Oscillations
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Transcript Oscillations
Topic 4: Waves
4.1 – Oscillations
Essential idea: A study of oscillations underpins many
areas of physics with simple harmonic motion
(SHM) a fundamental oscillation that appears in
various natural phenomena.
Topic 4: Waves
4.1 – Oscillations
Nature of science: Models: Oscillations play a great
part in our lives, from the tides to the motion of the
swinging pendulum that once governed our
perception of time. General principles govern this
area of physics, from water waves in the deep
ocean or the oscillations of a car suspension
system. This introduction to the topic reminds us
that not all oscillations are isochronous. However,
the simple harmonic oscillator is of great
importance to physicists because all periodic
oscillations can be described through the
mathematics of simple harmonic motion.
Topic 4: Waves
4.1 – Oscillations
Understandings:
• Simple harmonic oscillations
• Time period, frequency, amplitude, displacement and
phase difference
• Conditions for simple harmonic motion
Applications and skills:
• Qualitatively describing the energy changes taking
place during one cycle of an oscillation
• Sketching and interpreting graphs of simple harmonic
motion examples
Topic 4: Waves
4.1 – Oscillations
Guidance:
• Graphs describing simple harmonic motion should
include displacement–time, velocity–time,
acceleration–time and acceleration–displacement
• Students are expected to understand the significance
of the negative sign in the relationship: a = -x.
Data booklet reference:
•T=1/f
Topic 4: Waves
4.1 – Oscillations
International-mindedness:
• Oscillations are used to define the time systems on
which nations agree so that the world can be kept
in synchronization. This impacts most areas of our
lives including the provision of electricity, travel and
location-determining devices and all
microelectronics.
Theory of knowledge:
• The harmonic oscillator is a paradigm for modelling
where a simple equation is used to describe a
complex phenomenon. How do scientists know
when a simple model is not detailed enough for
their requirements?
Topic 4: Waves
4.1 – Oscillations
Utilization:
• Isochronous oscillations can be used to measure time
• Many systems can approximate simple harmonic
motion: mass on a spring, fluid in U-tube, models of
icebergs oscillating vertically in the ocean, and
motion of a sphere rolling in a concave mirror
• Simple harmonic motion is frequently found in the
context of mechanics (see Physics topic 2)
Topic 4: Waves
4.1 – Oscillations
Aims:
• Aim 6: experiments could include (but are not limited
to): mass on a spring; simple pendulum; motion on
a curved air track
• Aim 7: IT skills can be used to model the simple
harmonic motion defining equation; this gives
valuable insight into the meaning of the equation
itself
Topic 4: Waves
4.1 – Oscillations
Oscillations
Oscillations are vibrations which repeat themselves.
v=0
v
=
0
EXAMPLE: Oscillations
v=0
v = vmax
v=0
EXAMPLE: Oscillations can
be driven internally, like a
mass on a spring.
FYI In all oscillations,
v = 0 at the extremes…
and v = vmax in the
middle of the motion.
v = vmax
can be driven externally,
like a pendulum in a
gravitational field.
x
Topic 4: Waves
4.1 – Oscillations
Oscillations
Oscillations are vibrations which repeat themselves.
EXAMPLE: Oscillations can be
very rapid vibrations such as in
a plucked guitar string or
a tuning fork.
Topic 4: Waves
4.1 – Oscillations
Time period, amplitude and displacement
Consider a mass on a
spring that is displaced
4 meters to the right
x
and then released.
x0
We call the maximum displacement x0 the amplitude.
In this example x0 = 4 m.
We call the point of zero displacement the equilibrium
position. Displacement x is measured from equilibrium.
The period T (measured in s) is the time it takes
for the mass to make one full oscillation or cycle.
For this particular oscillation, the period T is
about 24 seconds (per cycle).
Topic 4: Waves
4.1 – Oscillations
Time period and frequency
The frequency f (measured in Hz or cycles / s) is
defined as how many cycles (oscillations, repetitions)
occur each second.
Since period T is seconds per cycle, frequency must
be 1 / T.
f=1/T
T = 1 / f relation between T and f
EXAMPLE: The cycle of the previous example repeated
each 24 s. What are the period and the frequency of the
oscillation?
SOLUTION:
The period is T = 24 s.
The frequency is f = 1 / T = 1 / 24 = 0.042 Hz
Topic 4: Waves
4.1 – Oscillations
Phase difference
We can pull the mass to the right and then release it to
begin its motion:
Start
stretched
x
The two motions are half a cycle out of phase.
Start
compressed
x
Or we could push it to the left and release it:
Both motions would have the same values for T and f.
However, the resulting motion will have a phase
difference of half a cycle.
Topic 4: Waves
4.1 – Oscillations
Phase difference
PRACTICE: Two identical mass-spring systems are
started in two different ways. What is their phase
difference?
