1. Alsulami, Falcone, Filippenko 1
Project 1: Forced Vibrations
Mark Falcone, Raed Alsulami, Simon Filippenko
March 4, 2016
Professor Henriksen

2. Alsulami, Falcone, Filippenko 2
1) Problem 1 is a mass­spring system much like we have
encountered in class. The challenge in this problem
lies in dealing with the and in the equation, inω ω0
addition to solving this problem entirely in terms of
variables.
y cos(ωt)y′′ + ω0
2 = m
F0
d = 0
No damping effect is given, meaning that d=0, and in
terms of writing this as a differential equation, b
also becomes 0.
b2
= 0
a and c are both positive and real
ac se complex number equationb2
− 4 < 0 → U
−α = b
2a = 0
e cos(βt) e sin(βt)yg = c1
0 + c2
0 + yp
Next, beta must be found, as in the quadratic formula.
β = 2a
√b −4ac2
= 2
√0−4ω0
2
= 2
2ω0
= ω0
cos(ω t) sin(ω t)yg = c1 0 + c2 0 + yp
Next, the constants and must be found using thec1 c2
given information and the particular solution.
cos(ωt) sin(ωt)yp = A + B
− ωsin(ωt) ωcos(ωt)yp′ = A + B
− ω cos(ωt) ω sin(ωt)yp′′ = A 2 − B 2
Y and its respective derivatives can be applied to the
original equation stated.
ω cos(ωt) ω sin(ωt) [Acos(ωt) sin(ωt)] cos(ωt)− A 2 − B 2 + ω0
2 2
+ B = m
F0
There are no sine terms on the right side of the
equation, therefore the sine terms can be eliminated.
ω cos(ωt) ω cos(ωt) cos(ωt)− A 2 + A 2 = m
F0

3. Alsulami, Falcone, Filippenko 3
can be factored out, and the equation can beos(ωt) c
simplified algebraically.
Aω ω ][ 0
2 − A 2 = m
F0
A =
F0
m(ω −ω )0
2 2
Now that A is a known value, it can be plugged into
the particular solution .yp
cos(ωt)yp =
F0
m(ω −ω )0
2 2
This particular solution can be used in the previously
used equation for the general solution to give a final
result of:
cos(ω t) sin(ω t) cos(ωt)yg = c1 0 + c2 0 +
F0
m(ω −ω )0
2 2
2)While Problem 2 is separate from 1, no new equations
are introduced. This problem takes the final result
from Problem 1 and applies new conditions to it to
explore the mathematical implications that altering
and applying initial conditions can have on such
equations. First to see what happens with an initial
velocity and displacement of 0 (y(0)=y’(0)=0), then to
explore the possibility of .ω0 ≈ ω
2­A)
y(t)= cos(w t) sin(w t) cos(ωt)C1 0 + C2 0 +
F0
m(ω −ω )0
2 2
y(0)=y’(0)=0
y(0)= cos(0) sin(0) cos(0)C1 + C2 +
F0
m(ω −ω )0
2 2
= , so C1 + 0 +
F0
m(ω −ω )0
2 2 = 0 −C1 =
F0
m(ω −ω )0
2 2
y’(0)=­ w sin(w t) w cos(w t) sin(ωt)C1 0 0 + C2 0 0 −
F0
m(ω −ω )0
2 2
= ­ w sin(0) w cos(0) sin(0)C1 0 + C2 0 −
F0
m(ω −ω )0
2 2
= 0+ , so wC2 0 − 0 = 0 C2 = 0
y(t)= cos(ω t)−
F0
m(ω −ω )0
2 2 0 + sin(ω t)
F0
m(ω −ω )0
2 2
= (cos(ω t) os(ω t))
F0
m(ω −ω )0
2 2 − c 0

4. Alsulami, Falcone, Filippenko 4
2­B)
Here is the solution using the trigonometric identity:
y(t)= (cos(ω t) os(ω t))
F0
m(ω −ω )0
2 2 − c 0
by applying the trigonometric identity we get:
cos𝛼­cos𝛽=­2sin 𝛼+𝛽/2 * sin 𝛼­𝛽/2
y(t)= *(
F0
m(ω −ω )0
2 2 sin wt t/2 in wt t/2)− 2 + w0 * s − w0
(t) cos(ω t) sin(ω t) cos(ωt)y = c1 0 + c2 0 +
F0
m(ω −ω )0
2 2
(0) (0)y = y′ = 0
(0) cos(0) sin(0) cos(0)y = c1 + c2 +
F0
m(ω −ω )0
2 2
cos(0)= c1 + 0 +
F0
m(ω −ω )0
2 2 = 0
−c1 =
F0
m(ω −ω )0
2 2
(0) − ω sin(ω t)y′ = c1 0 0 + c2 cos(ω t) sin(ωt)ω0 0 −
F0
m(ω −ω )0
2 2
− ω sin(0) ω cos(0) sin(0)= c1 0 + c2 0 −
F0
m(ω −ω )0
2 2
ω= 0 + c2 0 − 0 = 0
c2 = 0
(t) − cos(ω t) cos(ωt)y =
F0
(ω −ω )0
2 2 0 +
F0
m(ω −ω )0
2 2
(cos(ωt) os(ω t))=
F0
m(ω −ω )0
2 2 − c 0
(t) (cos(ωt) os(ω t))y =
F0
m(ω −ω )0
2 2 − c 0
Apply trigonometric identity
os(α) os(β) − sin inc − c = 2 2
α+β
* s 2
α−β
(t) (− sin( )sin( ))y =
F0
m(ω −ω )0
2 2 2 2
ωt+ω t0
2
ωt−ω t0
(sin( )sin( ))=
2F0
m(ω −ω )0
2 2 2
ω t+ωt0
2
ω t−ωt0

