Chapter 10: Simple Harmonic Motion
Oscillations are back-and forth motions. Sometimes the word
vibration is used in place of oscillation; for our purposes, we can
consider the two words to represent the same class of motions.
In the ideal case of no friction, free oscillations are a sub-class of
periodic motions; that is, in the absence of friction, all free
oscillations are periodic motions, but not all periodic motions
are oscillations. For example, uniform circular motion (which we
studied earlier in this course) is periodic, but not considered an
oscillation.
Uniform circular motion can be modelled by sine or cosine
functions of time (think back to the unit circle in high-school
math when you were learning trigonometry). (Sine and cosine
functions are collectively known as sinusoidal functions, or
sinusoids for short.) If the restoring force that causes oscillation
is a linear function of position, then the resulting oscillatory
motion can also be modelled by a sinusoidal function of time.
There is a close relationship between uniform circular motion
and oscillatory motion caused by a linear restoring force; we
won't explore this, but check the textbook (Section 10.2) if
you're interested.
What is a restoring force? What is a linear restoring force?
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Notice that the formula for Hooke's law is represented
by the graph; the magnitude of the restoring force is
proportional to the magnitude of the displacement
from equilibrium, and in the opposite direction. The
constant of proportionality is the stiffness constant of
the spring.
Q: What are the units of k? What are some typical
values for the stiffness constant for coil springs in your
experience (ones in your car's shock absorbers, in your
ball-point pen, attached to your aluminum door, etc.)?
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ball-point pen, attached to your aluminum door, etc.)?
Here is an example position-time diagram for an
oscillation:
Notice that the position-time diagram for the
oscillation resembles the graph of a sinusoidal
function. You'll get a chance to see that this must be
so for an oscillator that is subject to a linear restoring
force (Hooke's law) later in the chapter. Now is a good
time to review sinusoidal functions, so let's do it:
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Exercises
1. Determine the amplitude, period, frequency, and angular
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1. Determine the amplitude, period, frequency, and angular
frequency for each function. In each case, time t is measured in
seconds and displacement x is measured in centimetres.
(a)
(b)
c.
(d)
2. Sketch a graph of each position function in parts (a), (b), and (c)
of Exercise 1.
3. Calculus lovers only! Determine a formula for the derivative of
the sine function, and a formula for the derivative of the cosine
function, valid for angles measured in degrees.
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Exercises
4. Consider an oscillation with position function
x = 20 cos (4t)
where x is measured in cm and t is measured in s.
(a) Determine the positions on the x-axis that are turning
points.
(b) Determine the times at which the oscillator is at the
turning points.
(c) Determine the times at which the oscillator is at the
equilibrium position.
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equilibrium position.
(d) Determine the times at which the speed of the
oscillator is at (i) a maximum, and (ii) a minimum.
(e) Determine the times at which the acceleration of the
oscillator is at (i) a maximum, and (ii) a minimum.
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Note that the period (and therefore also both the frequency
and angular frequency) does not depend on the amplitude of
the oscillation. This is interesting. Does this match with your
experience?
Exercises
5. What would the position-time graphs look like for an oscillator
that is released from several different starting amplitudes?
6. Consider an oscillator of mass 4 kg attached to a spring with
stiffness constant 200 N/m. The mass is pulled to an initial
amplitude of 5 cm and then released.
(a) Determine the angular frequency, frequency, and period of
the oscillation.
(b) Write a position-time function for the oscillator.
7. A block of mass 3.2 kg is attached to a spring. The resulting
position-time function of this oscillator is x = 23.7 sin(4.3t),
where t is measured in seconds. Determine the stiffness of the
spring.
Elastic potential energy
An argument similar in spirit to the one above for gravitational
potential energy shows that the formula for elastic potential
energy is (see page 285ff in Section 10.3 of the textbook for
details; calculus lovers, antidifferentiate Hooke's law)
where x is the displacement of the spring from its equilibrium
position and k is the stiffness constant of the spring.
Where is potential energy stored? Answering this question leads
to the concept of a force field; we'll discuss force fields in PHYS
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to the concept of a force field; we'll discuss force fields in PHYS
1P22/1P92.
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Exercises
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8. Consider a block of mass 5.1 kg attached to a spring. The
position-time function of this oscillator is x = 8.2 sin(2.7t),
where x is measured in cm and t is measured in seconds.
