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
2
d
sin  
d
cos 
d
cos  
d
 sin 
2
2
 cos d   sin d   sin  cos d 
0
0
2
0
0
2
1
1
2
2
cos

d


sin
d 


2 0
2 0
1
2
Identities (see appendix A for more)
sin 2   cos 2   1
cos      cos  cos  sin  sin 


cos   cos   2 cos
sin
2
2
cos 2  
e
 i
1 1
 cos 2
2 2
 cos   jsin 
j
 j
e

e
Re e j   cos  
2  j
j
e e
 j
Im e    sin   
2j
Why Study Harmonic Motion
http://www.falstad.com/mathphysics.html
http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html
Relation to circular motion
x  Acos      Acos  t  
Or
x  Ae
j 
 Ae
j t 
2

T
Math Prereqs
 "Time Average"
f t
T
1
  f  t  dt
T0
Example:
T
T
1
2

1
1 1
1
2


 2  

2
2
cos  t    cos  t  dt     cos  2 t   dt 
T0
T 0 2 2
2
 T 
 T 
 T 
Horizontal mass-spring
 f  ma
Hooke’s Law:
fs  sx
d2 x
sx  m 2
dt
d2x s
 x0
2
dt
m
• Good model!
–
–
–
–
Force is linear
Mass is constant
Spring has negligible mass
No losses
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
x  A1 cos o t
Check
x  A1 cos o t
d2x s
 x0
2
dt
m
d2 x
2


A

1 o cos o t
2
dt
s
A1 cos o t  A1 cos o t  0
m
2
o
s
2
  o  cos o t  0
m

The restriction on the solution
s
 
m
2
o
o
1 s
fo 

2 2 m
1 2
m
T 
 2
f o o
s
Any Other Solutions?
x  A1 cos o t
x  A2 sin o t
x  A1 cos o t  A 2 sin o t
Or
A1
x  A cos  cos o t  Asin  sin o t

A2
A
x  Acos  o t  
Definitions
x  Acos  o t  
• Amplitude - (A) Maximum value of the displacement (radius of
circular motion). Determined by initial displacement and velocity.
• Angular Frequency (Velocity) - o 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.
The constants – Phase Angle
Case I:
x  t  0  x o
x  x o cos  o t  
u  x oo sin  o t  
xo
x o o
x o
 x o o

a   x o o2 cos o t  

Note phase
relationship
between x, u, and a
x o o2
 x o o2
Case II:
x  t  0  0
uo
o

0
u  t  0  0
uo
o
u  t  0  u 0


2
uo
A
o
General Case
x  t  0  x o
u  t  0  u 0
x  A1 cos o t  A 2 sin o t
A1  x o
u  A1o sin o t  A2o cos o t
uo
A2 
o
A1
 uo 
2
A  xo   
 o 

A2
A
2
 u o 
  tan 


x
 o o
1
Energy in the SHO
1
1 2 1 2 1
2
E  E K  E P  mu  sx  sA  mU 2
2
2
2
2
s
2
2
u
A

x


m
Average Energy in the SHO
x  Acos  o t  
EP
1
1 2
1 2
2
2
 s x  sA cos  o t     sA
2
2
4
dx
u
  A o sin  o t   
dt
EK
1
1
1
1 2
2
2 2
2
2 2
 m u  mo A sin  o t     mo A  sA
2
2
4
4
EK  EP
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
o
Im
x  Ae
jo t
A  a  jb  Ae j
jb
o t

•
•
•
•
Re
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?
A cos  o t  
u  jo Ae jo t  jo x
a  o2 Ae jo t  o2 x
a
Damped Oscillations
Dashpot
Dissipative forces
dx
f r  R m
dt
Equation of Motion:
d2x
dx
m 2  Rm
 sx  0
dt
dt
d 2 x R m dx
2



ox  0
2
dt
m dt
Solution Guess:
x  Ae
t
x  Ae
Check
t
Rm
t
2
t
 Ae 
Ae  o Ae  0
m
t
2
 2 Rm
2
t





