Transcript Chapter 2 - Motion in One Dimension

Lesson 1 - Oscillations
• Harmonic Motion
Circular Motion
• Simple Harmonic
Oscillators
– Linear Horizontal/Vertical
Mass-Spring Systems
• Energy of Simple
Harmonic Motion
Math Prereqs
d
sin  
d
cos 
d
cos  
d
 sin 
2
2
 cos d   sin d 
0
0
0
2
2
1
1
2
2
cos

d


sin
d 


2 0
2 0
1
2
Identities
sin 2   cos 2   1
cos      cos  cos  sin  sin 


cos   cos   2 cos
sin
2
2
1 1
cos    cos 2
2 2
2
e i  cos   i sin 
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 
Harmonic
Relation to circular motion
x  Acos      Acos  t  

2
T
Horizontal mass-spring
 F  ma
Hooke’s Law:
Fs  kx
kx  m block
d2x
dt 2
d2x
k

x0
2
dt
mblock
Frictionless
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  Acos  t  
Definitions
x  Acos  t  
• 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.
• Period - (T) Time for a particle/system to complete one cycle.
• Frequency - (f) The number of cycles or oscillations completed in
a period of time
• Phase - t   Time varying argument of the trigonometric
function.
• Phase Constant -  Initial value of the phase. Determined by
initial displacement and velocity.
The restriction on the solution
 
2
k
m block

1
k
f

2 2 mblock
mblock
2
T
 2

k
The constant – phase angle
x  t  0  A
0
v  t  0  0
x  Acos  t  
v  Asin  t  
a  A2 cos  t  
x  t  0  0
v  t  0   v0


2
Energy in the SHO
1
1 2 1
2
E  mv  kx  kA 2
2
2
2
k 2
2
v
A

x


m
Average Energy in the SHO
x  Acos  t  
1
1
1
2
2
2
U  k x  kA cos  t     kA 2
2
2
4
dx
v
 A sin  t   
dt
1
1
1
1
2
2 2
2
2 2
K  m v  m A sin  t     m A  kA 2
2
2
4
4
K  U
Example
• A mass of 200 grams is connected to a light spring that has
a spring constant (k) 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
Damped Oscillations
“Dashpot”
Fdamping  bv
dx
 kx  b
 ma
dt
Equation of Motion
Solution
d2x
dx
m 2  b  kx  0
dt
dt
x  Aet cos  t  
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
Damped frequency oscillation
b

2m
k
b2
 

m 4m 2
b  4mk
2
B - Critical damping (=)
C - Over damped (>)
Giancoli 14-55
• A 750 g block oscillates on the end of a spring
whose force constant is k = 56.0 N/m. The mass
moves in a fluid which offers a resistive force F =
-bv where b = 0.162 N-s/m.
– What is the period of the motion? What if there had
been no damping?
– What is the fractional decrease in amplitude per cycle?
– Write the displacement as a function of time if at t = 0,
x = 0; and at t = 1.00 s, x = 0.120 m.
Forced vibrations
dx
kx  b  F0 cos t  ma
dt
Fext  F0 cos t
2
d x
dx
m 2  b  kx  F0 cos t
dt
dt
x  A0 sin  t  0 
Resonance
x  A0 sin  t  0 
k
0 
m
Natural frequency
F0
A0 
m

2

2 2
0

b 2 2
 2
m
 m  2  02  

0  tan 1 


b


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.
m0
Q
b
 1

0 Q

Tacoma Narrows Bridge
Tacoma Narrows Bridge (short clip)