Ch 14 Notes: Oscillations and Waves -
Q: What constitutes a vibration, really?
A: Anything that switches to and fro; back and forth; side to side; off and on; loud and
soft; up and down, etc., is vibrating. A wiggle in time.
Q: What is a wave, really?
A: A wiggle in both space and time, extending from one point to another.
Q: What does it mean for something to oscillate?
A: Any, and all, of that stuff in the first “A” above. ☺
DEF: Sound = the propagation of vibration through a material, like a solid, liquid or a
gas.
Q: Can sound travel in a vacuum?
A: Nope; there is no medium through which it can travel.
DEF: Light = the propagation of electric and magnetic vibrations; a vibration of pure
energy.
Q: Can light travel in a vacuum?
A: Yep. Light can actually travel through many media, but it needs none!
14.1 Simple Harmonic Motion
Q: What does frequency mean? What is its symbol? What are its common units? What
does period mean? What is its symbol? What does “wavelength” mean? What is its
symbol? What are its common units? What does amplitude mean? What is its symbol?
What are its common units? What does “node” mean?
A: How many complete wave cycles pass a point in a given second. ν . Hertz = Hz. Your
book uses f, which many prefer for obvious reasons. The time it takes for one complete
wave cycle to pass a given point. Though closely related to frequency, frequency is
actually a “count” where period is a “time” to make the count. Τ. Frequency and period
are inversely related. The horizontal length of one complete wave cycle = the distance
from crest-to-crest (or trough-to-trough). λ The vertical height of a wave. a. cm. The
“zero” of a wave, or where it crosses the x-axis.
Can you tell the difference between whether the same exact note is played on a
versus a
? That’s because each of them has their own “quality” or
“timbre.” Often what we do, or do not, like about someone’s voice is the “quality” of their
voice. The reason we can detect a difference in what instrument is playing, or who is
singing, is due to the fact that those sounds are comprised of the superimposition of many
different tones differing in frequency. These various tones that make up the note we hear
are called partial tones, or just partials. The lowest of these frequencies, called the
fundamental frequency, determines the actual pitch of the note.
DEF: Harmonics = partial tones whose frequencies are whole-number multiples of each
other. A tone with twice the frequency of the harmonic is called the second harmonic. A
tone with three times the frequency of the harmonic is called the third harmonic, and so
on. It’s the variety of the partials in a note that give it its quality, or timbre. Though
we’ve been yabbering about sound, these definitions are applicable elsewhere, as well.
One of the simplest types of oscillatory motion is called simple harmonic motion. Like
you see in Fig 14-1 below, in equilibrium, the spring exerts no force on the mass. When
the object is displaced x amount from its original position, the spring exerts a force, as
given by Hooke’s Law. (‘Member it from Ch 4? ☺) F = - kx
Figure 14-1
Q: What does “k” really mean? What up with the “negative” sign?
A: k is the force constant of the spring, which reflects its elasticity (or stiffness). The
minus sign tells you it’s a restoring force.
If you substitute N2 for the left-hand side of Hooke’s Law, you get - kx = ma.
Or, a = - k x but a is d2x = - k x
m
dt2
m
14-2
☺
So what this really means is that a is proportional to the displacement, but in the opposite
direction. That’s cool, even if you didn’t think so initially. ☺
So, if the net force is proportional and opposite to the direction of motion of an object, it
will oscillate.
As stated above, f = 1 / Τ.
Q: What is ω?
A: Normally, it is angular velocity, but in this unit, it will also be understood as angular
frequency, where the units are radians per second, and the dimensions are inverse time.
You can relate x and t for the mass of a spring by using
The “phase” of the argument in eq 14-4 is ωt + δ where δ is the phase constant, or the
value of the phase when t = 0. The maximum displacement from equilibrium is called the
“amplitude,” A, of the function.
NOTE: Whether a function is described in terms of cosine or sine depends on the phase of
oscillation, though you can convert cos to sin by the addition of π/2.
If you take the derivative of eq 14-4 wrt time, you get
Differentiating again wrt time gives
a = dv = d2x = - ω2A cos (ω
ωt + δ)
2
dt dt
14-6
☺
If you go back to 14-4, and substitute for x, you get
Since 14-2 and 14-7 are both in terms of a, we can set them equal to each other to get
ω = √ (k/m)
14-8
Given the initial position, x0 and the initial velocity, v0, if you let t = 0 (14-4), you can find
A and δ
x0 = A cos δ
14-9
Also, if t = 0 in v = dx/dt = - A ω (sin ωt + δ), you get
v0 = - A ω sin δ
14-10
As you already know, the period for sin and cos are both 2π
π. Another way of looking at
this is that sin and cos repeat when the phase is increased by 2π
π.