(a) Determine the total mechanical energy of the oscillator.
(b) Determine the stiffness constant of the spring.
(c) Determine the maximum potential energy.
(d) Determine the maximum kinetic energy.
(e) Determine the positions at which the kinetic energy
and the potential energy of the oscillator are equal.
Pendulum motion
Recall from earlier in these lecture notes that for a block on
the end of a spring, applying Newton's law to the block
results in
This is an example of a differential equation, and to solve a
differential equation means to determine a position function
x(t) that satisfies the equation. You'll learn how to do this in
second-year physics (PHYS 2P20), and second-year math
(MATH 2P08), but for now you can verify that position
functions of the form
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and
both satisfy the differential equation above, provided that 
satisfies a certain condition (the same condition that was
observed earlier in the notes).
From a different perspective, we can also infer that if a
physical phenomenon is modelled by a differential equation of
the form given above, the phenomenon is an example of
simple harmonic motion.
Let's consider a simple pendulum. First draw a free-body
diagram:
In the radial direction, applying Newton's second law gives:
In the tangential direction, applying Newton's second law
gives:
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This looks very similar to the differential equation
written earlier that represents simple harmonic motion.
But not exactly; if the sine theta were replaced by just
theta, then the equation would have the form of the
SHM differential equation.
Note that for small angles, the value of sine theta is
about equal to theta, provided that theta is measured in
radians:
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Construct a table of values using a calculator and
you'll see for yourself that sine theta is nearly equal
to theta for small angles, provided that theta is
measured in radians. The approximation becomes
increasingly better the smaller the angle is.
Also note that
as long as theta is measured in radians; this is an
example of a power series, which you'll learn about
later in MATH 1P06 or MATH 1P02, if you are
taking either of these two courses. For small values
of theta, higher order terms are very small, and so
this formula embodies the approximation given
above.
Thus, for a pendulum with a small amplitude (so
that the angle is always small), the motion is
approximately SHM, described by the differential
equation
Substituting the model
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into the differential equation leads to expressions
for the period and frequency:
Substituting this expression into the differential
equation gives:
Thus
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and therefore
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Swinging your arms while running or walking; note
how the period of the swing is modified by
changing the effective length of your arms:
Damped oscillations
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Damped oscillations
Cars have shock absorbers to make the ride smoother.
A shock absorber consists of a stiff spring together with
a damping tube. (The damping tube consists of a piston
in an enclosed cylinder that is filled with a thick (i.e.,
viscous) oil.)
Without the damping tube, a car would oscillate for a
long time after going over a bump in a road; the
damping tube helps to limit both the amplitude and
duration of the oscillations.
When the damping tube doesn't work anymore, the
car tends to oscillate for a long time after going over a
bump, which is annoying. The same thing happens with
a screen door when its damping tube malfunctions.
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(There is a spring attached to the door, but it is not
shown in the photograph.)
The position function for a damped oscillator is
modified as follows:
You can think of this position function as representing
a sort of sinusoid, but one with a variable amplitude;
the amplitude is the constant A times the exponential
factor, which steadily decreases as time passes.
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The damping tube in a screen door is adjustable. If
the resistance is too great, then the door will take a
long time to shut after it is opened. If the resistance
is too little, then the door will swing back and forth
many times before shutting. There is an ideal
medium amount of resistance ("critical damping")
which works best; you'll learn more about this, and
see how these three cases (overdamping, critical
damping, and underdamping) follow naturally from
different classes of solutions to the appropriate
differential equation, in PHYS 2P20 and MATH 2P08.
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Resonance
Examples of driven oscillations:
•
•
child sloshing back and forth in a bathtub
adult pushing a child on a playground swing
Some systems have a natural oscillating frequency; if you drive
the system at its natural oscillating frequency, its amplitude
can increase dramatically. This phenomenon is called
resonance.
In many situations, one tries to avoid resonant oscillations. For
example, that annoying vibration in your dashboard when you
are driving on the highway at a certain speed … is very
annoying. More seriously, soldiers are trained to break ranks
when they march across a bridge, because if their collective
march is at the same frequency as the natural frequency of
the bridge, then there is danger that they could collapse the
bridge. (This is an ancient custom, from an age when many
bridges were made of wood.)
