Second Order Differential Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Example 55
A 16 lb object stretches a spring 8/9 ft by itself. There is no
damping and no external forces acting on the system. The spring
is initially displaced 6 inches upwards from its equilibrium position
and given an initial velocity of 1 ft/sec downward. Find the
displacement at any time t, u(t).
Solution
We first need to set up the IVP for the problem. We need to find
m and k. This is the British system so we’ll need to compute the
mass.
m=
16
1
W
=
=
g
32
2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Now, lets find k. We can use the fact that mg = kL to find k.
We’ll use feet for the unit of measurement for this problem.
k=
16
mg
=
= 18
L
8/9
We can now set up the IVP.
1 00
1
u + 18u = 0; u(0) = − (6inches), u 0 (0) = 1
2
2
Now, the natural frequency, is
s
18
ω0 =
=6
1/2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
The general solution, along with its derivative, is then,
u(t) = c1 cos(6t) + c2 sin(6t)
u 0 (t) = −6c1 sin(6t) + 6c2 cos(6t)
Applying the initial conditions gives
1
1
− = u(0) = c1 =⇒ c1 = −
2
2
1 = u 0 (0) = 6c2 =⇒ c2 =
1
6
The displacement at any time t is then
1
1
u(t) = − cos(6t) + sin(6t)
2
6
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Now, lets convert this to a single cosine function. First let’s get
the amplitude, R.
s 2 √
1
1 2
10
+
=
= 0.52705
−
R=
2
6
6
Now let’s get the phase shift.
1/6
−1
δ = tan
= −0.32175
−1/2
From the above equations, we have two angles
δ1 = −0.32175;
δ2 = δ1 + π = 2.81984
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
We need to decide which of these phase shifts is correct, because
only one will be correct. To do this recall that
c1 = Rcos(δ) = −1/2 < 0;
c2 = Rsin(δ) = 1/6 > 0
This means that the phase shift must be in Quadrant II and so the
second angle is the one that we need. Thus, the displacement at
any time t is.
√
10
−1 −1
u(t) =
cos(6t − δ); δ = tan
+π
6
3
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Free, Damped Vibrations
We are still going to assume that there will be no external forces
acting on the system, with the exception of damping of course. In
this case the differential equation will be
mu 00 + γu 0 + ku = 0
where m, γ, and k are all positive constants. Upon solving for the
roots of the characteristic equation we get the following.
p
−γ ± γ 2 − 4mk
r1,2 =
2m
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
We will have three cases here.
1. −
γ 2 − 4mk = 0
In this case we will get a double root out of the characteristic
equation and the displacement at any time t will be.
γt
γt
u(t) = c1 e − 2m + c2 te − 2m
Notice that as t → ∞ the displacement will approach zero.
This case is called critical damping and will happen when the
damping coefficient is,
γ 2 − 4mk = 0 =⇒ γ =
Dr. Marco A Roque Sol
√
4mk = γCR
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
The above value of the damping coefficient is called the critical
damping coefficient and denoted by γCR .
2. −
γ 2 − 4mk > 0.
In this case let’s rewrite the roots
r1,2 =
−γ ±
p
γ 2 − 4mk
γ
=−
2m
2m
s
1±
4mk
1− 2
γ
!
Also notice that from our initial assumption that we have,
s
4mk
4mk
4mk
< 1 =⇒ 1 − 2 < 1 =⇒
1− 2 <1
γ2
γ
γ
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
This means that the quantity in the parenthesis is guaranteed to
be positive and so the two roots in this case are guaranteed to be
negative. Therefore the displacement at any time t is
u(t) = c1 e
q
γ
1+ 1− 4mk
− 2m
2 t
γ
+ c2 e
q
γ
− 2m
1− 1− 4mk
2 t
and will approach zero as t → ∞.
This case will occur when
γ 2 − 4mk > 0 =⇒ γ >
√
4mk = γCR
and is called over damping.
Dr. Marco A Roque Sol
Ordinary Differential Equations
γ
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
3. −
γ 2 − 4mk < 0.
In this case we will get complex roots out of the characteristic
equation.
p
γ 2 − 4mk
−γ
±
=λ±i µ
r1,2 =
2m
2m
where the real part is guaranteed to be negative and so the
displacement is
!
p
p
−γ
4mk − γ 2
4mk − γ 2
t
t) + c2 sin(
t)
u(t) = e 2m
c1 cos(
2m
2m
u(t) = = Rcos(µt − δ)
Since λ < 0 the displacement will approach zero as t → ∞ . We
will get this case will occur when
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
γ 2 − 4mk < 0 =⇒ γ <
√
4mk = γCR
and is called under damping.
Example 56
Take the spring and mass system from the example 55 and consider
there is a damping force to it that will exert a force of 17 lbs when
the velocity is 2ft/s. Find the displacement at any time t, u(t).
