Download Chapter 11: Simple Harmonic Motion

Chapter 11
LECTURE NOTES
SIMPLE HARMONIC MOTION
Consider a mass m at rest at the end of an un-stretched spring. The location of the mass is the
equilibrium position x = 0.
m
x= 0
m
x=A
Now, stretch the spring as shown, moving the mass to x = A. If the mass is then released from
rest, it will oscillate (in the absence of friction) forever between x = ± A.
When the spring is stretched an amount x, it feels a restoring force F = -kx that tries to return the
system to its equilibrium position. Note if x is to the left or negative, the same equation applies,
the force being to the right. This equation F = -kx is Hooke’s law. Any system described by this
type of equation is called a simple harmonic oscillator; the back and forth oscillations are simple
harmonic motion.
The variable k is called the spring constant and for a range of displacements is indeed constant.
We find k by subjecting a spring to a known force and noting how far it stretches; then k = | F/x |
in N/m.
In simple harmonic motion (SHM) the maximum displacement from equilibrium (in m) is the
amplitude A. The spring above oscillates between ± A. One cycle of the motion is “there and
back again”; from x = A to x = -A and back to x = A for example. The time to complete one
cycle is the period T, measured in seconds. The number of cycles per second is the frequency f,
measured in Hz. Note for a spring hanging vertically, if we measure x from the equilibrium
position of the mass at rest (mg/k lower than the spring alone), all we say applies as in the
horizontal case.
Energy
Recall the elastic potential energy stored in a stretched or compressed spring is PEelastic =
1 2
kx .
2
1
kA2. When the mass is at x = 0, PEmin = 0. The mass
2
moves quickly near equilibrium and slowly near x = ± A. Without friction, the total mechanical
1
1
energy is conserved so E = mv2 + kx2.
2
2
Note: E is proportional to the square of the amplitude.
When the mass is at x = ± A, PEmax =
Chapter 11
LECTURE NOTES
Note: At x = 0, all the energy is kinetic and
In general, v =
1
1
mv 2  kA2 so v = ± A k / m
2
2
k 2
(A  x 2 )
m
Relation between SHM and Circular Motion
The motion of the mass at the end of the spring can be considered a projection or shadow of the
motion of a mass revolving about a circle of radius A at constant velocity v.
v
A

x
We have position x = A cos  = A cos (wt) and
from calculus
v = -A  sin (  t)
a = -A  2 cos (  t) and
F = ma = = -mA  2 cos(  t)
where A corresponds to the amplitude of our
motion and (recall v = r  for circular motion)
 = v/A.
The time for the mass to revolve about the circle or cycle back and forth is the period T =
2A 2

.
v

1

Since f =
is the frequency of the motion =
we have  = 2  f. We call  the angular
2
T
frequency. At x = 0, the velocity v is all in the x direction as the mass rotates about the circle.
k
k
So v = ±
. This connects our physical spring and imaginary
A  A implying  =
m
m
2
m
circular motions. Using this  we find T =
is the period (in s) in terms of the mass
 2

k
(in kg) at the end of the spring with spring constant k (in N/m).
 k 
.
Also x = A cos (  t) = A cos  t

m


The formulas for v, a, and F can similarly replace  by
k
.
m
Example: A mass m = 1.00kg oscillates at the end of a spring (k = 39.4 N/m) on a horizontal ,
frictionless surface. At time t = 0s, the mass is at x = +2.00m. Its equilibrium position is x = 0.
Analyze the motion. We note the amplitude A = 2.00m; the mass will oscillate between
x = ±2.00m.
Chapter 11
LECTURE NOTES
k
k
= 6.28/s, we have f = 1.00 Hz and T = 1.00s. vmax =
A = 12.6m/s, amax =
m
m
k
1
A = 78.8m/s2, Fmax = 78.8N, and E = kA2 = 78.8J. Thus at t = 0s the mass is at x = +2.00m;
2
m
the PE stored in the stretched spring is 78.8J. The velocity of the mass and its KE is zero, the
acceleration is
–78.8 m/s2 (left) and the force – 78.8N (left). One fourth of a cycle later at t = .25s, the mass is
at equilibrium x = 0. Its velocity is now maximum (left) at –12.6 m/s, its KE is 78.0J, and the PE
stored in the spring, the acceleration a, and the force F are all 0. The motion is very regular and
symmetric. At = .75s the mass is back at equilibrium moving right; and the motion is back to
start at t = 1.00s.
Since

The Pendulum
This system can be described as a simple harmonic oscillator for small oscillations in the absence
of air resistance.

