Transcript Simple Harmonic Motion

Warmup
A uniform beam 2.20m long with mass
m=25.0kg, is mounted by a hinge on a wall as
shown. The beam is held horizontally by a wire
that makes a 30° angle as shown. The beam
supports a mass M = 280kg suspended from its
end. Determine the components of the force FH
that the hinge exerts and the components
tension, FT in the supporting wire.
30°
M
Harmonic Motion
Simple Harmonic Motion
When a vibration or
oscillation repeats itself,
back and forth, over the
same path, the motion is
periodic
Any vibrating system in
which the restoring force is
directly proportional to the
negative of the displacement
is said to exhibit simple
harmonic motion

Such a system is often
called a simple harmonic
oscillator
A mass oscillating on the end
of a spring is an example of a
simple harmonic oscillator
The Motion of an Oscillating
Object
The motion of an
object undergoing
simple harmonic
oscillation is
sinusoidal in nature
x(t) = A sin (2πt/T)
The Total Energy of a
Vibrating System is Constant
KE + PE = constant
If the maximum amplitude of the motion is x0
then the energy at any point x is given by:
½ mv2 + ½ kx2 = ½ kx02
From this we can solve for velocity:
│v│= √ [(x02 –x2)(k/m)]
From Hooke’s law, F = -kx and F =ma,
therefore
a = -(k/m) x
Reference Circle
Point P moves with
constant velocity v0
around the circle
Point A is the projection
of point P on the x axis
The motion of A back
and forth about point O
is second harmonic
motion
The time for P to go
around the circle is T
The velocity of point A is
v = -v0 sin θ
Once around
in time T
θ v0
v
P
r= x0
O θ
Displacement x
A
Reference Circle
The period T is:
T = 2πr0/v0 = 2πx0/v0
But v0 is the maximum speed of point A

Maximum speed occurs when x = 0
Therefore, v0 = x0 √(k/m)
Which makes T = 2π√(m/k)
And, since f = 1/T,
f = (1/2π)√(k/m)
Period and Frequency of an
Oscillator
The period of a simple harmonic
oscillator is given by:
T = 2π√(m/k)
Therefore, since f = 1/T,
f = (1/2π)√(k/m)
Simple Pendulums
For a pendulum the
restoring force is
F = -mg sin θ,
But for small
displacements,
F = -mg sin θ ~ -mgθ
But since x = Lθ we have
F ~ (mg/L) x
Therefore the motion is
essentially harmonic
Simple Pendulums
But if a pendulum is an harmonic oscillator,
then k = mg/L, therefore
T = 2π√ (m/mg/L) = 2π√(L/g)
and
F = 1/T = 1/(2π)√(g/L)
The frequency of a pendulum does not
depend on the mass of the pendulum bob!

Consider the case of large and small children on a
swing—the period remains the same
In Real Life Most Harmonic
Motion is Damped
Natural oscillations
decrease with time
due to friction and
other losses
Sometimes the
damping is so great
that the motion does
not even appear to
be harmonic
A = overdamped
B = critically damped
C = underdamped
A shock absorber is a
damped oscillator
Shock Absorbers Keep a Car
From Oscillating
Links
Java simulation—spring
Java simulation-oscillator
Simple Harmonic Motion and
Uniform Circular Motion
Do Now (10/8/13):
Suppose that a pendulum has a period
of 1.5 seconds. How long does it take
to make a complete back-and-forth
vibration? Is this 1.5 s period pendulum
longer or shorter in length than a 1 s
period pendulum?
Practice:
Brainstorm answers to the Conceptual
Questions in Chapter 13 (10 min)
Complete the Multiple Choice questions
in Chapter 13 before the end of class.
Do Now (10/9/13):
Pass in your HW then wait quietly for
instructions