Ch 3'8: Mechanical

1
Ch 3.8 Mechanical Electrical Vibrations

• Applications mechanical and electrical
  oscillations.
• We will study the motion of a mass on a spring in
  detail.

2
Spring Model

• What if the mass is initially displaced?
• u(t) displacement of mass from equilibrium
  position.
• f net force acting on mass.
• Newtons 2nd Law - Force Mass x Acceleration
• In determining f, four separate forces to
  consider
• Weight w mg (downward)
• Spring force Fs - k(L u) (up or
  down)
• Damping force Fd(t) - ? u? (t) (up or
  down)
• External force F (t)
  (up or down)

3
Spring Model Differential Equation

• Taking into account these forces, Newtons Law
  becomes
• Recalling that mg kL, this equation reduces to
• or
• where the constants m, ?, and k are positive.

4
Spring Model Differential Equation

• Equation
• where the constants m, ?, and k are positive.
• We can prescribe initial conditions also
• It follows from Theorem 3.2.1 that there is a
  unique solution to this initial value problem.
• Physically, if mass is set in motion with a given
  initial displacement and velocity, then its
  position is uniquely determined at all future
  times.

5
Example 1

• A 4 lb mass stretches a spring 2". The mass is
  displaced an additional 6" and then released and
  is in a medium that exerts a viscous resistance
  of 6 lb when velocity of mass is 3 ft/sec.
  Formulate the IVP that governs motion of this
  mass

6
Example 1

• DE is
• and IVP is
• Graph of solution

7
Spring Model Undamped Free Vibrations

• DE for spring motion

8
Spring Model Undamped Free Vibrations

• DE for spring motion
• Suppose there is no external driving force and no
  damping.

9
Spring Model Undamped Free Vibrations

• DE for spring motion
• Suppose there is no external driving force and no
  damping. Then F(t) 0 and ? 0, and our
  equation becomes

10
Spring Model Undamped Free Vibrations

• DE for spring motion
• Suppose there is no external driving force and no
  damping. Then F(t) 0 and ? 0, and our
  equation becomes
• The general solution to this equation is

11
Spring Model Undamped Free Vibrations

• Using trigonometric identities, the solution can
  be rewritten
• where
• amplitude
  phase
• frequency
  period

12
Example 2

• A 10 lb mass stretches a spring 2". The mass is
  displaced an additional 2" and then set in motion
  with initial upward velocity of 1 ft/sec.
  Determine position of mass at any later time.
  Also find period, amplitude, and phase of the
  motion.

13
Example 2

• Simplifying
• Solve
• or

14
Example 2 Period, Amplitude, Phase

• natural frequency
• period
• amplitude
• phase ?

15
Spring Model Damped Free Vibrations

• With damping but no external driving force

16
Spring Model Damped Free Vibrations

• With damping but no external driving force
• What is effect of ? ?

17
Spring Model Damped Free Vibrations

• With damping but no external driving force
• What is effect of ? ?
• characteristic equation

18
Spring Model Damped Free Vibrations

• With damping but no external driving force
• What is effect of ? ?
• characteristic equation
• Three cases

19
Spring Model Damped Free Vibrations

• With damping but no external driving force

20
Spring Model Damped Free Vibrations

• With damping but no external driving force
• In all three cases,

21
Damped Free Vibrations Small Damping

• Of the cases for solution form, the last is most
  important, which occurs when the damping is
  small

22
Damped Free Vibrations Small Damping

• Of the cases for solution form, the last is most
  important, which occurs when the damping is
  small
• We examine this last case. Recall
• and hence
• (damped oscillation)

23
Damped Free Vibrations Quasi Period

• Compare ? with ?0 , the frequency of undamped
  motion
• Thus, small damping reduces oscillation frequency
  slightly.
• Similarly, quasi period is defined as Td 2?/?.
  Then
• Thus, small damping increases quasi period.

For small ?
24
Damped Free Vibrations

• Mass creeps back to equilibrium position for
  solns (1) (2), but does not oscillate about it,
  as for small ? in solution (3).

25
Ch 3.9 Forced Vibrations

• We continue the discussion of the last section,
  and now consider the presence of a periodic
  external force

26
Transient Solution and Initial Conditions

• For the DE and general solution,
• uC(t) enables us to satisfy any initial
  conditions.

27
Transient Solution and Initial Conditions

• For the DE and general solution,
• uC(t) enables us to satisfy any initial
  conditions.
• With increasing time, the energy put into system
  by initial displacement and velocity is
  dissipated through damping force. The motion
  then becomes the response U(t) of the system to
  the external force F0cos?t.

28
Transient Solution and Initial Conditions

• For the DE and general solution,
• uC(t) enables us to satisfy any initial
  conditions.
• With increasing time, the energy put into system
  by initial displacement and velocity is
  dissipated through damping force. The motion
  then becomes the response U(t) of the system to
  the external force F0cos?t.
• Without damping, the effect of the initial
  conditions would persist for all time.