L

mg
The weight of the pendulum “bob” mg can be
broken into components mg cos  along the
line of the string and mg sin  which is in the
direction of the velocity of the bob (tangent to
the arc). This restoring force F = -mg sin 
is approximately -mg  where  in radians is
a small angle less than 15° (less than 5° or .1
rad given even better results).
From the large triangle we see the displacement x is about the actual arc displacement L  and
mg
 mg 
can be substituted for k in our previous
 = x/L. Thus F = -mg sin  ≈ - 
 x and
L
 L 
L
m
becomes T = 2 
. This is an
g
k
approximate result for small  . Note T does not depend on the mass of the bob, is proportional
to L and inversely to proportional to g .
analysis of simple harmonic motion. Thus period T = 2 
Note: A pendulum of length L = 1.00m has a period of T = 2.00s on earth at sea level.

L  1 2  9

1  sin    sin 4    ...
Note: A more exact treatment gives T = 2 

g 4
 2  64
2

Waves
Energy can be transported from one place to another by waves. Waves on water, for example
transport energy but the water molecules merely move up and down while the waves passes; the
water does not travel with the wave.
Chapter 11
LECTURE NOTES
A water wave is a transverse wave and can be pictured by a sine curve with a crest and trough.
2
+A
0
5
1
x
3
-A
4
The greatest displacement is at points 2 and 4, the crest and trough; we say the amplitude of a
wave that lies between y = ± A is A (in m). The horizontal distance x between points 1 and 5 is
the wavelength  . Actually, this is the distance between any two points of adjacent waves that
are in phase (same amplitude and motion). The frequency f (in Hz) is the number of waves that
pass an observer each second. The period T (in s) is the time required for 1 wave to pass the

observer. We have T = 1/f. If the wave travels at velocity v, we have v   f .
T
Waves can also be longitudinal, where the particles of the medium vibrate along (instead of
perpendicular to) the direction of travel. Sound waves are of this type. We will produce both
longitudinal and transverse waves in our lab.
The intensity of a wave, defined as the power per unit area, is proportional to the square of the
wave's amplitude.
Waves can reflect from an obstacle. The angle of reflection will equal the angle of incidence.
Both angles are customarily measured to the normal or perpendicular to the surface. Waves that
reflect from a rigid barrier are inverted in phase.
The principle of superposition states when waves pass through the same space at the same time
the resulting displacement is the algebraic sum of the separate displacements.
Waves that interact in this way are exhibiting interference.
In constructive interference, the waves add to produce a greater disturbance.
In destructive interference, a crest meets a trough and all is calm.
When a wave passes from one medium into another (or a region of the same material with
different properties) the wave may exhibit refraction. If the velocities are v1 and v2 in the two
regions, a wave entering the boundary at an angle  1 from a normal will exit at  2
sin  2 v 2

where
. This is the law of refraction.
sin 1 v 1
Waves also exhibit diffraction when they bend around obstacles similar in size to the
wavelength. The angular spread of the waves in radian is about  =  /L where L is the size of
the obstacle.
Chapter 11
LECTURE NOTES
Waves on a string
If a string is stretched tightly between two fixed ends (as in stringed musical instruments), waves
can be excited to travel back and forth along the string. Only certain waves survive for long.
These are waves that vibrate at a resonance frequency or whose length just fits the string so its
back and forth oscillations produce a standing wave.
For a string of length L fixed at both ends, its lowest tone occurs when there is one “loop” so
1
L = .1 or  1  2L . We call this frequency f1 the fundamental or first harmonic. The fixed
2
ends are nodes and the center point of maximum displacement an antinode.
The string will also resonate at a second harmonic f2 frequency (called the first overtone) when
 2 = L. There are 3 nodes and 2 antinodes. We find f2 =2f1.
The third harmonic f3 (or second overtone) occurs when  3 =
2
3
L (or L contains  3 ). We find
3
2
f3 = 3f1.
n n
; the resonating standing waves are those where the length of the
2
2L
string contains an integral number of their half wavelengths. We write  n 
.
n
In general, we find L =
FT
where v (in m/s) is the
m/L
wave velocity, FT (in N) the tension of the string and m/L, the mass per unit length of the string
(in kg/m).
Now, the velocity of a wave on a string can be shown to be v =
Since v = f  we find fn =
v
nv

 nf 1 for the resonant frequencies or f n  n
 n 2L
FT m
L
.
2L
It is interesting that current theories picture subatomic particles as resonating strings connecting
modern physics with its origin in the Greek study of music 2500 years ago.