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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.
• 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.
kxF
An Example of a SHO: The spring
X = 0 m
FG
FN
NFFF GN 0
Equilibrium Position
An Example of a SHO: The spring
An Example of a SHO: The springEPEKETotal ...
22
2
1
2
1kxmvETotal
2
2
1.. kAEP Max
2
2
1.. MaxMax mvEK
MaxMax EKEP ....
22
2
1
2
1MaxmvkA
22 Am
kvMax
222
2
1
2
1
2
1kAkxmv
222 kAkxmv 222 kxkAmv
)1()(2
22222
A
xkAxAkmv
)1(2
222
A
x
m
kAv
)1(2
222
A
xvv Max
MaxTotal EPE ..
)1(2
2
A
xvv Max
An Example of a SHO: The spring
An Example of a SHO: The spring
For all SHO:
])2cos[()( itfAtx
])2sin[()( max itfvtv
])2cos[()( max itfata
Am
ka max
m
kAAfv 2max
Where:
Ai i when x0
Sinusoidal Curve Generator
)2cos()( tfAtx )2sin()( max tfvtv
)2cos()( max tfata
0 ,x i isoA Let:
* is measured in radians
)cos()( Atx
)sin()( max vtv
)cos()( max ata
tftT
t 22
Since ,
The Simple Pendulum
FG
FT
NFFF GT 0
Equilibrium Position
0x
m = mass of bob
L = length of cord
FG = weight, mg
FT= tension
The Simple Pendulum
0x FG
Lx
FT
m = mass of bob
L = length of cord
FG = weight, mg
FT= tension
is measured along the arcx
x
The Simple Pendulum
0x FG
Lx
FT
m = mass of bob
L = length of cord
FG = weight, mg
FT= tension
is measured along the arcx
xThe Assumption: For sufficiently small , sin
FGYFGX
FGY=mg cos(θ)
FGX=-mg sin(θ)
mgmgFGX )sin(
xL
mgFGX This is in the format of Hooke’s
law, and thus we have a SHO
T of a Simple Pendulum
kxF (definition of a SHO)
xL
mgFGX (restoring force of a pendulum)
L
mgk
k
mT 2 (eqn. 11-7)
g
L
Lmgm
T 22 (Period of a simple Pendulum)
Pendulum Lab Expected Results:
21
21
2 LgT
21
21
)2( LgT
2/1axy
21
2 ga
g
LT 2 (Period of a simple Pendulum)
s
m
s
m
Tv 8.2
0.3
5.8
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
v=?
1Tf
11 33.)0.3( ssf
fv Eqn. (11-12)
s
msmv 8.233.5.8 1
Oar. . .
TTv
1
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= 330 m/s
?
fv Eqn. (11-12)
f
v
s
sm
Hzsm
f
v1
26.1262
330
ms
s
m
ss
m
s
sm
1
11
m
s
sm
Hzsm
f
v26.1
126.1
262
330
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 high=1600 kHz
fAM low=550 kHz
fFM low=88.0 MHz
fFM high=108 MHz
f
v
v= 3x108 m/s
λAM high f= 188 m
λAM low f= 545 m
λFM low f= 2.78 m
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?
t
xv
m = 0.55 kg
L = 30 m = Δx
FT = 150 N
Δt=?v
xt
LmF
v T Eqn. 11-13
v = 90.5 m/s
s
smm
v
xt 33.
5.90
30
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.
Area
PowerI
Let the wave have power P.
210 2 r
PI km
2220 8)2(2 r
P
r
PI km
r10 km = 10,000 m
r20 km=20,000 m = 2r10 km
4
812
8
2
2
2
20
10
rPrP
I
I
km
km
I10km is 4 times greater than I20km
?20
10 km
km
I
I
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
rA
1
2
21
1
10
10
20
10
km
km
km
km
r
r
A
A?
20
10 km
km
A
A
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?
I50 km= 2.0 x 106 J/m2s
r50 km = 50,000 mArea
timeEnergy
Area
PowerI
250505050 2 kmkmkmkm rIAreaIP
25050 kmkm rIP
sm
J
r
rI
r
P
Area
PI
km
kmkm
kmkmkm
2
92
1
25050
211
1 100.52
2
2
r1 km= 1000 m
?1 kmI
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
r50 km = 50 kmArea
timeEnergy
Area
PowerI
25050 2 kmkm rIP
sm
JI km
2
91 100.5
r1 km= 1 km )(1 AreaItimeEnergy
km
Area = 10.0 m2
Wattsmsm
Jtime
Energy 1022
9 105)10(100.5
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.
rA
1In other words, show that:
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 and ,r
PI
2 2AI
r
PA
22
The only variable is r, so r
A1
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
2nn
L
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
Ln
2
2nn
L
Standing Waves; ResonanceA little algebra to find resonant frequencies:
v
f
11 )2
()( n
Ln
L
n
n 2
1
fv Eqn. (11-12)
L
nvfn 2
n
Ln
2 Eqn. (11-19)
For the first harmonic, n=1:
L
vf
21 so: 1nffn
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
fn=2 = ?
fn=3 = ?
fn=4 = ?
1nffn
fn=2 =880 Hz
fn=3 = 1320 Hz
fn=4 = 1760 Hz
(derived on pg. 336)
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:
0
log10I
I 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:xb y yxb )(logIf then
Logarithms are discussed in your text, Appendix A, pg. 1046
Logarithm identities:
)log()log()log( yxxy
)log()log( xyx y
)log()log()log( yxy
x
)log()log( xy yx y
xxy )log(
)log(
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?