Simple Harmonic Motion and
Damping
& Uncertainty Analysis Review
Session 6
Physics 2BL
Summer 2009
Outline
• Experiment 3 intro
• Physics of damping and SHM
• Experiment 3 objectives
Random and independent?
No
Yes
• End of ruler screwy
• Estimating between
• Reading meter from
marks on ruler or
the side
meter
(speedometer
• Releasing object from
effect)
‘rest’
• Scale not zeroed
• Mechanical vibration
Reaction time delay
o Judgment
o Problems of definition
o Calibration
o Zero
Random & independent
errors:
q = Bx
q = x+ y−z
δq = (δx) + (δy ) + (δz )
2
2
δq = B δx
2
δq
q
q = x× y ÷ z
δx
x
q = q ( x, y , z )
2
δq
⎛ δx ⎞ ⎛ δy ⎞ ⎛ δz ⎞
= ⎜ ⎟ + ⎜⎜ ⎟⎟ + ⎜ ⎟
q
⎝ x⎠ ⎝ y⎠ ⎝ z⎠
2
=
2
2
⎛ ∂q ⎞ ⎛ ∂q ⎞ ⎛ ∂q ⎞
δq = ⎜ δx ⎟ + ⎜⎜ δy ⎟⎟ + ⎜ δz ⎟
⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠
2
2
Experiment 3
• Model for shock absorber in car
• Goals: develop and demonstrate
critically damped system
• Test validity of model for damping
• check out setup, take preliminary data,
write proposal
• do experiment, write up results
Simple Harmonic Motion
• Position oscillates if
force is always
directed towards
equilibrium position
(restoring force).
• If restoring force is ~
position, motion is
easy to analyze.
FNet
FNet
Springs
• Mag. of force from
spring ~ extension
(compression) of
spring
• Mass hanging on
spring: forces due to
gravity, spring
• Stationary when
forces balance
FS = −kx
FG = −mg
FG = FS
mg = kx
m12
x = x1
m2
x = x2
Simple Harmonic Motion
x = x0
3
dipslacement
• Spring provides
linear restoring force
⇒ Mass on a spring
is a harmonic
oscillator
2
x =0
1
0
-1
F = −kx
d2x
m 2 = −kx
dt
x(t) = x 0 cos ωt
-2
-3
0
10
20
30
40
50
60
time
T=
2π
ω
ω=
k
m
Damping
• Damping force opposes
motion, magnitude
depends on speed
• For falling object,
constant gravitational
force
• Damping force
increases as velocity
increases until damping
force equals
gravitational force
• Then no net force so no
acceleration (constant
velocity)
Terminal velocity
• What is terminal velocity?
• How can it be calculated?
Falling Mass and Drag
At steady state:
From rest:
Fdrag = F gravity
bvt = mg
y(t) = vt[(m/b)(e-(b/m)t – 1) + t]
Terminal Velocity
For velocity:
.
y(t) = vt[1 - e-(b/m)t]
Experimental Setup for Falling
Mass and Drag
How do you measure velocity?
Damped Harmonic Motion
Damped SHM
• Consider both
position and velocity
dependant forces
• Behavior depends
on how much
damping occurs
during one
‘oscillation’
⎛ b ⎞ ⎛ k
b2 ⎞
⎟.
x = x 0 exp⎜−
t ⎟ cos⎜ t
−
⎝ 2m ⎠ ⎜⎝ m 4m 2 ⎟⎠
d 2x
dx
m 2 = − kx − b
dt
dt
⎛ b ⎞ ⎛
k
b2 ⎞
⎟
x = x0 exp⎜−
t ⎟ exp⎜⎜ it
−
2 ⎟
⎝ 2m ⎠ ⎝ m 4m ⎠
or
⎛⎛⎛ b
22
⎞ ⎞⎞
b
k
b
k
⎜
exp⎜⎜⎜−−
±+
−− ⎟⎟ t⎟⎟t
xx== xx00 exp
22
⎜⎜ 22m
4m
4m mm⎠ ⎟⎠⎟⎠
⎝⎝⎝ m
Relative Damping Strength:
Weak damping
b2
k
−
m 4m2
⎞
⎟⎟ .
⎠
3
dipslacement
⎛
⎛ b ⎞
t ⎟ cos ⎜⎜ t
x = x 0 exp ⎜ −
⎝ 2m ⎠
⎝
k
b2
<<
m
4m2
2
1
0
6
-1
4
weak damping
2
-2
0
0
-2
2
4
6
(underdamped)
-4
-6
time
8
10
0
100
200
300
400
time
500
600
Relative Damping Strength:
Strong damping
b2
k
>>
4m 2
m
strong damping
(overdamped)
3
dipslacement
2
⎛ b
⎞
b
k
⎟t
+
−
x = x0 exp⎜ −
2
⎜ 2m
⎟
m
m
4
⎝
⎠
2
1
0
0
5
10
15
time
20
25
Relative Damping Strength:
Critical damping
b2
k
=
2
4m
m
critical damping
⎞
⎟t
⎟
⎠
3
dipslacement
2
⎛ b
k
b
+
−
x = x0 exp⎜ −
2
⎜ 2m
m
m
4
⎝
2
1
0
-100
bcrit = 2 mk
0
100 200 300 400 500 600 700
time
Comparison of the various
types of damping
dipslacement
3
overdamped
critically
damped
2
1
0
-1
underdamped
-2
-100
0
100 200 300 400 500 600 700
time
Plotting Graphs
Give each graph a title
Determine independent and dependent variables
Determine boundaries
Include error bars
Experimental setup
spring
photogate
scale (mm)
cylinder
photogate
timer
piston
valve
holes
Experiment 3: achieve critical
damping
• Show/test method
– Determine spring constant, predict critical
damping coefficient
– Determine how damping coefficient
depends on air flow (valve position)
• easy at terminal velocity
• how do you know it’s vterminal?
– Set damping to critical level
Demonstrate critical damping:
show convincing evidence that
critical damping was achieved
• Demonstrate that damping is critical
– No oscillations (overshoot)
– Shortest time to return to equilibrium
position
Error propagation
(1) kspring = 4π2m/T2
σkspring = εkspring * kspring
εkspring = εm2 + (2εT)2
(2) kby-eye = m(g∆t*/2∆x)2
σby-eye = εby-eye * kby-eye
εby-eye =
(2ε∆t*)2 + (2ε∆x)2 + εm2
Reminder
• Experiment 3 proposal/preliminary data
• Finally experiment 3 write-up
• No lecture Monday, July 20