chapter 1 - vibrations harmonic motion/ circular motion simple harmonic oscillators –linear,...
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Chapter 1 - Vibrations
• Harmonic Motion/ Circular Motion
• Simple Harmonic Oscillators– Linear, Mass-Spring Systems
– Initial Conditions
• Energy of Simple Harmonic Motion
• Damped Oscillations
• Driven/Forced Oscillations
Math Prereqs
dcos
d
dsin
d
cos
sin
2 2 2
0 0 0
cos d sin d sin cos d
0
2 22 2
0 0
1 1cos d sin d
2 2
1
2
Identities (see appendix A for more)
cos cos 2cos sin2 2
2 2sin cos 1
cos cos cos sin sin
2 1 1cos cos 2
2 2
ie cos jsin j j
j e eRe e cos
2
j jj e e
Im e sin2j
Why Study Harmonic Motion
http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html
http://www.falstad.com/mathphysics.html
Relation to circular motion
x A cos A cos t
2
T
j j tx Ae Ae
Or
Math Prereqs
T
0
1f t f t dt
T
"Time Average"
2 2cos t
T
T T2
0 0
1 2 1 1 1 2 1cos t dt cos 2 t dt
T T T 2 2 T 2
Example:
Horizontal mass-spring
• Good model!– Force is linear– Mass is constant– Spring has negligible mass– No losses
f maHooke’s Law: sf sx
2
2
d xsx m
dt
2
2
d x sx 0
dt m
Frictionless(1D constraint)
Solutions to differential equations
• Guess a solution• Plug the guess into the differential equation
– You will have to take a derivative or two• Check to see if your solution works. • Determine if there are any restrictions (required
conditions).• If the guess works, your guess is a solution, but it
might not be the only one.• Look at your constants and evaluate them using
initial conditions or boundary conditions.
Our guess
1 ox A cos t
Check
1 ox A cos t 2
21 o o2
d xA cos t
dt
2
2
d x sx 0
dt m
21 o o 1 o
sA cos t A cos t 0
m
2o o
scos t 0
m
The restriction on the solution
2o
s
m
oo
1 sf
2 2 m
o o
1 2 mT 2
f s
Any Other Solutions?
1 ox A cos t
2 ox A sin t
1 o 2 ox A cos t A sin t
A
A1
A2
o ox A cos cos t Asin sin t
ox A cos t
Or
Definitions
• Amplitude - (A) Maximum value of the displacement (radius of circular motion). Determined by initial displacement and velocity.
• Angular Frequency (Velocity) - Time rate of change
of the phase. Natural Angular Frequency
• Period - (T) Time for a particle/system to complete one cycle.
• Frequency - (fo) The number of cycles or oscillations completed
in a period of time. Natural Frequency
• Phase - t Time varying argument of the trigonometric function.
• Phase Constant - Initial value of the phase. Determined by initial displacement and velocity.
ox A cos t
The constants – Phase Angle ox t 0 x u t 0 0 0
x t 0 0 0u t 0 u 2
o ox x cos t o o ou x sin t
o
2o oa x cos t
ox
ox
o ox
o ox
2o ox
2o ox
o
o
u
o
o
u
Case I:
Case II:
o
o
uA
Note phase relationshipbetween x, u, and a
General Case
ox t 0 x 0u t 0 u
1 o 2 ox A cos t A sin t 1 oA x
1 o o 2 o ou A sin t A cos t o2
o
uA
A
A1
A2
2
2 oo
o
uA x
1 o
o o
utan
x
Energy in the SHO
2 2 2 2K P
1 1 1 1E E E mu sx sA mU
2 2 2 2
2 2su A x
m
Average Energy in the SHO
2 2 2 2P o
1 1 1E s x sA cos t sA
2 2 4
2 2 2 2 2 2 2K o o o
1 1 1 1E m u m A sin t m A sA
2 2 4 4
ox A cos t
o o
dxu A sin t
dt
K PE E
Example
• A mass of 200 grams is connected to a light spring that has a spring constant (s) of 5.0 N/m and is free to oscillate on a horizontal, frictionless surface. If the mass is displaced 5.0 cm from the rest position and released from rest find:
• a) the period of its motion, • b) the maximum speed and • c) the maximum acceleration of the mass.• d) the total energy• e) the average kinetic energy• f) the average potential energy
Complex Exponential Solution
• Check it – it works and is simpler. • Phase relationships are more obvious.• Implied solution is the real part• Are there enough arbitrary constants? What are they?