Solution
So, the only difference between this example and the previous
example is damping force. So let’s find the damping coefficient
√
17 = γ(2) =⇒ γ = 2/17 = 8.5 > γCR =
Dr. Marco A Roque Sol
4km = 6
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
So it looks like we have got over damping this time around so we
should expect to get two real distinct roots from the characteristic
equation and they should both be negative. The IVP for this
example is
1
1 00 17 0
u + u + 18u = 0; u(0) = − ,
2
2
2
The roots of the characteristic equation are
√
−17 ± 145
r1,2 =
2
u 0 (0) = 1
The general solution for this example is
u(t) = c1 e
√
−17+ 145
2
Dr. Marco A Roque Sol
+ c2 e
√
−17− 145
2
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
and after applying initial conditions, the particular solution is
u(t) = −0.52e
√
−17+ 145
t
2
Dr. Marco A Roque Sol
+ 0.020e
√
−17− 145
t
2
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Example 57
Take the spring and mass system from the example 55 and consider
there is a damping force to it that will exert a force of 12lbs when
the velocity is 2ft/s. Find the displacement at any time t, u(t).
Solution
The damping coefficient is given by
12 = γ(2) =⇒ γ = 12/2 = 6 = γCR
So it looks like we have got critical damping this time. The IVP
for this problem is
1 00
u + 6u 0 + 18u = 0;
2
Dr. Marco A Roque Sol
1
u(0) = − ,
2
u 0 (0) = 1
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
The roots of the characteristic equation are
r1,2 = −
6
=6
(2)(1/2)
The general solution for this example is
u(t) = c1 e −6t + c2 te −6t
and after applying initial conditions, the particular solution is
1
u(t) = − e −6t − 2te −6t
2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Example 58
Take the spring and mass system from the example 55 and consider
there is a damping force to it that will exert a force of 5lbs when
the velocity is 2ft/s. Find the displacement at any time t, u(t).
Solution
The damping coefficient is given by
5 = γ(2) =⇒ γ = 5/2 = 2.5 < γCR
So it looks like we have got under damping this time. The IVP for
this problem is
1 00 5 0
u + u + 18u = 0;
2
2
Dr. Marco A Roque Sol
1
u(0) = − ,
2
u 0 (0) = 1
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
The roots of the characteristic equation are
√
−5 ± 119
r1,2 =
2
The general solution for this example is
!
√
119
− 25 t
u(t) = e
c1 cos
t + c2 sin
2
√
119
t
2
!!
and after applying initial conditions, the particular solution is
√
u(t) = e
− 5t
2
−0.5cos
119
t
2
Dr. Marco A Roque Sol
√
!
− 0.046sin
119
t
2
Ordinary Differential Equations
!!
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Lets convert this to a single cosine as we did in the undamped case.
q
R = (−0.5)2 + (.046)2 = 0.502
or
δ = tan
−1
=
−0.046
−0.5
= 0.09
δ = 0.09 + π = 3.23
This means δ must be in the Quadrant III ( why ?) and so the
second angle is the one that we want. The displacement is then
!
√
−5t
119
u(t) = 0.502e 2 cos
t − 3.23
2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Undamped, Forced Vibrations
We will first take a look at the undamped case. The differential
equation in this case is
mu 00 + ku = F (t)
This is just a nonhomogeneous differential equation and we know
how to solve these. The general solution will be
u(t) = uc + UP
There is a particular type of forcing function that we should take a
look at since it leads to some interesting results. Lets suppose that
the forcing function is a simple periodic function of the form
F (t) = F0 cos(ωt) or
Dr. Marco A Roque Sol
F (t) = F0 sin(ωt)
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
For the purposes of this discussion we will use the first one. Using
this, the IVP becomes,
mu 00 + ku = F0 cos(ωt)
The solution of the associate homogeneus , as pointed out above,
is just
uc (t) = c1 cos(ω0 t) + c2 sin(ω0 t)
where ω0 is the natural frequency.
We will need to be careful in finding a particular solution. The
reason for this will be clear if we use undetermined coefficients.
With undetermined coefficients our guess for the form of the
particular solution would be,
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
UP (t) = Acos(ωt) + Bsin(ωt)
Now, this guess will have problems if ω = ω0 . So, we will need to
look at this in two cases.
