Score
possible
1
(30)
2
(15)
3
(10)
4
(20)
5
(20)
total
(100)
1) Consider a general input-output model with two compartments as indicated below. The compartments
contain volumes V1, J’~ and solute amounts x1 (/), x2 (t) respectively. The flow rates (volume per time) are
indicated by r1, i = 1 ..6. The two input concentrations (solute amount per volume) are c1, c5.
r
,C
5—
~H.
5
~
rfc~L*
X51*)
hi Suppose r2
=
=
100, r1
=
=
=
200, r6
=
300 hur
Explain why the volumes
V~ (1), 172(t) remain constant.
~(t)z ~t~—ç_rq
v rl
~
i~
—
-
~
j~ Using the flow rates above, incoming concentrations
V1
=
V~
=
~
=
0.05, c5
=
(5 points)
V’4’
v
C~o
~
‘F
0 -~-, volumes
100 m3 , show that the amounts of solute x1 (t) in tank 1 and x2 (t) inj~jc2satis~—_...
x~’(t)
—3
x2’(t)
=
1
X1
2 -4
x2
10
+
~
lb
0
ints)
vc#r~
~-+y~)~a
~
0
2qt
~t5-t-r1~
Jrçl-ç)~
zQ42o~jj
~
~
jfl
2
j~) Find the solution to the homogeneous system of differential equations
[x1’(t)
F
1
x~
1
j[
[x2’(t)
\c—2
-J
-‘
2 -4
R~h
j[~2 j
(10 points)
L
Itc~e
I
~
L~j
~
1
As)CX~z-)
z~”47~4~tt3
~ I\o
2l~o
iF
-~
i_i
jj~) Find the general solution to the inhomogeneous system of differential equations in lb.
[xi’(t)
I 1Fx1]rio
l[-3
j[
[x2’(O
1±1
2 -4 j[x2]
[o
Hint: you will need to find a particular solution as part of your work.
(10 points)
—~
rol
_J
r~
LoVL 9JL~j
f
r101
I I
Lo.J~
r~0i
~ 9JL41 L~)
i~3
‘0
C
~ r~(t ~
H
(
4-c~C
~
~
Ltl
3
We used the principle of conservation of energy and the method of linearization to derive the second
order linear differential equation which describes undamped unforced pendulum motion. This differential
equation is
L~0’’(t) +g0(t)=0
where 0(t) is the angle from vertical, L is the length of the (massless) rod or string, m is the mass at the
end of the string, and g = 9.8
is the acceleration of gravity on earth.
~ Derive the linear model above, using the fact that kinetic plus potential energy is constant for any
solution to this conservative system. This will yield a non-linear differential equation which you can
linearize to the one above, under the assumption that 0(t) is near zero.
(10 points)
kE i—FE
U
L
TEtmn,U~Cl€~’~
~~
t~3
(L-L
a
fE
~
—
~O ‘(+1
~ ~s~w9~]
jt
ç’1fl~
L&’R)
o~d—
eoLcia~
-~
L
Vt- L&oss
~
~-
~-
3SJblU
C
~-
o
2k). If you wished to create a pendulum for which the natural period was 2-stc~fiflV~cle, wliat l~ngth L
would you choose? (A symbolic answer suffices the correct decimal value for L is close to 1 meter.)
Note that you can answer this question from the differential equation given at the start of this problem.
(5 points)
-
‘FE
.1
%;h1t
ot— ~
It—
4
3)~
A focus in this course is a careful analysis of the mathematics and physical phenomena exhibited in
forced and unforced mechanical (or electrical) oscillation problems. Using the mass-spring model, we’ve
studied the differential equation for functions x(t) solving
in x” + cx’ + kx = F0 cos w t)
(
with
in,
lçw> 0; c,F0 ≥ 0.
Explain what values (or ranges of values) for c, Ic F0, o lead to the phenomena listed below. (If you use
in this discussion make sure to explain what it is as well.) What form will the key parts of the
solutions x(t) have in those cases, in order that the physical phenomena be present? (We’re not expecting
the precise formulas for these parts of the solutions, just what their forms will be.)
~flh pure resonance
F)o
(3points)
-b
cto
(~~tocu~0i
~~LLsatktik)
~
Li) t Li)
C
~ beating
~
143$
~
(3points)
~
~o
‘~p__
~f)
~
-t-L~t
ad practical resonance.
e-)o Sf c%O, t~zt~’J0
tA~L%t
~
Q~$
~
~
~c
C
~c
“
I
ca~,st~
‘Rtcac&s€
~s~&
(4points)
~
1—s
tfrfl~0t) (s~(~9-~°1>)
—
~‘
LcVit.
~
~
~ ~4
tnJ\
C~c ~
occw~A~’~’~
C
1
5
)
4).
Consider the following configuration of 2 masses held together with two springs (a 2 ear train), with
positive displacements from equilibrium for each mass measured to the right as usual.
~“I
-~
x,tt)
-4
*2j4)
4~).
Use Newton’s Law and Hooke’s usual linearization to derive the system of 2 second order differential
equations governing the masses’ motion.
/
(5 points)
~s
—L
4h).
—
What is the dimension of the solution space to this problem? Explain.
A
n
(5 points)
-
a
~ca~cjt&~t? ~3C~ e~~t~&_
\150
‘-I
~ ~ bL~
~ NP
4g).
Assume that all
=
=
~ti
+ ~h(
~,
~
~(6)~
~‘(Ø,
~
m. Show that in this case the system in (3a) reduces to
x1’’(t)
k
—1
1
x2’’(t)
m
1
—1
~i
(5 points)
~7
(_C~~
~
t_t
(ç—<)
~
~
—
4-U
6
4≤fl
Find the general solution to the differential equation in 3c, repeated here for your convenience. Then
describe the general motion as a superposition of two “fundamental modes”. Hint: One of the modes is
not oscillatory, since this “train’ is not tethered to any wall.
[x~’’(t)
—1 1
lk[
[x2”(r)
1
1
jrn
-1 jjx2j
(10 points)
[‘~
tvdç
I
~L]
\ ~
~4L)
7\J0,-2
tic
~i
I.
i,
L ~
i
-j
Vt
~
-2~
1,-
-j
r—n ~
~ L’j’
(c~tc~L)v.
(1)
L
.i—
(c~e.oc~~4~
i~c1cc~4jt)
tfl~
~))
~
V-O
a_k
‘~‘
~
cLoo
7
Draw the phase portrait for the first order system, and classi& it as one of the following: saddle, nodal
source, nodal sink, center, spiral sink, spiral source.
x11(t)
4 3 xI
—
x21(t)
0 6
—
Hint: Use the eigendata and the form of the general solution.
(20 points)
1-?
~2)
?flW (-q-\)(t-X)
10-).
C
ca”cC
vw43 ‘at~_9, (~
LS
Cc
50
~
‘at -9
30
0
C
~ID
V
—Ic’
3 0
coo
-J
\1:
(3
(cak~
~lc~’i~
Vt
at~te’.4
si~
r
3
Li
~
t
L~
~co
e
a
4
2-
çe.
It,
cX10
E
~r’~
L~~’S~
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