Download Problem Set #7 solution

Math 203-04
Problem Set #7 – Solution
9 April 2001
Exercises:
1. Consider the undamped forced spring equation
mu00 + ku = F0 cos ωt
(1)
s
where m, k, and F0 are positive constants. Let ω0 =
k
, and suppose ω = ω0 .
m
(a) Verify that
U (t) =
F0
t sin ω0 t
2mω0
is a solution to equation (1). (That is, verify equation (7) on p. 202 of B & D.)
(b) Using the same notation, find a particular solution to the differential equation
mu00 + ku = F0 sin ωt
with ω = ω0 . Verify that your solution is correct.
(c) Compare the phase angles of the solutions in parts 1a and 1b above to the phase
angles of their respective forcing functions.
Solution:
(a) Given U (t) =
F0
t sin ω0 t, we have
2mω0
U 0 (t)
=
and
U 00 (t)
=
=
F0
(sin ω0 t + ω0 cos ω0 t)
2mω0
F0
(2ω0 cos ω0 t − ω02 t sin ω0 t)
2mω0
Ã
!
F0
k
2ω0 cos ω0 t − t sin ω0 t
2mω0
m
because ω02 =
Thus
k
.
m
Ã
mU 00 + kU 0
!
F0
k
=
2mω0 cos ω0 t − m t sin ω0 t + kt sin ω0 t
2mω0
m
F0
=
(2mω0 cos ω0 t)
2mω0
= F0 cos ω0 t
as required.
(b) We know the solution will have the form At cos ω0 t + Bt sin ω0 t, and having just
worked through part 1a, we hazard a guess that B = 0. Suppose, then, that
U (t) = At cos ω0 t. Then we have
U 0 (t)
= A(cos ω0 t − ω0 t sin ω0 t)
and
00
U (t) = A(−2ω0 sin ω0 t − ω02 t cos ω0 t)
Ã
!
k
= A −2ω0 sin ω0 t − t cos ω0 t .
m
We must choose A to satisfy
F0 sin ω0 t = mU 00 + kU
Ã
!
k
= A −2mω0 sin ω0 t − m t cos ω0 t + kt cos ω0 t
m
= −2Amω0 sin ω0 t.
Equating coefficients, we get
A = −
F0
,
2mω0
so our solution is
U = −
F0
t cos ω0 t.
2mω0
F0
sin ω0 t, lags 90◦ behind the forcing function,
2mω0
F0 cos ω0 t. This follows from the identity
(c) In 1a, the response function,
µ
π
cos ω0 t −
2
¶
= cos ω0 t cos
π
π
+ sin ω0 t sin
2
2
= sin ω0 t.
Similarly, in 1b, the response −
F0 sin ω0 t, because
µ
π
sin ω0 t −
2
F0
t cos ω0 t lags 90◦ behind the forcing function
2mω0
¶
= sin ω0 t cos
= − cos ω0 t.
π
π
− cos ω0 t sin
2
2
2. B & D p. 206, problem 17. Use the results given on p. 203; don’t re-invent the wheel.
Solution:
1
(a) The given problem fits the pattern of equation (1) on p. 193 with m = 1, γ = ,
4
k = 2, and F0 = 2. We also find that
s
√
k
= 2.
m
ω0 =
The steady-state solution is given by equation (11) on p. 203. With the given
constants, we have
2
U (t) = q
(2 − ω 2 )2 +
ω2
16
cos(ωt − δ),
ω
.
4(2 − ω 2 )
(b) The amplitude of the steady-state solution is simply the coefficient of the cosine
term in the expression above, that is,
where δ = tan−1
2
A = q
(2 − ω 2 )2 +
ω2
16
.
(c) Here is a graph showing A as a function of ω. Note that we plot only positive
values of ω, because for one thing, only positive frequencies make sense, and for
another, A is an even function of ω.
A
5
4
3
2
1
1
2
4
3
omega
5
(d) According to equation (12) on p. 203, the maximum amplitude occurs when ω
satisfies
γ2
2m2
1
= 2− .
32
ω 2 = ω02 −
So a forcing frequency of
s
ω =
63
32
gives the maximum response.
The amplitude of the maximum response is (according to equation (13) on the
same page)
q
F0
γω0 1 − (γ 2 /4mk)
= q
8
2(1 − 1/128)
64
= √
.
127
3. Consider a series circuit with a 10 microfarad capacitor (a microfarad is 10−6 farad),
a 0.25 henry inductor, and a 10 ohm resistor.
(a) Assume the impressed voltage is zero, the initial charge on the capacitor is
10−5 coulomb, and the initial current is zero.
Find Q(t), the charge on the capacitor, as a function of time.
(b) Now assume an impressed voltage (varying with time) given by
E(t) = 20 cos ωt.
For what value of ω will the circuit’s steady-state response be the greatest? What
is the maximum voltage drop across the resistor when this value of ω is used?
Give exact values and decimal approximations.
