Simple Harmonic Oscillator (SHO)
• Any situation where the force exerted on a mass
is directly proportional to the negative of the
object’s position from an equilibrium point is a
simple harmonic oscillator.
F  kx
• English: A simple harmonic oscillator is an
object that gets pushed toward some point, and
the farther the object is from that point, the
stronger the force pushing it back.
An Example of a SHO: The spring
X=0m
FN
F  F
N
 FG  0 N
Equilibrium Position
FG
An Example of a SHO: The spring
An Example of a SHO: The spring
ETotal  K .E.  P.E
ETotal 
P.E.Max
1 2 1 2
mv  kx
2
2
1
 kA2
2
K .E.Max 
1
2
mvMax
2
P.E.Max  K .E.Max
1 2 1
2
kA  mvMax
2
2
vMax 
2
ETotal  P.E.Max
1 2 1 2 1 2
mv  kx  kA
2
2
2
k 2
A
m
mv2  kx2  kA2
mv2  kA2  kx2
x2
mv  k ( A  x )  kA (1  2 )
A
2
2
2
2
kA2
x2
v 
(1  2 )
m
A
2
v  vMax
2
2
v  vMax
x2
(1  2 )
A
x2
(1  2 )
A
An Example of a SHO: The spring
An Example of a SHO: The spring
For all SHO:
x(t )  A cos[( 2f  t )  i ]
v(t )  vmax sin[( 2f  t )  i ]
a(t )  amax cos[( 2f  t )  i ]
Where:
vmax
k
 2Af  A
m
amax
k
 A
m
Sinusoidal Curve Generator
i  0 when x i  A
Let:
x i  A, so i  0
x(t )  A cos( 2f  t )
v(t )  vmax sin( 2f  t )
a(t )  amax cos(2f  t )
Since   t 
2
 t  2f  t ,
T
x(t )  A cos(  )
v(t )  vmax sin(  )
a(t )  amax cos(  )
*  is measured in radians
The Simple Pendulum
m = mass of bob
L = length of cord
FT
FG = weight, mg
FT= tension
x0
F  F
T
FG
 FG  0 N
Equilibrium Position
The Simple Pendulum
m = mass of bob
L = length of cord
FT
FG = weight, mg
FT= tension
x  L
x0
FG
x is measured along the arc
The Simple Pendulum
m = mass of bob
FGY=mg cos(θ)
L = length of cord
FGX=-mg sin(θ)
FT
FG = weight, mg
FT= tension
x  L
FGX
x0
The Assumption:
FGY
FG
x is measured along the arc
For sufficiently small  ,   sin 
FGX  mg sin(  )  mg
mg
FGX  
x This is in the format of Hooke’s
law, and thus we have a SHO
L
T of a Simple Pendulum
F  kx
FGX
(definition of a SHO)
mg

x
L
(restoring force of a pendulum)
mg
k
L
T  2
m
T  2
k
m
mg
 2
L
L
g
(eqn. 11-7)
(Period of a simple Pendulum)
Pendulum Lab Expected Results:
L
T  2
g
T  2g
1
T  (2g
2
(Period of a simple Pendulum)
1
L
1
2
y  ax1/ 2
2
)L
1
2
a  2g
1
2
Pg. 343, #34: A fisherman notices that wave crests pass the bow of his
anchored boat every 3.0 s. He measures the distance between two
crests to be 8.5 m. How fast are the waves traveling?
T=3.0 s
λ=8.5 m
Eqn. (11-12)
m

1
v  8.5Tm .33s  2.8
1
T
v=?
f T
v  f
1
f  (3.0s) 1  .33s 1
 8.5m
m
vOar.
 ..
 2.8
T 3.0 s
s
s
Pg. 343, #35: A sound wave in air has a frequency of 262
Hz and travels with a speed of 330 m/s. How far apart
are the wave crests (compressions)?
f=262 Hz
v  f
Eqn. (11-12)
v= 330 m/s
 ?
m
s  m 1  m s m
1
s s s 1
s
v

f
m
m
m
m
330
330
vv
s
s
s
s
  
11..26
26  1.26m
11
ff 262
262Hz
Hz
ss
Pg. 344, # 36: AM radio signals have frequencies
between 550 kHz and 1600 kHz and travel with a
speed of 3x108 m/s. On FM, the frequencies range
from 88.0 MHz to 108 MHz and travel at the same
speed. What are the wavelengths of these signals?
fAM low=550 kHz
fAM high=1600 kHz
v

f
λAM low f= 545 m
λAM high f= 188 m
v= 3x108 m/s
fFM low=88.0 MHz
λFM low f= 2.78 m
fFM high=108 MHz
λFM high f= 3.41 m
Pg. 344, #39: A cord of mass 0.55 kg is stretched between two
supports 30 m apart. If the tension in the cord is 150 N, how long
will it take a pulse to travel from one support to the other?
m = 0.55 kg
L = 30 m = Δx
FT = 150 N
Δt=?
v
FT
Eqn. 11-13
m
L
v = 90.5 m/s
x
v
t
x
30m
t 