29
Rewriting Forced Response

• Using trigonometric identities,
• can be rewritten as
• Where

30
Amplitude Analysis of Forced Response

• The amplitude R of the steady state solution
• depends on the driving frequency ?.

31
Amplitude Analysis of Forced Response

• The amplitude R of the steady state solution
• depends on the driving frequency ?. For
  low-frequency excitation we have
• where (?0)2 k /m.

32
Amplitude Analysis of Forced Response

• The amplitude R of the steady state solution
• depends on the driving frequency ?. For
  low-frequency excitation we have
• where (?0)2 k /m.
• For high frequency excitation,

33
Maximum Amplitude of Forced Response

• Thus

34
Maximum Amplitude of Forced Response

• Thus
• At an intermediate value of ?, the amplitude R
  may have a maximum value. To find this frequency
  ?, differentiate R and set the result equal to
  zero. Solving for ?max, we obtain

35
Maximum Amplitude of Forced Response

• Thus
• At an intermediate value of ?, the amplitude R
  may have a maximum value. To find this frequency
  ?, differentiate R and set the result equal to
  zero. Solving for ?max, we obtain

36
Maximum Amplitude of Forced Response

• Thus
• At an intermediate value of ?, the amplitude R
  may have a maximum value. To find this frequency
  ?, differentiate R and set the result equal to
  zero. Solving for ?max, we obtain
• where (?0)2 k /m. Note ?max lt ?0, and ?max
  is close to ?0 for small ?. The maximum value of
  R is

37
Maximum Amplitude for Imaginary ?max

• We have
• and
• where the last expression is an approx. for
  small ?.
• If ? 2 /(mk) gt 2 ?max is imaginary. In
  this case,
• Rmax F0 /k, which occurs at ? 0, and R is
  a monotone decreasing function of ?.
• Recall that critical damping occurs when ? 2
  4mk.

38
Resonance

• From the expression
• we see that Rmax? F0 /(? ?0) for small ?.

39
Resonance

• From the expression
• we see that Rmax? F0 /(? ?0) for small ?.
• Thus for lightly damped systems, the amplitude R
  of the forced response is large for ? near ?0,
  since ?max ? ?0 for small ?.

40
Resonance

• From the expression
• we see that Rmax? F0 /(? ?0) for small ?.
• Thus for lightly damped systems, the amplitude R
  of the forced response is large for ? near ?0,
  since ?max ? ?0 for small ?.
• This is true even for relatively small external
  forces, and the smaller the ? the greater the
  effect.

41
Resonance

• From the expression
• we see that Rmax? F0 /(? ?0) for small ?.
• Thus for lightly damped systems, the amplitude R
  of the forced response is large for ? near ?0,
  since ?max ? ?0 for small ?.
• This is true even for relatively small external
  forces, and the smaller the ? the greater the
  effect.
• This phenomena is known as resonance. Resonance
  can be either good or bad, depending on
  circumstances for example, when building bridges
  or designing seismographs.

42
Analysis of Phase Angle

• The phase angle ? given in the forced response
• is characterized by the equations

43
Analysis of Phase Angle

• The phase angle ? given in the forced response
• is characterized by the equations
• If ? ? 0, then cos? ? 1, sin? ? 0, and hence ?
  ? 0. Thus the response is nearly in phase with
  the excitation.

44
Analysis of Phase Angle

• The phase angle ? given in the forced response
• is characterized by the equations
• If ? ? 0, then cos? ? 1, sin? ? 0, and hence ?
  ? 0. Thus the response is nearly in phase with
  the excitation.
• If ? ?0, then cos? 0, sin? 1, and hence ?
  ? ? /2. Thus response lags behind excitation by
  nearly ? /2 radians.

45
Analysis of Phase Angle

• The phase angle ? given in the forced response
• is characterized by the equations
• If ? ? 0, then cos? ? 1, sin? ? 0, and hence ?
  ? 0. Thus the response is nearly in phase with
  the excitation.
• If ? ?0, then cos? 0, sin? 1, and hence ?
  ? ? /2. Thus response lags behind excitation by
  nearly ? /2 radians.
• If ? large, then cos? ? -1, sin? 0, and hence
  ? ? ? . Thus response lags behind excitation by
  nearly ? radians, and hence they are nearly out
  of phase with each other.

46
Example 1 Forced Vibrations with Damping

• Consider the initial value problem

47
Example 1 Forced Vibrations with Damping

• Consider the initial value problem
• Then ?0 1, F0 3, and ? ? 2 /(mk) 1/64
  0.015625.
• The unforced motion of this system

48
Example 1 Forced Vibrations with Damping

• Recall that ?0 1, F0 3, and ? ? 2 /(mk)
  1/64 0.015625.
• The solution for the low frequency case ? 0.3
  is graphed below, along with the forcing
  function.
• Specifically, R ? 3.2939 gt F0/k 3, and ? ?
  0.041185.

