Embed Size (px)
Citation preview
Ch 3.7: Mechanical & Electrical Vibrations
• Two important areas of application for second order linear equations with
constant coefficients are in modeling mechanical and electrical oscillations.
• We will study the motion of a mass on a spring in detail.
• An understanding of the behavior of this simple system is the first step in
investigation of more complex vibrating systems.
Spring – Mass System
• Suppose a mass m hangs from a vertical spring of original length l. The
mass causes an elongation L of the spring.
• The force FG of gravity pulls the mass down. This force has magnitude mg,
where g is acceleration due to gravity.
• The force FS of the spring stiffness pulls the mass up. For small elongations
L, this force is proportional to L. That is, Fs = - kL (Hooke’s Law).
• When the mass is in equilibrium, the forces balance each other : 0,Sw F
Spring Model
• We will study the motion of a mass when it is acted on by an external force
(forcing function) and/or is initially displaced.
• Let u(t) denote the displacement of the mass from its equilibrium position at
time t, measured downward.
• Let f be the net force acting on the mass. We will use Newton’s 2nd Law:
• In determining f, there are four separate forces to consider:
– Weight: w = mg (downward force)
– Spring force: Fs = - k(L+ u) (up or down force, see next slide)
– Damping force: Fd(t) = - u (t) (up or down, see following slide)
– External force: F (t) (up or down force, see text)
)()( tftum
Spring Model:
Spring Force Details
• The spring force Fs acts to restore a spring to the natural position, and is
proportional to L + u. If L + u > 0, then the spring is extended and the spring
force acts upward. In this case
• If L + u < 0, then spring is compressed a distance of |L + u|, and the spring
force acts downward. In this case
• In either case,
( ), 0.sF k L u L u
uLkuLkuLkFs
( )sF k L u
Spring Model:
Damping Force Details
• The damping or resistive force Fd acts in the opposite direction as the motion
of the mass. This can be complicated to model. Fd may be due to air resistance,
internal energy dissipation due to action of spring, friction between the mass
and guides, or a mechanical device (dashpot) imparting a resistive force to the
mass.
• We simplify this and assume Fd is proportional to the velocity.
• In particular, we find that
– If u > 0, then u is increasing, so the mass is moving downward. Thus Fd
acts upward and hence Fd = - u, where > 0.
– If u < 0, then u is decreasing, so the mass is moving upward. Thus Fd
acts downward and hence Fd = - u , > 0.
• In either case, ( ) ( ), 0.dF t u t
Spring Model:
Differential Equation
• Taking into account these forces, Newton’s Law becomes:
• Recalling that mg = kL, this equation reduces to
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 the mass is set in motion with a given initial
displacement and velocity, then its position is uniquely determined at all future
times.
(Question) How do we compute the constants m, , and k ?
( ) ( ) ( ) ( ) ( ) ( ) ( )s dmu t mg F t F t F t mg k L u t u t F t
( ) ( ) ( ) ( )mu t u t ku t F t
0 0(0) , (0)u u u v
(Example) An object weighing 4 lb-force 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-force when
the mass has a velocity of 3 ft/sec. Formulate the initial value
problem (IVP) that governs the motion of this object.
- w = mg
-
- Initial conditions ?
232 ft/secg
( ) ( ), 0.dF t u t SF kL
Example 1:
Find Coefficients (1 of 2)
• A mass weighing 4 lb-force 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-force when the object has a velocity of 3 ft/sec. Formulate
the IVP that governs the motion of this object:
• Find m:
• Find :
• Find k:
2
2
4lb 1 lbsec
32ft / sec 8 ft
ww mg m m m
g
6lb lbsec 2
3ft / sec ft
dF
u
4lb 4lb lb, = 24
2in 1/ 6ft ft
sS
Fk F w k k k
L
0 0( ) ( ) ( ) ( ), (0) , (0)mu t u t ku t F t u u u v
Example 1: Find IVP (2 of 2)
• Thus our differential equation becomes
and hence the initial value problem can be written as (unit: ft)
1( ) 2 ( ) 24 ( ) 0
8u t u t u t
( ) 16 ( ) 192 ( ) 01
(0) , (0) 02
u t u t u t
u u
Example 1: Find IVP (2 of 2)
• Thus our differential equation becomes
and hence the initial value problem can be written as
• This problem can be solved using the methods of
Chapter 3.3 and yields the solution:
1( ) 2 ( ) 24 ( ) 0
8u t u t u t
( ) 16 ( ) 192 ( ) 01
(0) , (0) 02
u t u t u t
u u
8 81 1( ) cos(8 2 ) sin(8 2 )
2 2 2
t tu t e t e t
(Example 2) A mass weighing 10 lb-force stretches
a spring 2". The mass is displaced an additional 2" and
then set in motion with an initial upward velocity of 1
ft/sec. Determine the position of the mass at any later
time, and find the period, amplitude, and phase of the
motion.