Ae
0
o

m


Rm

2m
      
2
2
o
    jd
d  o2 2
x  Ae
  jd  t
 A1e
  jd  t

x  Aet e 
j d t 
 A 2e
e
  jd  t
 j d t 


 e t A1e jd t  A 2e  jd t
 Aet cos  d t   

Damped frequency oscillation
Rm

2m
s R 2m
d 

m 4m2
R 2m  4ms
B - Critical damping (=)
C - Over damped (>)
Aet
Relaxation Time
• Decay modulus, decay time, time constant,
characteristic time
• Time required for the oscillation to decrease
to 1/e of its initial value
1 2m
 
 Rm
Forced Vibrations
f t
d2x
dx
m 2  Rm
 sx  f  t 
dt
dt
f  t   Fcos t or Fe
jt
• Transient Solution – decays away with time constant, 
• Steady State Solution
Resonance
x  Ae

jt

A2 m  jAR m  As e jt  Fe jt
1
A
j
F
s

R m  j  m  


make small!!
s
m   0

s

 0
m
Natural frequency
s
0 
m
d  o2 2
Mechanical Input Impedance
• Think Ohm’s Law
f
Zm 
u
x  Ae
jt
1

j
Z
V
I
f  Fe jt
jt
Fe
s

R m  j  m  


Fe jt
u
s

R m  j  m  


s

Zm  R m  j  m    R m  jX m  Zm e j


s

Zm  R m2   m  


s

m  


  tan1 

 R m 


2
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.
f
u
Zm
u
f
x

j jZm
Electrical Analogs
d 2q
dq q
L 2 R
  V(t)
dt
dt C
V
s
f
Elec
Zelec
Mech
V
f
I
u
L
jL
m
jm
R
R
Rm
Rm
1/C
1/jC
s
s/j
m
Rm
d2x
dx
m 2  Rm
 sx  f  t 
dt
dt
How would you electrically model this?
s
f
m
Rm
u
f
um
u
1/s um
Rm
m
Transient Response
x  Ae
jt
1

j
Fe jt
s

R m  j  m  


F
x
sin  t   
Z m
s

Zm  R   m  


2
m
Which is transient,
which is steady state?
x  Ae
t
s

m  


  tan1 

R
m




F
cos  d t    
sin  t   
Zm
See front cover and figure 1.8.1 (pg 14)
2
Instantaneous Power
 i  fu
P  VI
• Think EE
Fe jt
u
s

R m  j  m  


f  Fe
F
u
cos  t   
Zm
jt
f  Fcos t
2
F
i 
cos t cos  t   
Zm
Average Power
  i
T
1 T
   i dt
T 0
2
1
F
 
cos t cos  t    dt
T 0 Zm
T
F2 R m
F2

cos  
2Zm
2Zm2
2
1 T F
 
cos 2 t cos   cos t sin t sin dt
0
T Zm
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
Q

 u  l
m0 0 0 
Q


Rm
2
2

Tacoma Narrows Bridge
Tacoma Narrows Bridge (short clip)
x  Aet cos  t  
v
dx
 Aet  sin  t     A    e t cos  t   
dt
 Aet  sin  t      cos  t    
d2 x
a  2  Aet 2 cos  t     Aet  sin  t     Aet  sin  t     A 2et cos  t   
dt


 Aet 2 sin  t      2  2  cos  t   
d 2 x b dx k

 x0
2
dt
m dt m


Aet 2 sin  t      2  2  cos  t    
Ae

b
t
2m
b
k
Aet  sin  t      cos  t      Aet cos  t     0
m
m

b 
b
k

 2
2
  2   sin  t            cos  t      0
m 
m
m



b

2m
2
k  b 
2

    0
m  2m 
k  b 
 


m  2m 
2