 .33s
v 90.5 m
s
Pg. 344, #44: Compare the (a.) intensities and (b.) the
amplitudes of an earthquake wave as it passes two
points 10 km and 20 km from the source.
Let the wave have power P.
r10 km = 10,000 m
r20 km=20,000 m = 2r10 km
I10km
?
I 20km
I10km 
I 20km
P
2r 2
P
P


2
2 (2r )
8r 2
Power
I
Area
I10km
I 20km
P
2
2
2

r

 4
P
1
8r 2 8
I10km is 4 times greater than I20km
Pg. 344, #44: Compare the (a.) intensities and (b.) the
amplitudes of an earthquake wave as it passes two
points 10 km and 20 km from the source.
Let the wave have power P.
r10 km = 10,000 m
r20 km= 20,000 m = 2r10 km
I10km = 4 I20km
A10km
?
A20km
1
A
r
1
A10km
r10km

2
1
A20km
2r10km
Pg. 344, #45: The intensity of a particular earthquake
wave is measured to be 2.0 x 106 J/m2s at a distance of
50 km from the source.
a.) What was the intensity when it passed a point
only 1 km from the source?
Energy
Power
time
I50 km= 2.0 x 106 J/m2s
I

Area
Area
r50 km = 50,000 m
r1 km= 1000 m
P  I 50km Area50km  I 50km  2r50km
I1km  ?
P  I 50km  r50km
2
I 50km  2r50km
P
P
J
9



 5.0 10 2
2
2
Area1km 2r1km
m s
2r1km
2
I1km
2
Pg. 344, #45: The intensity of a particular earthquake
wave is measured to be 2.0 x 106 J/m2s at a distance of
50 km from the source.
b.) What was the rate energy passed through an
area of 10.0 m2 at 1.0 km?
I50 km= 2.0 x
106
J/m2s
I
r50 km = 50 km
Power

Area
time
Area
Energy
r1 km= 1 km
time
P  I 50km  2r50km
J
9
I1km  5.0 10 2
m s
2
Area = 10.0 m2
Energy
Energy
time
 5.0 10
 I1km ( Area)
9
J
m2  s
(10m 2 )  5 1010Watts
Pg. 344, #46: Show that the amplitude A of circular water
waves decreases as the square root of the distance r
from the source. Ignore damping.
In other words, show that:
1
A
r
1. The same energy that passes through the small circle
each second must pass through the big circle each
second, so P for each whole circle is constant.
2. Since
P
I
2r
and
IA
P
A 
2r
2
The only variable is r, so
A
1
r
2
,
Pg. 344, #50:
Standing Waves; Resonance
• Standing waves occur on a string of length L
when the waves have a wavelength,λ, in which
L is a multiple of .5λ
nn
L
2
Standing Waves; Resonance
Only certain wavelengths can create standing
waves for a given length of cord, but there are
an infinite number of wavelengths that can
create a standing wave on any given cord.
nn
L
2
2L
n 
n
Standing Waves; Resonance
A little algebra to find resonant frequencies:
2L
n 
n
Eqn. (11-19)
v  f
Eqn. (11-12)
2 L 1
( n )  ( )
n
1
1
n

n 2 L
f 
v

For the first harmonic, n=1:
nv
fn 
2L
v
f1 
2L
so:
f n  nf1
Pg. 344, #51: If a violin string vibrates at 440 Hz as its
fundamental frequency, what are the frequencies of the
first four harmonics?
fn=1 = 440 Hz
f n  nf1
(derived on pg. 336)
fn=2 = ?
fn=2 =880 Hz
fn=3 = ?
fn=3 = 1320 Hz
fn=4 = ?
fn=4 = 1760 Hz
Sound Intensity
• The more powerful a wave is when it reaches the
ear, the more energy per unit time it delivers to the
ear, and as a consequence, a more powerful
sound is perceived as being louder.
• The human ear can perceive a wide range of
sound intensities:
– 10-12 W/m2 – threshold of hearing
– 1 W/m2 – threshold of pain
• Notice that the quietest sound a human can hear
is almost a trillion times less intense than a sound
that is so intense that it causes physical pain.
The (deci) bel
• The human ear can hear a wide range of
intensities, but can not distinguish
between small changes in intensity.
• We connect the measurable quantity,
Intensity, I, to the perceived quantity of
loudness, called intensity level, β, using
the equation:
I
  10 log
I0
Where I0 is the threshold
of hearing, 1x10-12 W/m2
Logarithms Math Crash-course:
Logarithms Math Crash-course:
Arithmetic scale
Logarithmic scale
Logarithms Math Crash-course:
If
b x
y
then
log b ( x)  y
Logarithms are discussed in your text, Appendix A, pg. 1046
Logarithm identities:
log( xy)  log( x)  log( y )
x
log( )  log( x)  log( y )
y
log( x y )  y log( x)
log( x)
log( x ) 
y
y
x
log( y )
y
log(x )
The (deci) bel continued
• A change in the intensity level is
associated with a change in a sound’s
loudness.
• A human ear can perceive a change in
intensity level of about 1 dB.
– How great of an increase in intensity is an
increase in intensity level of 1 dB?
– How great of an increase in intensity is an
increase in intensity level of 10 dB?