5. Alsulami, Falcone, Filippenko 5
2­C)The graph below displays the equation solved in
2­B aswell as the graph of (t) ± sin y =
2F0
m(ω 2−ω )0
2 2
(ω −ω)t0
For this graph
ω = 1
.1ω0 = 1
.5F0 = 0
0.3m =
3)Problem 3 discusses pure resonance. The objective is
to derive and graph a particular equation that
resonates, and whose amplitude grows linearly. This
means that as the t value approaches infinity, as will
the amplitude.
y cos(ω t)y′′ + ω0
2 = m
F0
0
Solve for alpha and beta as they are in the quadratic
equation.

6. Alsulami, Falcone, Filippenko 6
−α = b
2a = 0
β = 2a
√b −4ac2
= 2
√0−4ω0
2
= ω0
Apply those to the complementary equation:
, which then becomese cos(yc = c1
αt t) e sin(βt)β + c2
αt
cos(yc = c1 t) sin(ω t)ω0 + c2 0
This complementary solution is not linearly
independent from all parts of the particular solution,
so it must be multiplied by t to get:
Atcos(ω t) tsin(ω )yp= 0 + B 0
Next, the first and second derivatives must be taken,
and applied to the original equation.
cos(ω t) tω sin(ω t) sin(ω t) tω cos(ω t)yp′ = A 0 − A 0 0 + B 0 + B 0 0
− Aω sin(ω t) tω cos(ω t) Bω cos(ω t) tω sin(ω t)yp′′ = 2 0 0 − A 0
2
0 + 2 0 0 − B 0
2
0
y cos(ω t)yp′′ + ω0
2 = m
F0
0
Aω sin(ω t) tω cos(ω t) Bω cos(ω t) tω sin(ω t) (Atcos(ω t) tsin(ω t) cos(ω t)− 2 0 0 − A 0
2
0 + 2 0 0 − B 0
2
0 + ω0
2
0 + B 0 = m
F0
0
This equation can be solved for A and B.
B =
F0
2mω0
Now that B is known, it can be applied to the
particular solution.
in(ω t)yp =
F0
2mω0
+ s 0
Now that the particular solution has been found, it
can be applied to the general solution to get a final
answer of:
cos(ω t) sin(ω t) tsin(ω t)yg = c1 0 + c2 0 +
F0
2mω0 0
3­B)The graph below displays the equation solved in
3­A aswell as the graph of (t) tsin(ω t)yss =
F0
2mω0 0
For this graph
ω = 1
.1ω0 = 1
.5F0 = 0

7. Alsulami, Falcone, Filippenko 7
0.3m =
4)This problem once again explores pure resonance, but
focuses on the amplification factor M( ). In theω
equations and calculations portion, M( ) is maximizedω
by taking the derivative, setting it equal to 0, and
solving for . The graphical portion of this problemω
explores the relationship between the amplification
factor and the frequency.
(ω)M = 1
√m (ω −ω ) +d ω2
0
2 2 2 2 2
is the given equation, where . ω0 =
√k
m
Then use chain rule to differentiate:
(ω
dω 1
√d ω +m ( −ω )2 2 2 k
m
2 2
( )d
du
1
√u
du
dω
where ω ( ) and −u = d2 2 + m2 k
m − ω2 d
du
1
√u
= 1
2u2
3