Example 2: Find IVP (1 of 2)
• A mass weighing 10 lb-force stretches a spring 2". The mass is displaced an
additional 2" and then set in motion with an initial upward velocity of 1 ft/sec.
Determine the position of the mass at any later time, and find the period,
amplitude, and phase of the motion: = 0
• Find m:
• Find k:
• Thus our IVP is
ft
seclb
16
5
sec/ft32
lb10 2
2 mm
g
wmmgw
ft
lb60
ft6/1
lb10
in2
lb10 kkkLkFs
1)(,6/1)0(,0)(60)(16/5 tuututu
00 )0(,)0(,0)()( vuuutkutum
Example 2: Find Solution (2 of 2)
• Simplifying, we obtain
• To solve, use methods of Ch 3.3 to obtain
or
1)0(,6/1)0(,0)(192)( uututu
tttu 192sin192
1192cos
6
1)(
tttu 38sin38
138cos
6
1)(
Spring Model:
Undamped Free Vibrations (1 of 4)
• Recall our differential equation 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
0)()( tkutum
)()()()( tFtkututum
mk
tBtAtu
/
where
,sincos)(
2
0
00
Spring Model:
Undamped Free Vibrations (2 of 4)
• Using trigonometric identities, the solution
can be rewritten as follows:
where
• Note that in finding , we must be careful to choose the correct quadrant.
This is done using the signs of cos and sin .
•
mktBtAtu /,sincos)( 2
000
,sinsincoscos)(
cos)(sincos)(
00
000
tRtRtu
tRtutBtAtu
A
BBARRBRA tan,sin,cos 22
cos( ) cos cos sin sin
Spring Model:
Undamped Free Vibrations (3 of 4)
• Thus our solution is
where
• The solution is a shifted cosine (or sine) curve, that describes simple harmonic
motion, with period
• The circular frequency 0 (radians/time) is the natural frequency of the
vibration, R is the amplitude of the maximum displacement of mass from
equilibrium, and is the phase or phase angle (dimensionless).
tRtBtAtu 000 cos sincos)(
k
mT
2
2
0
mk /0
Spring Model:
Undamped Free Vibrations (4 of 4)
• Note that our solution
is a shifted cosine (or sine) curve with period
• Initial conditions determine A & B, hence also the amplitude R.
• The system always vibrates with the same frequency 0 , regardless of the initial
conditions.
• The period T increases as m increases, so larger masses vibrate more slowly.
However, T decreases as k increases, so stiffer springs cause a system to vibrate
more rapidly.
mktRtBtAtu /,cos sincos)( 0000
k
mT 2
Example 2: Find IVP (1 of 3)
• A mass weighing 10 lb-force stretches a spring 2". The mass is displaced an
additional 2" and then set in motion with an initial upward velocity of 1 ft/sec.
Determine the position of the mass at any later time, and find the period,
amplitude, and phase of the motion: = 0
• Find m:
• Find k:
• Thus our IVP is
ft
seclb
16
5
sec/ft32
lb10 2
2 mm
g
wmmgw
ft
lb60
ft6/1
lb10
in2
lb10 kkkLkFs
1)(,6/1)0(,0)(60)(16/5 tuututu
00 )0(,)0(,0)()( vuuutkutum
Example 2: Find Solution (2 of 3)
• Simplifying, we obtain
• To solve, use methods of Ch 3.3 to obtain
or
1)0(,6/1)0(,0)(192)( uututu
tttu 192sin192
1192cos
6
1)(
tttu 38sin38
138cos
6
1)(
Example 2:
Find Period, Amplitude, Phase (3 of 3)
• The natural frequency is
• The period is
• The amplitude is
• Next, determine the phase :
tttu 38sin38
138cos
6
1)(
rad/sec 856.1338192/0 mk
sec 45345.0/2 0 T
ft 18162.022 BAR
rad 40864.04
3tan
4
3tantan 1
A
B
ABRBRA /tan,sin,cos
Thus ( ) 0.182cos 8 3 0.409u t t
Spring Model: Damped Free Vibrations (1 of 8)
• Suppose there is damping but no external driving force F(t):
• What is effect of the damping coefficient on system?
• The characteristic equation is
• Three cases for the solution:
0)()()( tkututum
2
2
21
411
22
4,
mk
mm
mkrr
term.damping
thefrom expected as ,0)(limcases, threeallIn :Note
.02
4,sincos)(:04
;02/where,)(:04
;0,0where,)(:04
22/2
2/2
21
2 21
tu
m
mktBtAetumk
meBtAtumk
rrBeAetumk
t
mt
mt
trtr
Damped Free Vibrations: Small Damping (2 of 8)
• Of the cases for solution form, the last is most important, which occurs when
the damping is small:
• We examine this last case. Recall
• Then
and hence
(damped oscillation)
1 22
1 2
2 2
2
/
/2
4 0 : ( ) , 0, 0
( ) cos sin
4 0 : ( ) , / 2 0
4 0 : , 0t m
r t r t
t m
mk u t Ae Be r r
mk u t A Bt e
u t e A t
m
k tm B
sin,cos RBRA
teRtu mt cos)( 2/
mteRtu 2/)(
Damped Free Vibrations: Quasi Frequency (3 of 8)
• Thus we have damped oscillations:
• The amplitude R depends on the initial conditions, since
• Although the motion is not periodic, the parameter determines the mass
oscillation frequency.
