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Oscillations and Waves
An oscillation is a repetitive motion back and forth around a
central point which is usually an equilibrium position.
A special type of oscillation is called Simple Harmonic Motion
(SHM). This type of oscillation is for a mass moving around a
point of equilibrium which serves as the reference point for
displacement (DISTANCE FROM EQUILIBRIUM).
To be simple harmonic motion, the motion MUST be produced
by a force that increases LINEARLY with respect to
displacement but in the opposite direction.
For all SHM:
F=-kx =ma
F = restoring force
k is a constant value ( usually the spring constant)
x = displacement from equilibrium
The k does not HAVE to be a spring constant; it simply needs
to be a constant value.
The motion of the mass is a result of the conservation of
energy while the mass moves. Usually there is an exchange of
energy between elastic potential energy and kinetic energy.
The POTENTIAL energy of the oscillator is: US = ½ k x2
( The energy stored in the elastic medium )
The KINETIC energy of the oscillator is: K = ½ m v2
The TOTAL energy of the oscillator remains constant:
ETot = K + US = ½ m v2 + ½ k x2
As the mass moves to either side of equilibrium, it will stretch
or compress or the spring. The spring gains potential energy.
Whenever the spring gains potential energy, the mass must
lose kinetic energy to keep the total energy constant.
As the extension of the spring decreases, the spring loses
potential energy which leads to a gain in kinetic energy (speed).
The motion will extend the same displacement to either side
of the equilibrium point.
If you determine the end points of the motion, the equilibrium
point will be midway between the two.
Imagine a mass on a frictionless surface which is attached to a
horizontal spring so that it can move back and forth.
-A
Maximum ( - )
Displacement
+A
Equilibrium
Maximum ( + )
Displacement
The maximum displacement is called the amplitude ( A ).
F
F
vmax
v=0
+A
-A
Maximum ( - )
Displacement
v=0
Equilibrium
Maximum ( + )
Displacement
At the maximum displacement ( x = +/- A ), the mass will be at rest.
At the equilibrium position ( x = 0 ), the mass will be at maximum
speed.
The force (F) is always pulling toward equilibrium.
This means that as the mass moves toward equilibrium,
it’s speeding up. Kinetic energy increases when potential
energy decreases.
When it moves away from equilibrium, it’s slowing down.
Kinetic energy decreases when potential energy increases
The total energy of the oscillator can then be determined by
using one of the end points where the kinetic energy is zero:
ETot = ½ k A2
(Since kinetic energy is zero, the total energy is simply the elastic
energy)
Using the conservation of energy equation:
Etot = ½ m v2 + ½ k x2
½ k A2 =
½ mv2 + ½ k x2
Use the above relationship to determine an expression for the
speed ( v ) of the mass as a function of displacement ( x ).
k A2 - k x2 =
k ( A2 - x2 ) =
mv2
mv2
( k / m ) ( A 2 - x2 ) =
(k / m)
( A 2 - x2 ) = v
v2
(k / m) is extremely
important to simple
harmonic motion.
v =
(k / m)
( A 2 - x2 )
This equation shows us why the maximum speed is at
equilibrium ( x = 0 ) and the minimum speed is zero at the
maximum displacement ( x = A ).
The number of complete motions (oscillations) it makes per unit
time is called the frequency ( f ).
We use the standard unit of seconds for time. 1 oscillation per
second is given the unit of 1 Hertz ( Hz ).
f = 1/(2p)
(k / m)
The time it takes the mass to move one full motion ( from one
side to the other side AND BACK ) is called the period ( T ). The
period will be in units of seconds. T = 1 / f
T =(2p)
(m / k)
T = period
This is on your
equation sheet.
The period of a simple harmonic oscillator DOES NOT depend
on the amplitude! In the case of a spring oscillator, the period
depends on the mass and the spring constant.
Earlier, we said that (k / m) is extremely important to simple
harmonic motion.
(k / m)
= 2pf
f = frequency
(k / m)
= w
w = angular frequency
Summary of equations:
v =
( A 2 - x2 )
(k / m)
f = 1/(2p)
T =(2p)
(k / m)
(m / k)
A horizontal spring oscillator has 0.400 kg mass that slides
over a frictionless surface. The spring constant is 35.0 N/m,
and the maximum displacement is 0.200 m.