Start stretched
and then release
x
Start unstretched
with a push left
x
SOLUTION:
The phase difference is one-quarter of a cycle.
Topic 4: Waves
4.1 – Oscillations
Phase difference
PRACTICE: Two identical mass-spring systems are
started in two different ways. What is their phase
difference?
Start stretched
and then release
x
Start unstretched
with a push right
x
SOLUTION:
The phase difference is three-quarters of a cycle.
Topic 4: Waves
4.1 – Oscillations
Conditions for simple harmonic motion
EXAMPLE: A spring having a
spring constant of 125 N m-1 is
attached to a 5.0-kg mass, stretched +4.0 m as shown,
and then released from rest.
(a) Using Hooke’s law, show that the acceleration a of a
mass-spring system is related to the spring’s
displacement x by the proportion a -x.
SOLUTION:
Recall Hooke’s law: F = -kx (see Topic 2-3).
From Newton’s second law F = ma we then have
ma = -kx or a = -(k / m) x.
Thus, a -x.
x
Topic 4: Waves
4.1 – Oscillations
Conditions for simple harmonic motion
EXAMPLE: A spring having a
spring constant of 125 N m-1 is
attached to a 5.0-kg mass, stretched +4.0 m as shown,
and then released from rest.
(b) Tailor your equation to this example, and find the
acceleration of the mass when x = -2.0 m.
(c) What is the displacement of the mass when the
acceleration is -42 ms-2?
SOLUTION:
(b) a = -(k / m) x = -(125 / 5) x = -25x. Thus a = -25x so
a = -25(-2.0) = +50. ms-2.
(c) a = -25x -42 = - 25 x x = -42 / -25 = +1.7 m.
x
Topic 4: Waves
4.1 – Oscillations
Conditions for simple harmonic motion
A very special kind of oscillation that shows up often in
the physical world is called simple harmonic motion.
In simple harmonic motion (SHM), a and x are
related in a very precise way: Namely, a -x.
a -x
definition of SHM
PRACTICE: Show that a mass oscillating on a spring
executes simple harmonic motion.
SOLUTION:
x
We already did when we showed that a = -(k / m)x,
since this means that a -x.
F
Topic 4: Waves
4.1 – Oscillations
F
x x
0
F and x oppose
each other.
Conditions for simple harmonic motion
a -x
definition of SHM
The minus sign in Hooke’s law, F = -kx, tells us that if
the displacement x is positive (right), the spring force F
is negative (left).
It also tells us that if the displacement x is negative
(left), the spring force F is positive (right).
Any force that is proportional to the opposite of a
displacement is called a restoring force.
For any restoring force F -x.
Since F = ma we see that ma -x, or a -x.
All restoring forces can drive simple harmonic motion
(SHM).
x
Topic 4: Waves
4.1 – Oscillations
Conditions for simple harmonic motion
If we place a pen on the oscillating mass, and pull
a piece of paper at a constant speed past the pen,
we trace out the displacement vs. time graph of SHM.
x SHM traces out perfect sinusoidal waveforms.
t
Note that the period can be found from the graph:
Just look for repeating cycles.
Topic 4: Waves
4.1 – Oscillations
Qualitatively describing the energy changes taking
place during one cycle of an oscillation
Consider the pendulum to
the right which is placed in
position and held there.
Let the green rectangle
represent the potential energy
of the system.
Let the red rectangle represent
the kinetic energy of the system.
Because there is no motion yet, there is no kinetic
energy. But if we release it, the kinetic energy will grow
as the potential energy diminishes.
A continuous exchange between EK and EP occurs.
Topic 4: Waves
4.1 – Oscillations
Qualitatively describing the energy changes taking
place during one cycle of an oscillation
Consider the mass-spring
system shown here. The
mass is pulled to the right
and held in place.
Let the green rectangle represent the
potential energy of the system.
Let the red rectangle
FYI If friction and drag are
represent the kinetic
energy of the system. both zero ET = CONST.
A continuous exchange between EK and EP occurs.
Note that the sum of EK and EP is constant.
EK + EP = ET = CONST relation between EK and EP
x
Topic 4: Waves
4.1 – Oscillations
Qualitatively describing the energy changes taking
place during one cycle of an oscillation
EK + EP = ET = CONST relation between EK and EP
Energy
If we plot both kinetic
energy and potential
energy vs. time for either
system we would get the
following graph:
time
x
Topic 4: Waves
4.1 – Oscillations
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The displacement x vs. time t
for a 2.5-kg mass on a
spring having spring
constant k = 4.0 Nm-1
is shown in the
sinusoidal graph.
(a) Find the period and frequency of the motion.
SOLUTION: The period is the time for a complete cycle.
From the graph it is T = 6.0 ms = 6.010-3 s.
