Damped Harmonic Motion (DHM)
Look at drag force that is proportional to
velocity; b is the damping coefficient:
FD = −bv = −b
dx
dt
For small v
DHM
d 2x
dx
+ b + kx = 0
dt
dt 2
GDE:
m
Solution:
x = Ae (−αt ) cos(ω ′t + φ )
Draw FBD = KD
DHM
Solution:
d x
dx
+ kx = 0
GDE: m 2 + b
dt
dt
2
x = Ae
(−αt )
cos(ω ′t + φ )
DHM
d 2x
dx
+ b + kx = 0
dt
dt 2
GDE:
m
Solution:
x = Ae⎝
⎛− t ⎞
⎜
t L ⎟⎠
cos(ω ′t + φ )
A
Amplitude
2m 1
tL =
=
b α
Mean Lifetime
k
b2
−
m 4m 2
ω′ =
φ
f =
Angular frequency
Phase angle
2π
T=
ω′
1
T
SHM
DHM
Period
Frequency
1
A mass on a spring with a natural angular frequency ωo = 3.6 rad/s is placed in
an environment in which there is a damping force proportional to the speed of
the mass. If the amplitude is reduced to 0.35 times its initial value in 12.9 s,
what is the angular frequency of the damped motion?
A harmonic oscillator starts with an amplitude of 20.0 cm. After 10.0 s, the
amplitude decreases to 15.0 cm. If the linear damping coefficient is 2.00 N-s/m,
how much mass is oscillating?
Values of Damping
Forced Vibration
⎛− t ⎞
⎜
t L ⎟⎠
x = Ae⎝
k
b2
−
ω′ =
m 4m 2
cos(ω ′t + φ )
There is a oscillating force that is driving
the harmonic motion of the system.
DHM
Angular frequency
When
,
This is critical damping, and
the value of b for which this
occurs is bc:
When b > bc, overdamped
When b < bc, underdamped
= Fo cos(ω t )
2
Forced Vibration
GDE:
d 2x
dx
m 2 + b + kx = F0 cos(ω t )
dt
dt
Solution:
x = A sin (ωt + φ )
Forced Vibration
Solution:
x = A sin (ωt + φ )
d 2x
dx
m 2 + b + kx = F0 cos(ω t )
dt
dt
GDE:
(k − mω )A sin(ωt + φ ) + ωbA cos(ωt + φ ) − F cos(ω t ) = 0
2
0
Use:
cos(ωt + φ ) = cos(ωt ) cos(φ ) − sin (ωt )sin (φ ), sin (ωt + φ ) = sin (ωt ) cos(φ ) + cos(ωt )sin (φ )
[(k − mω )A sin(φ ) + ωbA cos(φ ) − F ]cos(ωt )
+ [(k − mω )A cos(φ ) − ωbA sin (φ )]sin (ωt ) = 0
2
0
2
A
Amplitude
ω Angular frequency
φ
⇒
Phase angle
and
= Fo cos(ω t )
Forced Vibration
d 2x
dx
+ b + kx = F0 cos(ω t )
dt 2
dt
GDE:
m
Solution:
x = A sin (ωt + φ )
tan (φ ) =
Resonance
[(k − mω )A cos(φ ) + ωbA sin(φ )] = 0,
⇒
2
[(k − mω )A sin (φ ) + ωbA cos(φ ) − F ] = 0
tan (φ ) =
k
ωo2 − ω 2
, ωo =
mωb
m
2
0
∴ A=
(
Fo
m 2 ω 2 − ωo2
)
2
+ b 2ω 2
A 3.0-kg mass is attached to a spring with a force constant of
48 N/m. The motion is subject to a damping force F =-bv,
where b = 4.0 kg/s. The mass is subjected to an oscillating
force with a driving frequency ω. At what value of ω will the
oscillations have their maximum amplitude?
k
ωo2 − ω 2
, ωo =
mωb
m
= Fo cos(ω t )
Tacoma Narrows Bridge
3