Determine the
frequency of the oscillator.
Determine the period of the oscillator.
Determine the maximum velocity of the oscillator.
Determine the maximum acceleration of the oscillator.
Another oscillator has an amplitude of 0.350 m and a
frequency of 2.00 Hz. Determine the maximum acceleration
of this oscillator.
If a mass is hanging from a vertical spring, it will also undergo
simple harmonic motion.
Although, there is gravitational potential energy as well as
elastic potential energy and kinetic energy, the equations of
motion are the same!
There will be a shift in the equilibrium position ( xo ) :
k xo =
mg
[ equilibrium; forces cancel ]
This shift compensates for the effects of gravity.
As long as the displacement is measured from the new
equilibrium position, there will be no change in the equations.
v =
( A 2 - x2 )
(k / m)
f = 1/(2p)
T =(2p)
(k / m)
(m / k)
A 0.375 kg mass is placed on a vertical spring. The spring
stretches 0.210 m to where the mass will hang at rest. The
mass is then displaced an additional 0.100 m ( 0.310 m total )
and released. What is the amplitude of the oscillator?
What is the spring constant?
Determine the period of the oscillator.
m = 0.375 kg
xo = 0.210 m
A = 0.100 m
k = 17.5 N/m
Determine the maximum acceleration of the oscillator.
Determine the maximum velocity of the oscillator.
Determine the speed of the mass when it is midway between
equilibrium and its amplitude ( a displacement of 0.050 m ).
A 0.200 kg mass is hanging from a vertical spring next to a
meter stick. The spring has a spring constant of 12.0 N/m, and
the mass is oscillating between the 45.0 cm mark on the meter
stick and the 75.0 cm mark. What is the amplitude of the
oscillations?
At what cm marking is the equilibrium
position?
75 cm
(highest pt)
The spring is at the 78.0 cm mark when there
is no mass on it. Use this and the previous
answer to find the spring constant?
45 cm
(lowest pt)
m = 0.200 kg
m = 0.200 kg
A = 0.150 m
K = 10.9 N/m
Determine the period of the oscillations.
75 cm
(highest pt)
What is the frequency of the oscillator?
45 cm
(lowest pt)
m = 0.200 kg
What is the maximum acceleration of the
oscillator?
PENDULUM MOTION
Pendulum motion is a type of harmonic motion. It is not truly
simple harmonic motion, but it approximates SHM very well for
small angles. The restoring force for the pendulum is
mg sin q.
q
L
BUT this force does not change linearly
with displacement, so it’s not truly SIMPLE
harmonic motion.
T
q
For small angles, however, sin q = q.
(in radians)
This gives us: F = - mg q
q=x/L
mg
SO:
F = - [ mg / L ] x
k = mg / L
SO:
2pf=
k/m
2pf =
g/L
Like a mass on a spring, the motion of the mass will extend
the same displacement to either side of the equilibrium point.
This maximum displacement is called the amplitude ( A ).
A
A
The Equilibrium Point
At the maximum displacement ( A ), the mass will be at rest as
its motion in one direction will stop before it moves in the other
direction.
This is also the point where the mass will have the
greatest force and acceleration.
At the equilibrium point, the mass will be moving the fastest.
The net force and acceleration will be zero.
The period and frequency are NOT affected by the amplitude!
The pendulum will have the same period when it’s moving just
a little as when it is moving a lot.
The period of a simple pendulum depends on the length of the
string and the acceleration of gravity.
For a simple pendulum the frequency can be described by the
following equation:
1
f =
g/L
2p
The period would then be:
f = frequency
T =
2p
T = period
L/g
g = gravity
L = the length of the pendulum
Determine the period of a pendulum that is 1.0 m long.
Determine the frequency of this pendulum.
How would the frequency be changed if this set-up were
brought to the moon where the gravity is 1/6 that of Earth?
A bowling ball is hung from the ceiling of a large room. When
the ball is set into harmonic motion by releasing it at an angle,
it moves forward and back to its starting point in 3.6 s. How
long is the string from which the ball is hanging?