Engineers must be careful to design bridges and tall buildings
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Engineers must be careful to design bridges and tall buildings
so that they don't have natural vibration frequencies;
otherwise an unlucky wind could cause dangerous largeamplitude vibrations.
http://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_(1940)
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Another good example of resonance is tuned
electrical circuits, such as the ones used in radio or
television reception.
Radio waves of many different frequencies are
incident on a radio receiver in your home; each tries
to "drive" electrical oscillations in an electrical
circuit. The natural frequency of the electrical circuit
can be adjusted so that it will resonate with only a
certain frequency of radio wave; this is how you
"tune in" to a certain radio station. The oscillations
due to the resonant frequency persist, while all the
other frequencies are rapidly damped.
Answers to exercises from earlier in the notes
1. Determine the amplitude, period, frequency, and angular
frequency for each function. In each case, time t is measured in
seconds and displacement x is measured in centimetres.
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frequency for each function. In each case, time t is measured in
seconds and displacement x is measured in centimetres.
(a)
(b)
(c)
(d)
Solution:
(a) A = 4 cm, T =  s, f = 1/ cycles/s,  = 2 rad/s
(b) A = 3 cm, T =  s, f = 1/2 cycles/s,  =  rad/s
(c) A = 1/2 cm, T = 4 s, f = 1/(4 cycles/s,  = 1/2 rad/s
(d) This is not a sinusoidal function, and not a periodic
function either. Therefore this function does not have an
amplitude, a period, a frequency, or an angular frequency; it
cannot serve as a model for a simple harmonic motion.
2. Sketch a graph of each position function in parts (a), (b), and (c)
of Exercise 1.
Solution:
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3. Calculus lovers only! Determine a formula for the derivative of
the sine function, and a formula for the derivative of the cosine
function, valid for angles measured in degrees.
[See me for a discussion on this point … it's a bit complicated.]
4. Consider an oscillation with position function
x = 20 cos (4t)
where x is measured in cm and t is measured in s.
(a) Determine the positions on the x-axis that are turning
points.
(b) Determine the times at which the oscillator is at the
turning points.
(c) Determine the times at which the oscillator is at the
equilibrium position.
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equilibrium position.
(d) Determine the times at which the speed of the
oscillator is at (i) a maximum, and (ii) a minimum.
(e) Determine the times at which the acceleration of the
oscillator is at (i) a maximum, and (ii) a minimum.
Solution:
(a) x = 20 cm
(b) t = 0 s, /4 s, /2 s, 3/4 s,  s, etc.
(c) t = 0 s, /8 s, 3/8 s, 5/8 s, 7/8 s, etc.
(d) The speed of the oscillator is maximum at the times in
Part (c) (i.e., when the oscillator passes through the
equilibrium position); the speed of the oscillator is
minimum (i.e., zero) at the times in Part (b) (i.e., when
the oscillator is at a turning point).
(e) The accelerator of the oscillator is maximum at the
times in Part (b) (i.e., when the oscillator is at a turning
point); the accelerator of the oscillator is minimum (i.e.,
zero) at the times in Part (c) (i.e., when the oscillator
passes through the equilibrium position).
5. What would the position-time graphs look like for an oscillator
that is released from several different starting amplitudes?
Solution: The stiffness constant of the spring and the value of
the mass are the same, so the frequency and period of each
motion are the same. The only difference is the amplitude.
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6. Consider an oscillator of mass 4 kg attached to a spring with
stiffness constant 200 N/m. The mass is pulled to an initial
amplitude of 5 cm and then released.
(a) Determine the angular frequency, frequency, and period of
the oscillation.
(b) Write a position-time function for the oscillator.
Solution:
7. A block of mass 3.2 kg is attached to a spring. The resulting
position-time function of this oscillator is x = 23.7 sin(4.3t),
where t is measured in seconds. Determine the stiffness of the
spring.
Solution:
8. Consider a block of mass 5.1 kg attached to a spring. The
position-time function of this oscillator is x = 8.2 sin(2.7t),
where x is measured in cm and t is measured in seconds.
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where x is measured in cm and t is measured in seconds.
(a) Determine the total mechanical energy of the oscillator.
(b) Determine the stiffness constant of the spring.
(c) Determine the maximum potential energy.
(d) Determine the maximum kinetic energy.
(e) Determine the positions at which the kinetic energy
and the potential energy of the oscillator are equal.
