy go?
Once you confirm that a system undergoes simple
harmonic motion, all the equations for a simple harmonic
oscillator apply.
The total mechanical energy of the system is the kinetic
energy of the object oscillating, plus the potential energy
associated with the restoring force.
You need not worry about other details of the system!
This is abstract, but powerful!
The Periodic and Sinusoidal Nature of SHM
Important: the force in SHM is not constant, so the
acceleration is not constant, so you can’t use the
equations of kinematics.
An object rotating in a circle (which is has a frequency
and a period) is mathematically analogous to the SHM of
a vibrating spring.
Click here to see
this in action.
Your text shows that T=2(m/k)1/2 and f=(1/2)(k/m)1/2.
Before, for an object
undergoing circular
motion,  was the
angular velocity and
T=2/ the time for a
revolution.
For circular motion, the angular velocity  is the angle
per second swept out by the radius vector, and is
measured in radians per second.
For SHM, the frequency f is the number of complete
cycles per second completed by the oscillator.
There are 2 radians in a circle, so if we represent a
cycle by one time around the circle, =2f, and there are
2 radians in a cycle. For SHM, the angular frequency 
also has units of radians per second.
It is no coincidence that angular velocity and angular
frequency have the same symbol and the same units.
Relationships between period, frequency, and angular
frequency:
m
T = 2π
k
1 k
f=
2π m
2π
T=
ω
k
For SHM, the angular frequency is ω = 2πf =
m
Using our analogy with circular motion:
x = A cos (t) = A cos (2f t)
v = -v0 sin (2f t) where v0 = 2Af
a = -a0 cos (2f t) where a0 = kA/m.
All the equations on this page are OSE’s!
Example 11-7. The cone of a loudspeaker vibrates in
SHM at a frequency of 262 Hz. The amplitude of the
center of the cone is A=1.5x10-4 m, and at t=0, x=A. (a)
What is the equation describing the motion of the center
of the cone. (b) What is the maximum velocity and
maximum acceleration? (c) What is the position of the
cone at t=1x10-3 s?
The Simple Pendulum
The restoring force is F = mg sin.
In the limit of small , this
becomes F = mg  (restoring).
This is a Hooke’s Law force and
SHM occurs.
Your text shows that the period is
OSE :
L
T = 2π
.
g
For SHM, the period does not depend on the amplitude.
For a pendulum with small , this is true, so a pendulum
exhibits SHM for small displacements.
For large  (greater than 15 degrees or so) the smallangle approximation is not valid and the period does
depend on the amplitude (max).
Example 11-8. (a) Estimate the length of the
pendulum in a grandfather clock that ticks once per
second. (b) What would be the period of a clock with a 1
meter long pendulum?