Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 3.8 Forced Mechanical Vibration due 03/25/2021 at 11:59pm MST
1. (1 point) This is an example of an Undamped Forced Os-
cillation where the phenomenon of Beats Occurs.
Find the solution of the initial value problem:
x00 +3.24x=5cos(2t),x(0) = x0(0) = 0
x(t) =
Graph the solution to confirm the phenomenon of Beats. Note
that you may have to use a large window in order to see more
than one beat.
What is the length of each beat?
Length =
Would you be able to explain why the beats phenomenon oc-
curs for this particular example?
Solution: The complementary solution has the form
xc=c1cos(1.8t) + c2sin(1.8t)
We assume a particular solution of the form
xp=Acos(2t) + Bsin(2t)
Then
x0
p=−2Asin(2t) + 2Bcos(2t)
and
x00
p=−4Acos(2t)−4Bsin(2t)
Substituting into the differential equation yields
−4Acos(2t)−4Bsin(2t)+3.24(Acos(2t) + Bsin(2t)) = 5cos(2t)
(−4A+3.24A)cos(2t)+(−4B+3.24B)sin(2t) = 5cos(2t)
Thus
−4A+3.24A=5
−4B+3.24B=0.
which gives A=−125
19 ,B=0.
Hence the particular solution is xp=−125
19 cos(2t)and the gen-
eral solution is
x(t) = xc+xp=c1cos(1.8t) + c2sin(1.8t)−125
19 cos(2t)
Substituting the initial conditions, yields c1= +125
19 and c2=0.
Thus the solution to the initial value problem is
x(t) = −125
19 (cos(2t)−cos(1.8t))
Using the identity cos(ωt)−cos(ωot) = −2sinω−ωo
2tsinω+ωo
2t,
with ω=2 and ωo=1.8, the solution can be written as
x(t) = 250
19 sin(0.1t)sin(1.9t).
The slowly varying amplitude A(t) = 250
19 sin(0.1t)determines
the length of the beats. The period of A(t)is 2π
0.1. Thus the
length of the beats is
π
0.1.
The beats phenomenon occurs because ω+ωo=2+1.8=3.8
is large in comparison to ω−ωo=2−1.8=0.2.
Answer(s) submitted:
•
•
(incorrect)
Correct Answers:
•-6.57894736842105*(cos(2*t)-cos(1.8*t))
•31.4159265358979
2. (1 point) This is an example of an Undamped Forced Os-
cillation where the phenomenon of Pure Resonance Occurs.
Find the solution of the initial value problem:
x00 +9x=24sin(3t),x(0) = x0(0) = 0
x(t) =
Graph the solution to confirm the phenomenon of Pure Reso-
nance.
Solution: The complementary solution has the form
xc=c1cos(3t) + c2sin(3t)
We assume a particular solution of the form
xp=t(Acos(3t) + Bsin(3t)
Then
x0
p=Acos(3t) + Bsin(3t) +t(−3Asin(3t) + 3Bcos(3t))
and
x00
p=−6Asin(3t) + 6Bcos(3t) + t(−9Acos(3t)−9Bsin(3t))
Substituting into the differential equation yields
−6Asin(3t)+6Bcos(3t)+t(−9Acos(3t)−9Bsin(3t))+9t(Acos(3t)+Bsin(3t)) = 24sin(3t)
−6Asin(3t) + 6Bcos(3t) = 24 sin(3t)
Thus
A=−4,B=0.
Hence the particular solution is xp=−4tcos(3t)and the general
solution is
x(t) = xc+xp=c1cos(3t) + c2sin(3t)−4tcos(3t)
Substituting the initial conditions, yields c1=0 and c2=4
3.
Thus the solution to the initial value problem is
x(t) = 4
3sin(3t)−4tcos(3t)
1
Note that the amplitude of the solution grows unbouded as
t→∞, which is consistent with the phenomenon of pure reso-
nance.
Answer(s) submitted:
•-4sin(4t)-4cos(4t)
(incorrect)
Correct Answers:
•-4 * t * cos(3*t) + 24*sin(3*t)/(2*3ˆ2)
3. (1 point) The solution to the Initial value problem
x00 +2x0+10x=5cos(7t),x(0) = 0,x0(0) = 0
is the sum of the steady periodic solution xsp and the transient
solution xtr. Find both xsp and xtr.
xsp =
xtr =
Solution: The roots of the characteristic equation r2+2r+
10 =0 are −1±3i. Thus the complementary solution has the
form
xc=e−t(c1cos(3t) + c2sin(3t))
We assume a particular solution of the form
xp=Acos(7t) + Bsin(7t)
Then
x0
p=−7Asin(7t) + 7Bcos(7t)
and
x00
p=−49Acos(7t)−49Bsin(7t)
Substituting into the differential equation yields
(−39A+14B)cos(7t) + (−39B−14A)sin(7t) = 5cos(7t)
Thus
−39A+14B=5
−39B−14A=0.
which gives A=−195
1717 ,B=70
1717 .
Hence the steady periodic solution is
xsp =−195
1717 cos(7t) + 70
1717 sin(7t)
and the general solution is
x(t) = xtr +xsp =e−t(c1cos(3t)+c2sin(3t))−195
1717 cos(7t)+ 70
1717 sin(7t)
Substituting the initial conditions in the general solution yields
c1= + 195
1717 and c2=−295
5151 .
Thus the transient solution is
xtr =e−t+195
1717 cos(3t)−295
5151 sin(3t)
The amplitude of the steady periodic solution is C1=
q−195
1717 2+70
1717 2=q25
1717 .
The steady periodic solution can be written as xsp =
q25
1717 cos(7t−α).
Since Aand Bare both negative, αis in quadrant 3. Thus
α=arctan(B/A) + π=arctan−14
39 +π
Answer(s) submitted:
•
•
(incorrect)
Correct Answers:
•0.120665983430988*cos(7*t-2.79694533797177)
•exp(-t)*5/1717*(--39*cos(3*t) - (10 + 7**2) *sin(3*t)/3)
2