Solution:
Ah, we don't know the value of the stiffness constant k,
so we should calculate it:
Thus,
(e) When the kinetic and potential energies are equal, each
is half the total energy. Therefore, the potential energy is
equal to half of the total energy at the positions of interest.
Thus,
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Additional Solved Problems
Example: An object in simple harmonic motion has an amplitude of 6.0 cm
and a frequency of 0.50 Hz. Draw a position-time graph showing two cycles of
the motion.
Solution: Should we use a sine model or a cosine model? It doesn't really
matter, so I'll use a sine model. Because the amplitude is 6.0 cm, the positiontime graph looks like this:
We just have to correctly label the horizontal axis now. We are given that the
frequency is 0.5 Hz, so we can determine the period:
Thus, one complete cycle of the position-time graph takes 2.0 s.
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We weren't asked for a formula, but using the general formula
we can write a specific formula for the position function of this motion:
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Example: Some passengers on an ocean cruise may suffer from motion
sickness as the ship rocks back and forth on the waves. At one position on the
ship, passengers experience a vertical motion of amplitude 1 m with a period
of 15 s. (a) To one significant figure, what is the maximum acceleration of the
passengers during this motion? (b) What fraction is this of g?
Solution: This time I'll use a cosine model; the reasoning would be the same
for a sine model.
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Because the cosine function takes values between -1 and 1, the maximum
value of the acceleration is related to the right side of the equation on the
previous line by
We are given that the amplitude A is 1 m. If we can determine the angular
frequency, then we will be able to calculate the maximum value of the
acceleration. But we are told the period, so we will be able to determine
the angular frequency:
Thus, the maximum value of the acceleration is
(b) In terms of the acceleration due to gravity, the maximum
acceleration of the ship's oscillation is
The maximum acceleration is only about 2% of the acceleration due to
gravity, so presumably it's not too bad on the ship.
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Example: (a) When the displacement of a mass on a spring is (1/2)A, what
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Example: (a) When the displacement of a mass on a spring is (1/2)A, what
fraction of the mechanical energy is kinetic energy and what fraction is
potential energy? (b) At what displacement, as a fraction of A, is the
mechanical energy half kinetic and half potential?
Solution: First determine a formula for the total energy of the oscillator. In
general,
However, this doesn't seem to help much, because we also need a way of
comparing how much of the total energy is of each type. What to do? Well,
note that when x = A, the oscillator momentarily stops, so its kinetic energy
is zero. Thus, the total energy at this position is
But remember, the total energy of the oscillator is conserved (there is no
friction). Thus, the total energy of the oscillator has the same value
throughout its entire motion. Thus, the ratio of the potential energy to the
total energy is
Thus, when the displacement is half of the amplitude,
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Thus, when the displacement is half of the amplitude,
Thus, when the displacement is half of the amplitude, the
potential energy is 1/4 of the total energy, and therefore the
kinetic energy is 3/4 of the total energy.
(b) Using the solution from Part (a),
Thus, the energy is shared equally between kinetic energy and potential
energy when the displacement is plus-or-minus about 70% of the amplitude.
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Example: A 1.0 kg block is attached to a spring with stiffness constant 16
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Example: A 1.0 kg block is attached to a spring with stiffness constant 16
N/m. While the block is sitting at rest, a student hits it with a hammer and
almost instantaneously gives it a speed of 40 cm/s. Determine (a) the
amplitude of the subsequent oscillations, and (b) the block's speed at the
position where x = (1/2)A.
Solution: From the given information we can immediately calculate the
angular frequency. I am guessing this might be useful, so let's do that first:
Because the block is at its equilibrium position and is given an initial speed,
a sine model is appropriate for its position function:
We are told that the oscillator's initial speed, which is also its maximum
speed, is 40 cm/s. Thus,
(b) As we learned in a previous problem, the total energy can be expressed
in terms of the amplitude as follows:
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(b) As we learned in a previous problem, the total energy can be expressed
in terms of the amplitude as follows:
To determine the block's speed when the amplitude is half of its maximum,
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Example: A 507 g mass oscillates with an amplitude of 10.0 cm on a spring
whose stiffness constant is 20.0 N/m. Determine (a) the period, (b) the
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whose stiffness constant is 20.0 N/m. Determine (a) the period, (b) the
maximum speed, and (c) the total energy.
