Single Degree of Freedom System
2.003J/1.053J
Dynamics and Control I, Spring 2007
Professor Thomas Peacock
5/2/2007
Lecture 20
Vibrations: Second Order Systems with One
Degree of Freedom, Free Response
1
Single Degree of Freedom System
Figure 1: Cart attached to spring and dashpot.
Figure by MIT OCW.
mx ˙ + kx = F ( t )
System response?
What is x ( t )?
Use 18.03
Background.
x ( t ) = Free
�
Response
�� �
Complementary Solution, when F ( t )=0
+
This lecture will cover the Free Response.
Response Due to Forcing
� ��
Particular Solution
�
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Free Response
Free Response
Look at k → 0
2
Figure 2: Cart with dashpot only.
Figure by MIT OCW.
mx ˙ = 0
Assume conditions x (0) = x
0 and ˙ (0) = v
0
.
mx ˙ = mv ˙ + cv = 0 v = v
0 e
( − ct/m ) already used ˙ (0) = v
0
Integrate v ( t ) once.
Using x (0) = x
0
, we obtain: x = x
0
+ mv
0 c
�
1 − e
− c m t
�
Figure 3: Solution to differential equation.
Solution attenuates to a steady state value.
Figure by MIT OCW.
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Free Response 3
Figure 4: Velocity profile of solution.
Velocity attenuates to zero.
Figure by
MIT OCW.
No oscillations.
Because k = 0, there was no restoring term.
Look at m → 0 or
Therefore: cx kx = 0 x − k c x x (0) = x
0 x ( t ) = x
0 e
− k c t
Figure 5: Solution to differential equation.
Position decays to zero.
Figure by
MIT OCW.
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Free Response 4
Figure 6: Velocity profile of solution.
Value attenuates to steady state value.
Figure by MIT OCW.
˙ = − kx c
0 e
− k c t
No oscillations in this system.
Dashpot force balances the spring force as x → 0, spring force → 0.
Vibrations require a restoring force (e.g.
spring) and inertia (e.g.
mass).
Full Free Response Problem
So let us consider the full problem: mx ˙ + kx = 0 (1)
Note that cx c > 0) is a damping term and is responsible for decay of oscillations.
Examination of Energy d dt
( T + V ) = d
�
1 dt 2 mx
2
2
1 kx
2
�
= mx x kx x ˙ ( mx kx ) = ˙ ( − cx = − ˙ 2
For c > 0: d dt
( T + V ) < 0
Damping.
Mechanical energy is dissipated.
For c < 0:
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Free Response 5 d dt
( T + V ) > 0
Energy input (Control system providing energy)
Solution of the Equation with Engineering Quantities
Rewrite mx ˙ + kx = 0 as: x ζω n x ω
2 n x = 0
ω
2 n
ζ =
= k m c
2 mω n
ω n
: Natural Frequency
ζ : Damping Ratio
To solve, we assume a solution of the form x = Ae
( λt )
Substitute in Equation
(2):
λ
2
+ 2 ζω n
λ + ω
2 n
= 0
�
λ = − ζω n
± ω n
ζ 2 − 1
When ζ
2
> 1 and ζ
2
< 1, the behavior is different.
Assume c ≥ 0.
( ζ ≥ 0) We have the following cases.
Case 1: Overdamped
ζ > 1 ⇒ λ
1
, λ
2
= Real Negative Numbers x = A
± e
“
− ζω n
±
√
ζ 2 − 1
”
→ 0 as t → ∞
Case 2: Critically Damped
ζ = 1 ⇒ λ
1
, λ
2
= − ω n
(2)
(3) x = ( A
1
+ A
2 t ) e
− ω n t
→ 0 as t → ∞ (4)
Equation
(4)
is the fastest approach to the set point.
That is why it is named critically damped.
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Free Response 6
Case 3: Underdamped
0 ≤ ζ < 1
λ
1
, λ
ω d
2
=
= − ζω n
± iω d
ω n
�
1 − ζ 2
Underdamped (Not enough damping to prevent oscillations).