49
Example 1 Forced Vibrations with Damping

• Recall that ?0 1, F0 3, and ? 2 /(mk)
  0.015625.
• The solution for the resonant case ? 1
• The steady-state amplitude is eight times the
  static displacement, and the response lags
  excitation by ? /2 radians. Specifically, R 24
  gt F0/k 3, and ? ? /2.

50
Example 1 Forced Vibrations with Damping

• Recall that ?0 1, F0 3, and ? 2 /(mk)
  0.015625.
• The solution for the relatively high frequency
  case ? 2
• The steady-state response is out of phase with
  excitation, and response amplitude is about one
  third the static displacement.
• Specifically, R ? 0.99655 ? F0/k 3, and ? ?
  3.0585 ? ?.

51
Undamped Equation General Solution for the Case
?0 ? ?

• Suppose there is no damping term. Then our
  equation is

52
Undamped Equation General Solution for the Case
?0 ? ?

• Suppose there is no damping term. Then our
  equation is
• Assuming ?0 ? ?, then the method of undetermined
  coefficients can be use to show that the general
  solution is

53
Undamped Equation Mass Initially at Rest

• If the mass is initially at rest, then the
  corresponding initial value problem is

54
Undamped Equation Mass Initially at Rest

• If the mass is initially at rest, then the
  corresponding initial value problem is
• Recall that the general solution to the
  differential equation is

55
Undamped Equation Mass Initially at Rest

• If the mass is initially at rest, then the
  corresponding initial value problem is
• Recall that the general solution to the
  differential equation is
• Using the initial conditions to solve for c1 and
  c2, we obtain

56
Undamped Equation Mass Initially at Rest

• If the mass is initially at rest, then the
  corresponding initial value problem is
• Recall that the general solution to the
  differential equation is
• Using the initial conditions to solve for c1 and
  c2, we obtain
• Hence

57
Undamped Equation Solution to IVP

• Thus our solution is

58
Undamped Equation Solution to IVP

• Thus our solution is
• To simplify the solution even further, let A
  (?0 ?)/2 and B (?0 - ?)/2. Then A B ?0t
  and A - B ?t. Using the trigonometric identity
• it follows that
• and hence

59
Undamped Equation Beats

• Using the results of the previous slide, it
  follows that

60
Undamped Equation Beats

• Using the results of the previous slide, it
  follows that
• When ?0 - ? ? 0, ?0 ? is much larger than ?0
  - ?, and sin(?0 ?)t/2 oscillates more rapidly
  than
• sin(?0 - ?)t/2.

61
Undamped Equation Beats

• Using the results of the previous slide, it
  follows that
• When ?0 - ? ? 0, ?0 ? is much larger than ?0
  - ?, and sin(?0 ?)t/2 oscillates more rapidly
  than
• sin(?0 - ?)t/2.
• Thus motion is a rapid oscillation with frequency
  (?0 ?)/2, but with slowly varying sinusoidal
  amplitude given by

62
Undamped Equation Beats

• This phenomena is called a beat.
• Beats occur with two tuning forks of
• nearly equal frequency.

63
Example 2 Undamped Equation,Mass Initially at
Rest

• Consider the initial value problem

64
Example 2 Undamped Equation,Mass Initially at
Rest

• Consider the initial value problem
• Then ?0 1, ? 0.8, and F0 0.5

65
Example 2 Undamped Equation,Mass Initially at
Rest

• Consider the initial value problem
• Then ?0 1, ? 0.8, and F0 0.5, and hence the
  solution is

66
Example 2 Undamped Equation,Mass Initially at
Rest

• The displacement of the springmass system
  oscillates with a frequency of 0.9, slightly less
  than natural frequency ?0 1.
• The amplitude variation has a slow frequency of
  0.1 and period of 20?.
• A half-period of 10? corresponds to a single
  cycle of increasing and then decreasing
  amplitude.

67
Example 2 Increased Frequency

• Recall our initial value problem

68
Example 2 Increased Frequency

• Recall our initial value problem
• If driving frequency ? is increased to ? 0.9

69
Example 2 Increased Frequency

• Recall our initial value problem
• If driving frequency ? is increased to ? 0.9,
  then the slow frequency is halved to 0.05 with
  half-period doubled to 20?.

70
Example 2 Increased Frequency

• Recall our initial value problem
• If driving frequency ? is increased to ? 0.9,
  then the slow frequency is halved to 0.05 with
  half-period doubled to 20?.
• The multiplier 2.77778 is increased to 5.2632,
  and the fast frequency only marginally increased,
  to 0.095.