Solution:
(b) The maximum speed occurs when x = 0:
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Example: On your first trip to planet X you happen to take along a 200 g
mass, a 40.0-cm-long spring, a metre stick, and a stopwatch. You're curious
about the free-fall acceleration on Planet X, where ordinary tasks seem easier
than on Earth, but you can't find this information in your visitor's guide. One
night you suspend the spring from the ceiling in your room and hang the mass
from it. You measure that the mass stretches the spring by 31.2 cm. You then
pull the mass down an additional 10.0 cm and release it. With your stopwatch
you measure that 10 oscillations take 14.5 s. Determine the value of the
acceleration due to gravity on Planet X.
Solution: Let g represent the acceleration due to gravity on Planet X. Placing
the mass on the spring stretches it by 31.2 cm, which is the same as 0.312 m.
This gives us a relation between g and the stiffness constant k of the spring,
which you can get by drawing a free body diagram and using Newton's second
law:
Now that we have a relation between g and k, we see that we can calculate g
provided that we can determine k independently of this equation. Re-reading
the statement of the problem, looking for unused information, it occurs to us
that perhaps we can use the fact that the oscillator completes 10 oscillations in
14.5 s to determine k. Indeed, this can be done:
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Therefore,
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Example: A 1.00 kg block is attached to a horizontal spring with stiffness
constant 2500 N/m. The block is at rest on a frictionless surface. A 10.0-g
bullet is fired into the block in the face opposite the spring, and it sticks.
(a) Determine the bullet's speed if the subsequent oscillations have an
amplitude of 10.0 cm.
(b) Could you determine the bullet's speed by measuring the oscillation
frequency? If so, how? If not, why not?
Strategy: When the bullet strikes the block, mechanical energy is not
conserved. That is, some of the initial kinetic energy of the bullet is converted
to sound, thermal energy, deformation of the bullet and block, and so on.
However, momentum is conserved in this process.
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However, momentum is conserved in this process.
Once the bullet has embedded into the block, mechanical energy is conserved
in the subsequent motion of the block + bullet system, because there is no
friction between the block and the surface on which it rests.
Thus: Use conservation of momentum to determine the speed of the bullet +
block just after the bullet is embedded. Then use conservation of mechanical
energy to relate the speed just determined to the amplitude of the
subsequent oscillations.
Solution: (a) The following diagrams are intended to make this strategy clear:
Momentum is conserved in the collision process; that is, the momentum just
before and just after the collision is conserved:
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After the bullet + block begin to oscillate, mechanical energy is conserved. Thus,
the initial kinetic energy of the system is equal to the maximum potential
energy stored in the spring:
Thus, the speed of the bullet is
(b) No. The frequency of the oscillation depends only on the stiffness
constant of the spring and the total mass of the bullet + block. The
oscillation frequency is independent of the bullet's speed, so
measuring the oscillation frequency can't possibly give us any insight
into the bullet's speed.
The greater the bullet's speed, the greater the amplitude of the
oscillation; thus, it's the amplitude of the oscillation that is connected
to the speed of the bullet.
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Example: A spring is hung from the ceiling. A 0.450-kg block is then
attached to the free end of the spring. When released from rest, the
block drops 0.150 m before momentarily coming to rest, after which it
moves back upward.
(a) Determine the spring's stiffness constant.
(b) Determine the block's angular frequency of oscillation.
Solution: See page 279 of the textbook, which explains clearly how to
think about the unstretched length of the spring and the amplitude of
the resulting oscillations. Carefully examine the following diagram:
When the block reaches the
lowest point of its motion,
the loss in the block's
gravitational potential
energy is balanced by the
gain in the spring's elastic
potential energy:
The block's angular frequency is therefore
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Example: A vertical spring with stiffness constant 450 N/m is
mounted on the floor. From directly above the spring, which is
unstrained, a 0.30-kg block is dropped from rest. It collides with and
sticks to the spring, which is compressed by 2.5 cm in bringing the
block to a momentary halt. Assuming air resistance is negligible,
determine the height (in cm) above the compressed spring from which
the block was dropped.
Solution:
By the principle of conservation of energy, the gravitational potential
energy of the block "before" is converted to elastic potential energy in
the spring "after." Thus,
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the spring "after." Thus,
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Example: A simple pendulum is swinging back and forth through a
small angle, its motion repeating every 1.25 s. How much longer
should the pendulum be made so that its period increases by 0.20 s?
Solution:
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Thus, the pendulum should be made 35% longer to increase its
period by 0.2 s.
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