When ζ
ω d
→ ω n
(Natural frequency).
→ 0, x =
�
A
1 e iω d t
+ A
2 e
− iω d t
� e
− ζω n t
Must have that A
1 and A
2 are complex conjugates because x is real.
x =[ A
1
(cos ω d t + i sin ω d t ) + A
2
(cos ω d t − i sin ω d t )] e
− ζω n t
=[(
�
A
1
+ A
2
) cos ω d t + i ( A
1
− A
2
) sin ω d t ] e − ζω n t
��
A
3
� � ��
A
4
�
A
1
+ A
2
= A
3 i ( A
1
− A
2
) = A
4 x = A
3
� cos ω d t +
A
A
4
3 sin ω d
� t e
− ζω n t x = A
3
� cos ω d t + tan φ sin ω d
� t e
− ζω n t x =
A
3 cos φ
� � cos ω d t cos φ + sin ω d t sin φ e − ζω n t
Note the trigonometric identity.
x ( t ) = Ce
− ζω n t cos( ω d t − φ ) e − ζω n t : Decaying in time cos( ω d t − φ ): Oscillatory Behavior
C and φ can be found from initial conditions.
C =
A
3 cos φ
A
φ = arctan
A
4
3
(5)
(6)
(7)
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Free Response 7
Equations
(6)
and
(7)
relate C and φ to A
3 and A
4
.
But 1 cos 2 φ
= 1 + tan 2
φ .
1 cos 2 φ
= 1 +
A
2
4
A 2
3
1 cos φ
=
�
A 2
3
+
A
3
A 2
4
⇒ C =
�
A 2
3
+ A 2
4
If 0 ≤ ζ < 1, the solution will show decaying oscillations.
How do we determine
( C and φ ) or ( A
3 and A
4
)?
Often easier to relate A
3 and A
4 to initial conditions.
Initial Conditions: x (0) = x
0
, x = v
0 x = [ A
3 cos ω d t + A
4 sin ω d t ] e
− ζω n t
At t = 0, x
0
= A
3
(using x (0) = x
0
) x − A
3
ω d sin ω d t + A
4
ω d cos ω d t ] e
− ζω n t
− ζω n
[ A
3 cos ω d t + A
4 sin ω d t ] e
− ζω n t
At t = 0:
Examine solution.
v
0
= A
4
ω d
− ζω n
A
3
= A
4
ω d
− ζω n x
0
A
4
= v
0
+ ζω n x
0
ω d
�
C = x 2
0
+
� v
0
+ ζω n x
0
� 2
ω d tan φ = v
0
+ ζω n x
0
ω d x
0 x ( t ) = Ce
− ζω n t cos( ω d t − φ )
(8)
(9)
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].

Free Response e − ζω n t : Decay cos( ω d t − φ ): Oscillating
8
Figure 7: Solution both decays and oscillates given the presence of exponential solution and sinusoidal solution.
Figure by MIT OCW.
Calculate Amplitude.
x ( t x (
+ t ) nτ d
)
= e e − ζω n t
[ − ζω n
( t + nτ d
)]
= e
ζω n nτ d ln
� x ( t ) x ( t + nτ d
)
�
= nζω n
τ d
= nζ
ω n
ω
2 π d
= nζ
ω n
ω n
2 π
�
1 − ζ 2
= nζ
�
2 π
1 − ζ 2
For ζ << 1: ln
� x ( t ) x ( t + nτ d
)
�
= 2 πnζ
Need ω n
, ζ to define system.
(10)
(11)
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].

Free Response
Example Experiment: Flexible Rod.
9
Figure 8: Flexible rod.
Figure by MIT OCW.
Measure frequency of oscillation: ω d
.
Measure amplitude over several periods to obtain x ( t ) x ( t + nτ d
)
.
This ratio is related to the damping ratio ζ by the equations
(10)
or
(11)
if ζ << 1.
With ω d and ζ , one can calculate the natural frequency ω n
.
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J
Dynamics and
Control I, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].