So, ωΤ = 2π
π, or ω = 2π
π/Τ
Every time t increases by Τ, the phase increases by 2π
π and one cycle of motion is
completed. Since frequency is the reciprocal of the period,
Substituting in eq 14-8, we get
So what this really means is that f increases with increasing k, and decreases with
increasing mass.
See Ex 14-1, p 429
You are on a boat, which is bobbing up and down. The boat’s vertical displacement is
given by
Y = (1.2 m) cos [(1 / 2s)(t) + π / 6]
a) Find the amplitude, angular frequency, phase constant, regular frequency, and period
of the motion.
b) Where is the boat at t = 1.0 s?
c) Find the velocity and accelerations as functions of time t.
d) Find the initial values of the position, velocity, and acceleration of the boat.
a) From 14-4, we know:
A = 1.2 m
ω = 1 /2 rad / s
f = ω/2π
π = 0.0796 Hz
δ = π/6 rad
Τ = 1/f = 12.6 s
b) y = (1.2 m) cos [(1 / 2s)(1.0 s) + π /6] = 0.624 m
c) vy = dy/dt = d/dt (A cos (ω
ωt + δ) = - ωA sin (ω
ωt + δ) = - ½ s(1.2 m) sin [(1/2s)(t) + π /6]
= - (0.6 m/s) sin [(1/2s)(t) + π /6]
ay = dvy/dt = d/dt [-ω
ω A sin (ω
ωt + δ)] = - ω2 A cos (ω
ωt + δ) = - ½ s(1.2 m) cos[(1/2s)(t) + π /6]
= - (0.3 m/s) cos [(1/2s)(t) + π /6]
d) Set t = 0 to find all of the initial values requested
y0 = (1.2 m) cos (π
π/6) = 1.04 m
vy,0 = (- 0.6 m/s) sin (π
π/6) = - 0.300 m/s
2
ay,0 = (-0.3 m/s ) cos (π
π/6) = - 0.260 m/s2
Q: What does an “amp” do?
Does an amp change the note?
A: Increases the amplitude of the sound. Nope, just the volume. ☺
NOTE:
What this really means is that the pitch, which corresponds to the frequency, does not
depend on how loudly the note is played. Rock on!! ☺
The visual proof of this is shown in Fig 14-4 below. Both objects are playing the same
note, but object 2 obviously has a higher amplitude than object 1, so it’s louder than
object 1. They both reach their respective equilibrium at the same time.
Figure 14-4
See ex 14-2, p 431
An object oscillates with an angular frequency ω = 8.0 rad/s. At t=0, the object is at 4.0
cm with an initial velocity v0 = - 25 cm/s.
a) Find the amplitude and phase constant for the motion.
b) Write x as a function of time
Use eq’s 14-4 and 14-5. If t = 0, x0 = A cos δ
and
v0 = - ωA sin δ
Since we don’t know A, and we do know of a relationship between sin and cos, if we divide
v0 by x0, we’ll get v0 = - ωA sin δ = - ω tan δ
x0
A cos δ
If we cross-multiply-solve (or just divide by - ω), we get
tan δ = -v0 / ω x0 so δ = tan -1 (- v0 / ω x0) = - ____- 25 cm/s____ = 0.663 rad
(8.0 rad/s)(4.0 cm)
You can find A using either the x0 or v0 eq’s. A = __x0___ = __4.0 cm__ = 5.08 cm
cos δ
cos 0.663
b) Plug back into eq 14-4 to write x as a function of t, and you get
x = (5.08 cm) cos [(8.0 s-1)t + 0.663]
Absolutely worth knowing:
Determining δ:

v0
 ω x0
δ = tan −1  −

y 
 or δ = sin −1  0  for vertical motion
 A

When δ = 0, eq’s 14-4, 14-5, and 14-6 become:
x = A cos ωt
14-13a
v = - ωA sin ωt
14-13b
a = - ω2 A cos ωt
14-13c
These are shown below in Fig 14-5
Figure 14-5
Analyzing these graphs conceptually, where x, v, and a are functions of time, and δ = 0, at
t=0, displacement is a maximum, v=0, and a = - ω2A. The velocity becomes negative as the
object moves back toward its equilibrium position. Does that mean the value has shrunk?
After ¼ period (t = Τ/4), the object is at equilibrium, x=0, a=0, and the speed is maxed =
ωA. At t = Τ/2, the displacement = -A, v=0 again, and a = +ω
ω2A. At t= 3Τ
Τ/4, x=0, a=0, and
v=+ω
ωA.
See Ex 14-3, p 432.
1st: Calculate ω from ω = k / m
2nd: Use the result to find f and T
3rd: Find A and δ from the initial conditions
4th: write x(t) using the above results
A 2-kg block is attached to a spring as shown.
The force constant of the spring is k = 196 N/m.
The block is held a distance 5 cm from the
equilibrium position and released at time t = 0.
a)
Find the angular frequency, ω, the
frequency, f, and the period, T.
b) Write x as a function of time.