Then f = 1 / T = 1 / 0.006 = 170 Hz.
Topic 4: Waves
4.1 – Oscillations
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The displacement x vs. time t
for a 2.5-kg mass on a
spring having spring
constant k = 4.0 Nm-1
is shown in the
sinusoidal graph.
(b) Find the amplitude of the motion.
SOLUTION:
The amplitude is the maximum displacement.
From the graph it is xMAX = 2.0 mm = 2.010-3 m.
Topic 4: Waves
4.1 – Oscillations
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The displacement x vs. time t
for a 2.5-kg mass on a
spring having spring
constant k = 4.0 Nm-1
is shown in the
sinusoidal graph.
(c) Sketch the graph of x vs. t for the situation where
the amplitude is cut in half.
SOLUTION:
For SHM, the period is independent of the amplitude.
Topic 4: Waves
4.1 – Oscillations
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The displacement x vs. time t
for a 2.5-kg mass on a
spring having spring
constant k = 4.0 Nm-1
is shown in the
sinusoidal graph.
(c) The blue graph shows an equivalent system in SHM.
What is the phase difference between the red and blue?
SOLUTION:
We see that it is T / 6 (= 360/ 6 = 60 = 2/ 6 rad).
v=0
v = vMAX v = 0
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
x
EXAMPLE: The displacement x vs. time t for a system
undergoing SHM is shown here.
x-black
v-red
(different
scale)
(+) t
( -)
(+)
( -)
(+)
Sketch in red the velocity vs. time graph.
SOLUTION: At the extremes, v = 0.
At x = 0, v = vMAX. The slope determines sign of vMAX.
v=0
v = vMAX v = 0
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
x
EXAMPLE: The displacement x vs. time t for a system
undergoing SHM is shown here.
x-black
v-red
(different
scale)
t
a-blue
(different
scale)
Sketch in blue the acceleration vs. time graph.
SOLUTION: Since a -x, a is just a reflection of x.
Note: x is a sine, v is a cosine, and a is a – sine wave.
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
the graph. The system consists
of a 0.125-kg mass on a spring.
(a) Determine the maximum
velocity of the mass.
SOLUTION:
When the kinetic energy is maximum, the velocity is
also maximum. Thus 4.0 = (1/ 2)mvMAX2 so that
4.0 = (1/ 2)(.125)vMAX2 vMAX = 8.0 ms-1.
x
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
ET
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
EK
the graph. The system consists
of a 0.125-kg mass on a spring.
EP
(b) Sketch EP and determine the
total energy of the system.
SOLUTION:
Since EK + EP = ET = CONST, and since EP = 0 when
EK = EK,MAX, it must be that ET = EK,MAX = 4.0 J.
Thus the EP graph will be the “inverted” EK graph.
x
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
the graph. The system consists
of a 0.125-kg mass on a spring.
(c) Determine the spring
constant k of the spring.
SOLUTION: Recall EP = (1/2)kx2.
Note that EK = 0 at x = xMAX = 2.0 cm. Thus
EK + EP = ET = CONST ET = 0 + (1/ 2)kxMAX2 so that
4.0 = (1/ 2)k 0.0202 k = 20000 Nm-1.
x
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
the graph. The system consists
of a 0.125-kg mass on a spring.
(c) Determine the acceleration
of the mass at x = 1.0 cm.
SOLUTION:
From Hooke’s law, F = -kx we get
F = -20000(0.01) = -200 N.
From F = ma we get -200 = 0.125a a = -1600 ms-2.
x
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(a) How do you know that the
mass is undergoing SHM?
SOLUTION:
In SHM, a -x. Since F = ma, then F -x also.
The graph shows that F -x. Thus we have SHM.
x
Topic 4: Waves
4.1 – Oscillations
x
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
F = -5.0 N
EXAMPLE: A 4.0-kg mass is
x = 1.0 m
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(b) Find the spring constant of
the spring.
SOLUTION: Use Hooke’s law: F = -kx.
Pick any F and any x. Then use k = -F / x.
Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(b) Find the total energy of the
system.
SOLUTION: Use ET = (1/2)kxMAX2. Then
ET = (1/2)kxMAX2 = (1/2) 5.0 2.02 = 10. J.
x
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(c) Find the maximum speed of
the mass.
SOLUTION: Use ET = (1/2)mvMAX2.
10. = (1/2) 4.0 vMAX2
vMAX = 2.2 ms-1.
x
Topic 4: Waves
4.1 – Oscillations
-2.0
0.0
2.0
Sketching and interpreting graphs of simple harmonic
motion examples
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(b) Find the speed of the mass
when its displacement is 1.0 m.
SOLUTION: Use ET = (1/2)mv 2 + (1/2)kx 2. Then
10. = (1/2)(4)v 2 + (1/2)(2)12
v = 2.1 ms-1.
x