oj tx Ae
oj to ou j Ae j x
oj t2 2o oa Ae x
o
ot
Re
Im
jA a jb Ae
a
jb
oA cos t
Dashpot
r m
dxf R
dt
2
m2
d x dxm R sx 0
dt dt Equation of Motion:
Solution Guess:
Damped Oscillations
Dissipative forces
22mo2
Rd x dxx 0
dt m dt
tx Ae
Check2 t t 2 tm
o
RAe Ae Ae 0
m
2 2 tmo
RAe 0
m
2 2o
tx Ae
mR
2m
2 2d o dj
d d d d dj t j t j t j t j tt1 2 1 2x Ae A e A e e A e A e
d dj t j tt tdx A e e e Ae cos t
Damped frequency oscillation
2m
d 2
Rs
m 4m
2mR 4ms
B - Critical damping (=)C - Over damped (>)
mR
2m
tAe
Relaxation Time
• Decay modulus, decay time, time constant, characteristic time
• Time required for the oscillation to decrease to 1/e of its initial value
m
1 2m
R
Forced Vibrations
j tf t Fcos t or Fe
2
m2
d x dxm R sx f t
dt dt
f t
• Transient Solution – decays away with time constant, • Steady State Solution
Resonance0
s
m Natural frequencyj tx Ae
m
1 FA
sj R j m
2 j t j tmA m jA R As e Fe
sm 0
0
s
m
make small!!
2 2d o
Mechanical Input Impedance
• Think Ohm’s LawV
ZI
m
fZ
u
j t
j t
m
1 Fex Ae
sj R j m
j tf Fe
j t
m
Feu
sR j m
jm m m m m
sZ R j m R jX Z e
22
m m
sZ R m
1
m
sm
tanR
Significance of Mechanical Impedance• It is the ratio of the complex driving force to the
resulting complex speed at the point where the force is applied.
• Knowledge of the Mechanical Impedance is equivalent to solving the differential equation. In this case, a particular solution.
m
fu
Z
m
u fx
j j Z
V
Electrical Analogs2
2
d q dq qL R V(t)
dt dt C
2
m2
d x dxm R sx f t
dt dt
m
s
Rm
Elec Zelec Mech
V f
I u
L jL m jm
R R Rm Rm
1/C 1/jC s s/jf
How would you electrically model this?
m
s
Rm
f
u um
f 1/s
Rm
m
u um
Transient Responsej t
j t
m
1 Fex Ae
sj R j m
m
Fx sin t
Z
22
m m
sZ R m
1
m
sm
tanR
td
m
Fx Ae cos t sin t
Z
See front cover and figure 1.8.1 (pg 14)
Which is transient, which is steady state?
Instantaneous Power
• Think EE P VIi fu
j t
m
Feu
sR j m
m
Fu cos t
Z
2
im
Fcos t cos t
Z
f Fcos t j tf Fe
Average Power
T
i iT 0
1dt
T
2
T
0m
1 Fcos t cos t dt
T Z
22m
2m m
F RFcos
2Z 2Z
2T 2
0m
1 Fcos t cos cos t sin t sin dt
T Z
Quality (Q) value
• Q describes the sharpness of the resonance peak
• Low damping give a large Q• High damping gives a small Q• Q is inversely related to the
fraction width of the resonance peak at the half max amplitude point.
0 0 0
m
mQ
R 2 2
0 0
u l
Q
Tacoma Narrows Bridge
Tacoma Narrows Bridge (short clip)
tx Ae cos t
t tdxv Ae sin t A e cos t
dt
2
t 2 t t 2 t2
d xa Ae cos t A e sin t A e sin t A e cos t
dt
t 2 2Ae 2 sin t cos t
tAe sin t cos t
2
2
d x b dx kx 0
dt m dt m
t t2 2 tb kAe Ae Ae 0
m mcos t cos t cos2 sin tt sin t
t 2 2
b
2mb k
cAe 0b
2 s oin tm
s tm m
22k b
0m 2m
2k b
m 2m
b
2m
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