1. ω 6= ω0
In this case our initial guess is okay since it wont be the
complementary solution. Upon differentiating the guess and
plugging it into the differential equation and simplifying we get,
mu 00 + ku = F0 cos(ωt)
m (Acos(ωt) + Bsin(ωt))00 + k (Acos(ωt) + Bsin(ωt)) = F0 cos(ωt)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
m −ω 2 Acos(ωt) − ω 2 Bsin(ωt) + k (Acos(ωt) + Bsin(ωt)) = F0 cos(ωt)
−mω 2 A + kA cos(ωt) + −mω 2 B + kB sin(ωt) = F0 cos(ωt)
Setting coefficients equal gives us,
cos(ωt)
−mω 2 + k A = F0 =⇒ A =
sin(ωt)
F0
k − mω 2
−mω 2 + k B = 0 =⇒ B = 0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
The particular solution is then
F0
F0
F0
cos(ωt)
cos(ωt) =
cos(ωt) =
k − mω 2
m (k/m − ω 2 )
m ω02 − ω 2
Note that we rearranged things a little. Depending on the form
that you’d like the displacement to be in we can have either of the
following.
u(t) = c1 cos(ω0 t) + c2 sin(ω0 t) +
u(t) = Rcos(ω0 t − δ) +
F0
cos(ωt)
m ω02 − ω 2
F0
cos(ωt)
m ω02 − ω 2
If we used the sine form of the forcing function we could get a
similar formula.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
2. ω = ω0
In this case we will need to add in a t to the guess for the
particular solution.
UP (t) = Atcos(ωt) + Btsin(ωt)
Differentiating our guess, plugging it into the differential equation
and simplifying gives us the following.
−mω02 + k Atcos(ωt) + −mω02 + k Btsin(ω)t + ...
... + 2mω0 Bcos(ωt) − 2mω0 Asin(ωt) = F0 cos(ωt)
but
−mω02 + k = m −ω02 + k/m = m −ω02 + ω02 = 0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
So, the first two terms actually drop out and this gives us
cos(ωt)
2mω0 B = F0 =⇒ B =
sin(ωt)
F0
2mω0
2mω0 A = 0 =⇒ A = 0
In this case the particular will be,
F0
tsin(ω0 t)
2mω0
The displacement for this case is then
u(t) = c1 cos(ω0 t) + c2 sin(ω0 t) +
Dr. Marco A Roque Sol
F0
tsin(ω0 t)
2mω0
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
depending on the form that you prefer for the displacement.
F0
tsin(ω0 t)
u(t) = Rcos(ω0 t − δ) +
2mω0
So, what was the point of the two cases here? Well in the first
case, our displacement function consists of two cosines and is nice
and well behaved for all time.
In contrast, the second case, will have some serious issues at t
increases. The addition of the t in the particular solution will mean
that we are going to see an oscillation that grows in amplitude as t
increases. This case is called resonance and we would generally
like to avoid this at all costs.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
In this case resonance arose by assuming that the forcing function
was,
F (t) = F0 cos(ω0 t)
We would also have the possibility of resonance if we assumed a
forcing function of the form.
F (t) = F0 sin(ω0 t)
We should also take care to not assume that a forcing function will
be in one of these two forms. Forcing functions can come in a wide
variety of forms. If we do run into a forcing function different from
the one that used here you will have to go through undetermined
coefficients or variation of parameters to determine the particular
solution.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Example 59
A 3 kg object is attached to spring and will stretch the spring 392
mm by itself. There is no damping in the system and a forcing
function of the form
F (t) = 10cos(ωt)
is attached to the object and the system will experience resonance.
If the object is initially displaced 20 cm downward from its
equilibrium position and given a velocity of 10 cm/sec upward find
the displacement at any time t.
Solution
Since we are in the metric system we wont need to find mass as its
been given to us. Also, for all calculations we will be converting all
lengths over to meters. The first thing we need to do is find k.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
mg
(3)(9.8)
=
= 75
L
0.392
Now, we are told that the system experiences resonance so let’s go
ahead and get the natural frequency.
r
r
k
75
=
=5
ω0 =
m
3
The IVP for this is then
k=
3u 00 + 75u = 10cos(5t);
u(0) = 0.2, u 0 (0) = −0.1
The complementary solution is the free undamped solution which
is easy to get and for the particular solution we can just use the
formula that we derived above. The general solution is then,
u(t) = c1 cos(5t) + c2 sin(5t) +
Dr. Marco A Roque Sol
10
tsin(5t)
(2)(3)(5)
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
Applying the initial conditions gives
1
1
1
u(t) = cos(5t) − sin(5t) + tsin(5t)
5
50
3
The last thing that we’ll do is combine the first two terms into a
single cosine.
s 2
1 2
−1
−1 −1/50
R=
+
= 0.201 δ1 = tan
= −0.099
5
50
1/5
δ2 = δ1 + π = 3.042
In this case c1 > 0 is positive and c2 < 0 . This means that the
phase shift needs to be in Quadrant IV and so the first one is the
correct phase shift this time.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Mechanical and Electrical Vibrations
Mechanical and Electrical Vibrations
The displacement then becomes,
r
1
1 101
−1 −1
cos 5t + tan
+ tsin(5t)
u(t) =
5 100
10
3
Dr. Marco A Roque Sol
Ordinary Differential Equations