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Diff. Eqns.
Problem Set 5
Solutions
1. Find the general solution of the nonhomogeneous equation
1
1
y 00 − y 0 + 2 y = −6t2 ,
t
t
t > 0.
(1)
(Hint: One solution of the homogeneous equation is y1 = t.)
Since we are given y1 (t), we can use reduction of order to solve this problem. We begin by
assuming our solution will be of the form
y(t) = v(t)y1 (t) = v(t)t.
(2)
Plugging this into (1), we get the following equation, and we can solve first for v 0 (t), then
integrate to get v(t):
1
1 0
v (t) t + v (t) + 2 v (t) t = −6t2
v 00 (t) t + 2v 0 (t) −
t
t
tv 00 (t) + v 0 (t) = −6t2
d
tv 0 (t) = −6t2
dt
Z
tv 0 (t) =
(3)
(4)
(5)
−6t2 dt
tv 0 (t) = −2t3 + c1
0
2
(6)
(7)
−1
v (t) = −2t + c1 t
Z
v (t) =
−2t2 + c1 t−1 dt
2
= − t3 + c1 ln t + c2 .
3
(8)
(9)
(10)
So our solution is
y (t) = v (t) y1 (t)
2 3
= − t + c1 ln t + c2 t
3
2
= − t4 + c1 t ln t + c2 t.
3
2. A mass weighing 3 lb
contracting the spring a
ward velocity of 2 ft/sec
of the mass at any time
the oscillation.
(11)
(12)
(13)
stretches a spring 2 in. If the mass is pushed upward,
distance of 2 in, and then set in motion with a downand if there is no damping, then find the position u(t)
t. Determine the frequency, period and amplitude of
We start with the equation governing this system,
mu00 + γu0 + ku = F (t).
(14)
We can calculate the parameters from the information given.
weight
3 lb · s2
3 lb
3
=
=
=
slugs
2
gravity
32 ft
32
32 ft/s
damping constant = γ = 0
3 lb
lb
weight
=
= 18
spring constant = k =
displacement
1/6 ft
ft
external force = F (t) = 0.
mass = m =
(15)
(16)
(17)
(18)
We are also given initial conditions u(0) = −2 in and u0 (0) = 2 ft/sec, so (14) becomes
3 00
u + 18u = 0,
32
u00 + 192u = 0.
u(0) = −
1
ft,
6
u0 (0) = 2 ft/sec
(19)
(20)
√
The characteristic equation is r2 + 192 = 0, with roots r = ±8 3i. Thus our solution is
√ √ (21)
u (t) = c1 cos 8 3t + c2 sin 8 3t .
We use the initial conditions to solve for c1 and c2 .
1
− = u(0) = c1
6
√
2 = u0 (0) = 8 3c2
(22)
1
⇒ c2 = √ .
4 3
So our equation for displacement of the spring is
√ √ 1
1
u (t) = − cos 8 3t + √ sin 8 3t .
6
4 3
(23)
(24)
To determine frequency, period and amplitude of the oscillation, we want to reformat our
solution from
u (t) = A cos (ω0 t) + B sin (ω0 t)
(25)
u (t) = R cos (ω0 t + δ) ,
(26)
A = R cos δ,
B = R sin δ
(27)
B
.
A
(28)
to
using the formulas
which give us
R=
p
A2 + B 2 ,
tan δ =
√
So in this√case we have A = −1/6, B = 1/(4 3), and the angular frequency of oscillation
is ω0 = 8 3 rad/sec. The period can be calculated from the angular frequency as
T =
2π
π
= √ sec.
ω0
4 3
(29)
Using (27), the amplitude is
p
R = A2 + B 2 =
s 2 √
1
1 2
7
√
+
=
−
ft.
6
12
4 3
(30)
We are not asked to calculate the phase δ.
3. A spring is stretched 5 cm by a force of 2 N. A mass of 2 kg is hung from the
spring and is also attached to a viscous damper that exerts a force of 2 N when
the velocity of the mass is 4 m/s. If the mass is pulled down 3 cm below its
equilibrium position and given an initial upward velocity of 5 cm/s, determine
the position u(t) of the mass at any time t.
We start with the equation governing this system,
mu00 + γu0 + ku = F (t).
(31)
We can calculate the parameters from the information given.
mass = m = 2 kg
force
2N
1 N·s
1 kg
damping constant = γ =
=
=
=
velocity
4 m/s
2 m
2 s
weight
2N
N
kg
spring constant = k =
=
= 40
= 40 2
displacement
0.05 m
m
s
external force = F (t) = 0 N.
(32)
(33)
(34)
(35)
We can also determine initial conditions from the information. So (31) becomes
1
2u00 + u0 + 40u = 0,
2
u(0) = 0.03 m,
u0 (0) = −.05 m/s.
(36)
The characteristic equation is 2r2 + 12 r + 40 = 0, which has roots r = −1/8 ±
The general solution is then
!
!!
√
√
1
1279
1279
u (t) = exp − t c1 cos
t + c2 sin
t
.
8
8
8
√
1279/8i.
(37)
To find c1 and c2 we use the initial conditions.
0.03 = u(0) = c1
√
−0.05 = u0 (0) =
(38)
1279
1
c2 − c1
8
8
⇒ c2 = −
37
√
≈ −0.010346.
100 1279
(39)
Thus our solution is
1
u (t) = exp − t
8
√
0.03 cos
!
1279
37
√
t −
sin
8
100 1279
√
1279
t
8
!!
(40)
4. The displacement u(t) of a mass-spring-damper system is governed by the
equation
mu00 + γu0 + ku = F0 cos ωt
(41)
where m is the mass, γ is the damping constant, k is the spring constant, F0 is
the amplitude of the applied force, and ω is the frequency of the periodic forcing.
(a) Consider first the undamped case with γ = 0. Find the value of ω for which
resonance occurs for the case m = 2, k = 10 and F0 = 3.
Resonance occurs when the natural frequency ωp
0 equals ω,
√ the frequency of the applied
force.√ The natural frequency is given by ω0 = k/m = 5. So resonance occurs when
ω = 5.
(b) Now consider a damped system with γ = 1. Determine the amplitude of the
forced oscillation for the case m = 3, k = 2, F0 = 1 and ω = 1. (Hint: See page
208.)
Page 208 gives us the following formula for the amplitude of the forced oscillation:
F0
R= q
,
2
m2 ω02 − ω 2 + γ 2 ω 2
ω02 = k/m.
(42)
First calculate ω02 = k/m = 2/3. Then
1
1
R= q
=√ .
2
32 (2/3 − 12 )2 + 12 · 12
(43)