• Thus is called the quasi frequency.
• Recall
sin,cos,sincos)( 2/ RBRAtBtAetu mt
mtmt eRtuteRtu 2/2/ )(cos)(
m
mk
2
4 2
Damped Free Vibrations: Quasi Period (4 of 8)
• Compare with 0 , the frequency of undamped motion:
• Thus, small damping reduces oscillation frequency slightly.
• Similarly, the quasi period is defined as Td = 2/. Then
• Thus, small damping increases quasi period.
kmkmmkkm
kmkm
km
mkm
km
mkm
km
81
81
6441
41
4
4
/4
4
/2
4
22
2
22
42
22
2
22
0
For small
kmkmkmT
Td
81
81
41
/2
/2 21
22/1
2
0
0
Damped Free Vibrations:
Neglecting Damping for Small 2/4km (5 of 8)
• Consider again the comparisons between damped and undamped frequency
and period:
• Thus it turns out that a small is not as telling as a small ratio 2/4km.
• For small 2/4km, we can neglect the effect of damping when calculating the
quasi frequency and quasi period of motion. But if we want a detailed
description of the motion of the mass, then we cannot neglect the damping
force, no matter how small it is.
2/12
2/12
0 41,
41
kmT
T
km
d
Damped Free Vibrations:
Frequency, Period (6 of 8)
• Ratios of damped and undamped frequency, period:
• Thus
• The importance of the relationship between 2 and 4km is supported by our
previous equations:
2/12
2/12
0 41,
41
kmT
T
km
d
dkmkm
T22
lim and 0lim
0,sincos)(:04
02/,)(:04
0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
Damped Free Vibrations:
Critical Damping Value (7 of 8)
• Thus the nature of the solution changes as passes through the value
• This value of is known as the critical damping value, and for larger values
of the motion is said to be overdamped.
• Thus for the solutions given by these cases,
we see that the mass creeps back to its equilibrium position for solutions (1)
and (2), but does not oscillate about it, as it does for small in solution (3).
• Soln (1) is overdamped and soln (2) is critically damped.
.2 km
)3(0,sincos)(:04
)2( 02/,)(:04
)1(0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
Damped Free Vibrations:
Characterization of Vibration (8 of 8)
• The mass creeps back to the equilibrium position for solutions (1) & (2), but
does not oscillate about it, as it does for small in solution (3).
• Solution (1) is overdamped and
• Solution (2) is critically damped.
• Solution (3) is underdamped
)3( (Blue)sincos)(:04
)2( Black) (Red,02/,)(:04
)1((Green)0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
Example 3: Initial Value Problem (1 of 4)
• Suppose that the motion of a spring-mass system is governed by the initial value problem
• Find the following:
(a) quasi frequency and quasi period;
(b) time at which mass passes through equilibrium position;
(c) time such that |u(t)| < 0.1 for all t > .
• For Part (a), using methods of this chapter we obtain:
where
0)0(,2)0(,0125.0 uuuuu
tettetu tt
16
255cos
255
32
16
255sin
255
2
16
255cos2)( 16/16/
)sin,cos (recall 06254.0255
1tan RBRA
Example 3: Quasi Frequency & Period (2 of 4)
• The solution to the initial value problem is:
• The graph of this solution, along with solution to the corresponding undamped
problem, is given below.
• The quasi frequency is
and quasi period is
• For the undamped case:
tettetu tt
16
255cos
255
32
16
255sin
255
2
16
255cos2)( 16/16/
998.016/255
295.6/2 dT
283.62,10 T
Example 3: Quasi Frequency & Period (3 of 4)
• The damping coefficient is = 0.125 = 1/8, and this is 1/16 of the critical
value
• Thus damping is small relative to mass and spring stiffness. Nevertheless the
oscillation amplitude diminishes quickly.
• Using a solver, we find that |u(t)| < 0.1
for t > 47.515 sec
22 km
Example 3: Quasi Frequency & Period (4 of 4)
• To find the time at which the mass first passes through the equilibrium
position, we must solve
• Or more simply, solve
016
255cos
255
32)( 16/
tetu t
sec 637.12255
16
216
255
t
t
Electric Circuits
• The flow of current in certain basic electrical circuits is modeled by second
order linear ODEs with constant coefficients:
• It is interesting that the flow of current in this circuit is mathematically
equivalent to motion of spring-mass system.
• For more details, see text.
00 )0(,)0(
)()(1
)()(
IIII
tEtIC
tIRtIL