ω = 9.90 rad/s (divide by 2 π)
f = 1.58 Hz; T = 0.635 s (reciprocals)
A = 5 cm;
δ=0
x = (5 cm) cos(9.90 s-1)t
See ex 14-4, p 432-433.
Consider an object on a spring whose position is given by x = (5.0 cm) cos (9.90 s-1 t).
a) What is the maximum speed of the object?
b) When does this max speed first occur?
c) What is the max of the acceleration?
d) When does the max acceleration first occur?
a) If the object is released at rest, δ = 0 and x, v, and a are given by eq’s 14-13a – c.
x = A cos ωt
v = dx/dt = - ωA sin ωt
the max speed occurs when |sin ωt|| = 1
so |v|| = ωA |sin ωt||
|v||max = ωA = (9.90 rad/s)(5.0 cm) = 49.5 cm/s
b) |sin ωt|| = 1 first occurs when ωt = π/2 so, at π/2, 3π
π/2, 5π
π/2, …
If you solve for t when ωt = π/2
c) To find a, take dv/dt.
then t = π/2ω
ω = ___π
π___ = 0.159 s
2(9.90 s-1)
A = dv/dt = - ω2A cos ωt
the max acceleration happens when cos ωt = -1
amax = ω2A (negatives cancel each other out) = (9.90 rad/s)2(5.0 cm) = 490 cm/s2 = 4.90 m/s2
NOTE: This is pretty much = ½ g
d) The max acceleration happens when |cos ωt|| = 1
Thus, t = π/ω
ω = π/(9.90 s-1) = 0.317 s
so
ωt = π
NOTE: a maxes out when t = the ½ multiples of Τ; 0, Τ/2, 2Τ
Τ/2, 3Τ
Τ/2, 4Τ
Τ/2, etc.
Simple Harmonic Motion and Circular Motion
Q: What keeps them moving in a circle?
A: Centripetal force, of course!
Figure 14-6 a
Figure 14-6 b
Q: Are particles moving due to a centripetal force the only kinds of particles moving in
simple harmonic motion?
A: No. Any motion that is oscillatory, and whose graph of the position of its shadow
versus time produces a sinusoidal wave is considered to move with simple harmonic
motion, or SHM.
14-2 Energy in Simple Harmonic Motion (SHM)
3 Examples of Objects that move with SHM
1. Mass on a Spring
This demonstration shows the motion of a mass connected to a horizontal spring. This motion was
started by displacing the mass from its resting position (when the spring is neither compressed nor
stretched) and then releasing it.
2. Pendulum
This demonstration shows the motion of a mass connected to a (massless and inelastic) spring. This
motion was started by displacing the mass from its resting (lowest) position and then releasing it.
3. Uniform Circular Motion
This demonstration shows the uniform motion of a particle along a circular trajectory. This type of
motion results when the particle is subject to a central force, e.g., from the tension in a string or the
gravity of a central mass.
Q: What Do These Motions Have in Common?
Compare these three motions shown together in this demonstration. What do they have in common?
(Hint: compare the motion of their shadows.)
If an object oscillates, its PE and KE vary with time, but their sum is constant.
In Ch 6, you first learned about PE (U).
If you substitute-solve with eq 6-21, where Uspring = ½ kx2, you get
We also learned that KE = ½ mv2. If we substitute ω2 = k/m, we get
The total ME = KE + PE, or E = K + U. From trig, we know sin2θ + cos2θ = 1. So if we
combine eq’s 14-15 and 14-16 above, we get
What this really means is
When an object is at its maximum displacement, all of the E is U. As it moves toward
equilibrium, K ↑’s and U ↓’s. As it reaches equilibrium, K is a max, and U = 0, just like
we learned about in 10th grade, with the bowling ball, and this year again, when we
studied K and U.
Figure 14-7 below illustrates the K:U relationship.
Figure 14-7
In Figure 14-8, U is graphed as a function of x. The total energy, E, is constant, and
therefore plotted as a horizontal line. This line intersects the U curve at x = A and x = -A.
These are the points at which oscillating objects reverse direction and head back toward
the equilibrium position. These are called the turning points. Any relation to calculus
here? ☺
Figure 14-8
NOTE: Because U ≤ Etotal, the motion is restricted to - A ≤ x ≤ +A
Absolutely worth knowing:
Velocity as a function of position: v = ±
(
k 2
A − x2
m
)
Determination
Try ex 14-5, p 435 – 436 now.
energy.
(It’s easy! ☺) Make sure you use the proper units for
The following applet shows the relation between simple harmonic motion (SHM), the
equation for a circle, spring motion, and the corresponding sinusoidal graph.
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=148
This next applet shows the relationship between SHM, tangential and centripetal velocity.
Call up the control panel on the lower left to operate the applet.
http://dev.physicslab.org/asp/applets/circularSHM/default.asp