Chapter 13
Simple Harmonic Motion
 A single sequence of moves that constitutes the
repeated unit in a periodic motion is called a cycle
 The time it takes for a system to complete a cycle is
a period (T)
Simple Harmonic Motion
 The period is the number of units of time per cycle;
the reciprocal of that—the number of cycles per unit
of time—is known as the frequency (f).
f 
1
T
 The SI unit of frequency is the hertz (Hz), where 1
Hz = 1 cycle/s = 1 s-1
Amplitude (A) is the maximum displacement of an
 object
in SHM
Simple Harmonic Motion
 One complete orbit (one cycle)
object sweeps through 2p rad
 f - number of cycles per second
 Number of radians it moves
through per second is 2pf - that's
angular speed (w)
2p
w  2pf 
T


w - angular frequency
Sinusoidal motion (harmonic) with a single frequency
- known as simple harmonic motion (SHM)
Displacement in SHM
x  A cos  xmax cos wt
Velocity in SHM
vx   Aw sin wt
vx   Aw sin wt   vmax 1  ( x / A) 2
vmax = Aw
Acceleration in SHM
a x   Aw 2 cos wt
a x  w x
2
amax   Aw
2
The acceleration of a simple harmonic oscillator is proportional
to its displacement
Example 1
 A spot of light on the screen of a computer is





oscillating to and fro along a horizontal straight line
in SHM with a frequency of 1.5 Hz. The total length
of the line traversed is 20 cm, and the spot begins
the process at the far right. Determine
(a) its angular frequency,
(b) its period,
(c) the magnitude of its maximum velocity, and
(d) the magnitude of its maximum acceleration,
(e) Write an expression for x and find the location of
the spot at t = 0.40 s.
Example 1
 Given: f = 1.5 Hz and A = 10 cm
 Find: (a) w, (b) T, (c) |vx(max)|, (d) |ax(max)|, (e) x
in general, and x at t = 0.40s
From Eq. (10.12)
(a) w = 2pf = 2p(1.5 Hz) = 9.4 rad/s = 3.0p rad/s
(b) T= 1/f= 1/1.5 Hz = 0.67s
(c) |vx(max)| = vmax = Aw = 2pfA
vmax = 2p(1.5 Hz)(0.10 m) = 0.94 m/s
Example 1
(d) From Eq. (10.17),
|ax(max)| = Aw2 = A(2pf)2= (0.10 m)(2p 1.5 Hz)2
= 8.9 m/s2
(e) x = A cos wt = (0.10 m) cos (9.4 rad/s)t
At t = 0.40 s
x = (0.10m) cos (3.77 rad) = (0.10 m)(-0.81) = -8.1 cm
Problem
 A point at the end of a spoon whose handle is
clenched between someone’s teeth vibrates in SHM
at 50Hz with an amplitude of 0.50cm. Determine
its acceleration at the extremes of each swing.
Equilibrium
 The state in which an elastic or
oscillating system most wants to be in
if undisturbed by outside forces.
Elastic Restoring Force
 When a system oscillates naturally
it moves against a restoring force
that returns it to its undisturbed
equilibrium condition
 A "lossless" single-frequency ideal
vibrator is known as a simple
harmonic oscillator.
An Oscillating Spring
 If a spring with a mass attached to
it is slightly distorted, it will
oscillate in a way very closely
resembling SHM.
 Force exerted by an elastically
stretched spring is the elastic
restoring force F, = -ks.
 Resulting acceleration ax = (k/m)x

F is linear in x; a is linear in x - hallmark of SHM
Frequency and Period
Simple harmonic oscillator
Shown every ¼ cycles for 2
cycles
Relationship between x, vx, t,
and T
Hooke’s Law
 Beyond being elastic, many materials deform in
proportion to the load they support - Hooke's Law
F s
Hooke’s Law

The spring constant or elastic
constant k - a measure of the
stiffness of the object being
deformed
F  ks

k has units of N/m
Hooke’s Law

k has units of N/m
Frequency and Period
 w0 - the natural angular frequency, the specific
frequency at which a physical system oscillates all
by itself once set in motion
k
natural angular frequency
w0 
m
 and since w0 = 2pf0
natural linear frequency
1
2p
k
m
T  2p
m
k
f0 
 Since T= 1/f0
Period
Resonance vs. Damping
• If the frequency of the disturbing force
equals the natural frequency of the system,
the amplitude of the oscillation will
increase—RESONANCE
• If the frequency of the periodic force does
NOT equal the natural frequency of the
system, the amplitude of the oscillation will
decrease--DAMPING
Example 2
 A 2.0-kg bag of candy is hung on a vertical,
helical, steel spring that elongates 50.0 cm
under the load, suspending the bag 1.00 m
above the head of an expectant youngster.
The candy is pulled down an additional 25.0
cm and released. How long will it take for
the bag to return to a height of 1.00 m above
the child?
Example 2
 Given: m = 2.0 kg, ∆L = 50.0 cm, and A = 25.0 cm.
Find: t when y = 0.
 To find T, we must first find k
 Initially the load (F = Fw) stretches the spring
F  k (L)  mg
mg (2.0 kg)(9.81m / s 2 )
k

 39.2 N / m
L
0.50 m
T  2p
m
2.0 kg
 2p
 1.4 s
k
39.2 N / m
¼T = 0.35 s
The Pendulum
 The period of a pendulum is
independent of the mass and
is determined by the square
root of its length
f0 
1
2p
T  2p
g
L
L
g
Example 3
 How long should a pendulum be if it is to have a
period of 1.00 s at a place on Earth where the
acceleration due to gravity is 9.81 m/s2?
 Given: T = 1.00 s and g = 9.81 m/s2. Find: L
 From T2 = 4π2 (L/g); hence
T 2 g (1.00 s) 2 (9.81m / s 2 )
L

 0.248 m
2
2
4p
4p
Problem 2
 What would the length of a pendulum need to be on
Jupiter in order to keep the same time as a clock on
Earth? gJupiter = 25.95m/s2
When “f” is known
1
f 
T
w  2pf 
2p
T
x  A cos  xmax cos wt
vmax = Aw
amax   Aw
2
When “f” is NOT known
F  ks
1
f0 
2p
k
m
m
T  2p
k
k
w0 
m
1
f0 
2p
g
L
L
T  2p
g