Solution:
(a) We’ll use equation (32) on p. 196:
LQ00 + RQ0 +
1
Q = E(t).
C
Using the given values for L, R, C, and E, we have
0.25Q00 + 10Q0 + 105 Q = 0.
The roots of the characteristic equation are
√
−10 ± 100 − 105
r =
0.5√
= −20 ± 60i 111
so that the general solution to (2) is
Q(t) = e−20t (c1 cos µt + c2 sin µt),
√
where µ = 60 111.
The initial conditions are Q(0) = 10−5 , Q0 (0) = 0. We compute
Q0 (t) = (µc2 − 20c1 )e−20t cos µt − (µc1 + 20c2 )e−20t sin µt
and then use the initial conditions to obtain the system
10−5 = c1
0 = µc2 − 20c1
(2)
from which we get
20
10−5
c2 = c1 = √
.
µ
3 111
−5
c1 = 10 ,
The solution to the given initial value problem is
Ã
−5 −20t
Q(t) = 10 e
!
1
cos µt + √
sin µt
3 111
√
where µ = 60 111.
(b) To find the frequency ωmax , we use equation (12) on p. 203. We have m = 0.25,
γ = 10, k = 105 and
ω02 =
k
= 4 × 105
m
Thus
100
2(0.25)2
= 4 × 105 − 800
2
ωmax
= 4 × 105 −
so that
ωmax =
√
399 200 ≈ 631.822.
According to equation (13) on p. 203, the amplitude of the maximum response
will be
Rmax =
q
F0
γω0 1 − (γ 2 /4mk)
1
= √ 5
.
10 − 102
Assuming the driving frequency is ωmax , the steady-state charge Q(t) on the capacitor is given by
Q(t) = Rmax cos(ωmax t − δ)
for some phase angle δ. Thus the current I(t) is given by
I(t) = Q0 (t) = −ωmax Rmax sin(ωmax t − δ)
so that the amplitude of the current is ωmax Rmax . Finally, the voltage drop across
the resistor (by Ohm’s law) is I times the value of the resistor, in this case, 10 Ω.
So the maximum voltage drop across the resistor is given by
VR = 10ωmax Rmax
√
10 399 200
√
=
99 900
≈ 19.99.
The maximum voltage drop across the resistor is just slightly smaller than the
amplitude of the impressed voltage.
(c) (Alternate solution.) We begin with the form of the steady-state solution given
on p. 203,
F0
U (t) = q
cos(ωt − δ).
m2 (ω02 − ω 2 )2 + γ 2 ω 2
Changing from the m, γ, and k parameters of the spring-mass system to the R,
L, and C parameters of the RLC circuit system, we get
E0
Q(t) = q
cos(ωt − δ)
L2 (ω02 − ω 2 )2 + R2 ω 2
(3)
1
where Q(t) is the charge on the capacitor as a function of t, ω02 =
and δ is
LC
some phase angle.
Let VR (t) denote the voltage drop across the resistor as a function of time. From
Ohm’s law, we know that
VR (t) = RI(t) = RQ0 (t),
because the current in the circuit is the time derivative of Q(t). Thus
RωE0
VR (t) = − q
L2 (ω02 − ω 2 )2 + R2 ω 2
sin(ωt − δ),
and the amplitude of VR is given by
RωE0
q
L2 (ω02 − ω 2 )2 + R2 ω 2
(4)
We want to choose ω so as to maximize this amplitude. To maximize expression
(4), we may as well maximize its square,
R2 ω 2 E02
.
L2 (ω02 − ω 2 )2 + R2 ω 2
(5)
Since everything in sight is positive, we can maximize expression (5) by minimizing
its reciprocal,
L2 (ω02 − ω 2 )2
γ2
L2 (ω02 − ω 2 )2 + R2 ω 2
=
+
.
R2 ω 2 E02
R2 ω 2 E02
R2 E02
(6)
Since we’re minimizing with respect to ω, and ω does not appear in the right-hand
term, we need only find the value of ω that minimizes
m2 (ω02 − ω 2 )2
.
R2 ω 2 E02
(7)
Now since everything in sight is positive, it’s clear that the minimum value of
expression (7) is 0, and that it is achieved when ω = ω0 .
Thus the maximum voltage drop across the resistor occurs when ω = ω0 , and the
amplitude of this drop is (from expression (4))
Rω0 E0
q
R2 ω02
= E0 .
In the problem at hand, L = 0.25 and C = 10−5 , so
s
1
LC
√
= 200 10
≈ 632.46 radians per second
≈ 100.66 Hz
ω0 =
and the maximum voltage drop across the resistor is E0 = 20 volts, the same as
the impressed voltage on the circuit.
Is this a coincidence? No, indeed.√We dimly recall from AC circuit theory that
when the driving frequency is 1/ LC, the capacitive and inductive reactances
cancel each other out, so the entire voltage drop in the circuit must be accounted
for by the resistor.