8. Alsulami, Falcone, Filippenko 8
d ω +m ( −ω )d
dω
2 2 2 k
m
2 2
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
Then differentiate the sum and factor out constants:
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
d ( (ω ))+m ( (( −ω ) ))2 d
dω
2 2 d
dx
k
m
2 2
Use the power rule to reduce to 2 :ωd
dω
2 ω
m ( ( −ω ))+2ωd2 d
dω
k
m
2 2
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
Then, using chain rule:
( )d
dω
k
m − ω2 2
= du
du2 du
dw
u = k
m − ω2
(u ) ud
du
2 = 2
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
−2d ω+2( −ω )( ( −ω ))m2 k
m
2 d
dω
k
m
2 2
Next, differentiate the sum one term at a time and
factor out the constants to yield:
−
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
2d ω+ ( )− (ω )2m ( −ω )2 d
dω
k
m
d
dω
2 2 k
m
2
The derivative of .k
m = 0
−
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
2d ω+2m ( −ω )(− (ω )2 2 k
m
2 d
dω
2
Then use the power rule:
(ω ) ω (ω ) ωd
dw
n = n n−1 → d
dω
2 = 2
This yields:
−
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
2d ω−2ω 2m ( −ω )2
* 2 k
m
2
This can be simplified to get:
(ω) −M′ =
2d ω−4ωm ( −ω )2 2 k
m
2
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
Then this must be set equal to 0 to find the maximum.
0 = −
2d ω−4ωm ( −ω )2 2 k
m
2
2(d ω +m ( −ω ) )2 2 2 k
m
2 2
2
3
The denominator cannot equal 0, so the denominator can
be ignored, and the numerator can be set equal to 0.
− d ω ωm ( )0 = 2 2
− 4 2 k
m − ω2

9. Alsulami, Falcone, Filippenko 9
− d m ω( )0 = 2 2
+ 4 2
ω
−ωk
m
2
d m ω( )2 2
= 4 2
ω
−ωk
m
2
( )d2
2m2 = ω ω
−ωk
m
2
− d2
2m2 + k
m = ω2
ωd = √k
m − d2
2m2
The graph below displays for the damping values(ω)M
of d=0.25,0.5,1,1.5, and 2. The other values used are
given as , to fit in a [0,2] x [0,4] window.m = k = ω0 = 1
For part c, the point on each line is plottedM(ω ), )( d ωd
to show how this value changes with d.
5) ​Applications of Forced Vibrations:
Forced vibrations have many applications in the real world.
From suspension systems in bikes and cars to the less widely
observed guitar strings. When guitar strings are “plucked” the
vibrations create sound,obviously the force they are plucked

10. Alsulami, Falcone, Filippenko 10
with determine factors such as length of the note, but other
factors like the tightness and thickness of the strings play a
heavy role in the notes being played. In addition, when a guitar
player, holds down a string at various locations the forced
vibrations are affected because the string is as free to move as
it was previously.
Appendix: Matlab Codes
function Diff1()
w=1;
w0=1.1;
F0=0.5;
m=0.3;
t=0:0.01:2*pi/((1.1­1)/2);
y=((2*F0)/(m.*(w0.^(2)­w.^(2)))).*sin(((w0­w).*t)/2).*sin(((w0+w).*t)/2);
plot(t,y);hold on;
y=((2*F0)/(m.*(w0.^(2)­w.^(2)))).*sin(((w0­w).*t)/2);
plot(t,y,'red')
y=­((2*F0)/(m.*(w0.^(2)­w.^(2)))).*sin(((w0­w).*t)/2);
plot(t,y,'red')
title('Problem 2c: Equation 5')
xlabel('t values')
ylabel('y(t) values')
function Diff2()
w0=1.1;
F0=0.5;
m=0.3;
t=0:0.01:2*pi/((1.1­1)/2);
y=(F0/(2.*m.*w0)).*t.*sin(w0.*t);
plot(t,y);hold on;
y=(F0/(2.*m.*w0)).*t;
plot(t,y);
y=­(F0/(2.*m.*w0)).*t;

11. Alsulami, Falcone, Filippenko 11
plot(t,y)
title('Problem 3b: Steady State')
xlabel('t values')
ylabel('y(t) values’)
function Diff3()
w0=1;
m=1;
k=1;
d=1/4;
t=0:0.001:2*pi/((1.1­1)/2);
y=1./(sqrt((m.^2).*(((w0.^2)­(t.^2)).^2)+(d.^2).*(t.^2)));
plot(t,y);hold on;
axis([0 2 0 4.5]);
d=1/2;
y=1./(sqrt((m.^2).*(((w0.^2)­(t.^2)).^2)+(d.^2).*(t.^2)));
plot(t,y)
d=1;
y=1./(sqrt((m.^2).*(((w0.^2)­(t.^2)).^2)+(d.^2).*(t.^2)));
plot(t,y)
d=3/2;
y=1./(sqrt((m.^2).*(((w0.^2)­(t.^2)).^2)+(d.^2).*(t.^2)));
plot(t,y)
d=3;
y=1./(sqrt((m.^2).*(((w0.^2)­(t.^2)).^2)+(d.^2).*(t.^2)));
plot(t,y)
d=[1/4,1/2,1,3/2,3];
wd=sqrt((k/m)­(d.^2)/(2*m.^2));
y=1./(sqrt((m.^2).*(((w0.^2)­(wd.^2)).^2)+(d.^2).*(wd.^2)));
plot(wd,y,'*')
title('Problem 4a+c: Equation 11 and 12')
xlabel('w values')
ylabel('M(w) values')