LECTURE #25 – SUMMARY
(1) Grouping together the “sin(ωt+δ)” terms gives:
(m ωα
+ m αω − b ω ) Ae − α t sin( ω t + δ ) = 0
α=
so
b
2m
(2) Grouping together the “cos(ωt+δ)” terms gives:
(− m ω
2
+ m α 2 − b α + k ) Ae − α t cos( ω t + δ ) = 0
so
ω2 = α 2 +
k bα
−
m m
k
= natural angular frequency of oscillation
m
b2
2
and ω = ω o −
= damped angular frequency
4m 2
Note: ω < ωo (damping reduces frequency)
where ω o =
Damped displacement:


b 2 
2
+
δ
cos   ω o −
t

4 m 2 
 

⇒ SHM !
x = A cos( ω t + δ )
x = Ae
Special case: no damping, b = 0:
 b 
−
t
 2m 
As b (damping) increases:
→ ω decreases and T increases,
slower (more sluggish) oscillation
→ α increases, amplitude decreases
more rapidly
2
If ω = 0, then ω o =
⇒
b = 2m ω o
b2
4m 2
x(t)
A
weakly damped motion:
sinusoidal oscillations within
envelope of a decaying exponential
t
-A
If b < 2m ω o , then the motion is underdamped.
→ oscillation occurs with an amplitude that decreases with time (as shown)
If b = 2m ω o , then the motion is critically damped.
→ the damping force is the same as the spring force and the system returns to
its equilibrium state with no oscillations
If b > 2m ω o , then the motion is overdamped.
→ the damping force is so large that the equation for x(t) is no longer a valid
solution of the equation of motion, and the system slowly returns to its
equilibrium state with no oscillations
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PHY 140Y – Foundations of Physics, Fall Term 2001 (K. Strong)
Lecture 25
Driven Harmonic Motion
Now add an external harmonic forcing to the system.
Fd = - b v
Fs = - k x
F (external)
F = Fo cos ω d t
Driving force:
where ωd = driving angular frequency
The equation of motion is now Fs + Fd + F = ma
x
2
m
x=0
d x
dx
+b
+ kx − Fo cos ω d t = 0
2
dt
dt
What is the solution to this equation?
→ on short time scales - complicated transient solution
→ on longer time scales - not complicated
If Fo cos ω d t is applied for long enough, then the response will eventually have
the frequency of the driving force, ωd. So, we expect a solution of the form
x = A cos( ω d t + δ )
… after the transients die off.
Calculate v and a, and substitute in the terms:
2
− m ω d A cos( ω d t + δ ) − b ω d A sin( ω d t + δ ) + kA cos( ω d t + δ ) − Fo cos ω d t = 0
Apply the trigonometric relations:
cos( ω d t + δ ) = cos( ω d t ) cos( δ ) − sin( ω d t ) sin( δ )
sin( ω d t + δ ) = sin( ω d t ) cos( δ ) + cos( ω d t ) sin( δ )
First, group together the cos( ω d t ) terms:
2
− m ω d A cos( δ ) − b ω d A sin( δ ) + kA cos( δ ) − Fo = 0
Next, group together the sin( ω d t ) terms:
2
m ω d A sin( δ ) − b ω d A cos( δ ) + kA sin( δ ) = 0
Solve for A and δ - messy but not hard!
The result: A =
A(ω d )
A max
full width at half maximum
0.5 A max
FWHM
ω
Fo
m 2 ( ω 2d − ω o2 ) 2 + b 2 ω 2d
This has a general form - a resonance curve.
ωd
ωmax
d
_
ωd
+
ωd
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PHY 140Y – Foundations of Physics, Fall Term 2001 (K. Strong)
Lecture 25