Section 3.8: 3, 6, 10, 17, 28
p
√
√
√
√
3. u = 4 cos 3t − 2 sin 3t. R = A2 + B 2 , which is 42 + (−2)2 = 16 + 4 = 20 = 2 5. δ =
√
tan−1 (B/A) = tan−1 (−1/2) = − tan−1 (1/2). So u = 2 5 cos(3t + tan−1 (1/2)) .
6. I’ll convert everything to kilograms and meters. To find the spring constant, I use the equation
(weight) = k(distance), which in this case is (0.1)(9.8) = k(0.05). So k = 19.6. There isn’t any damping
here, so the equation is
0.1x00 + 19.6x = 0, or x00 + 196x = 0.
The characteristic equation is r 2 + 196 = 0, which has roots r = ±14i, so the general solution is
x(t) = A cos 14t + B sin 14t.
The initial conditions are x(0) = 0 and x 0 (0) = 0.1. The first of these gives A = 0, and the second
gives 14B = 0.1, so B = 0.1/14 = 1/140, and the solution is x(t) = 1/140 sin 14t . This is with distance
measured in meters. (Since 1/140 = 5/700, it agrees with the answer in the back of the book, in which
distance is measured in centimeters.) To figure out when the mass first returns to its equilibrium position,
I need to find out when x(t) = 0. sin 14t = 0 when 14t = 0, π, 2π, . . . ; the first of these (after the initial
time t = 0) is thus 14t = π, or t = π/14 .
10. I’ll use pounds, feet, and seconds (and slugs!) for my units. The first sentence tells me that
16 = k(3/12), so k = 64; it also tells me that since the weight is 16 pounds, the mass is 16/g = 16/32 = 1/2
slugs. The second sentence tells me that γ = 2. Thus the equation under consideration is
1 00
x + 2x0 + 64x = 0, or x00 + 4x0 + 128x = 0.
2
(1)
√
√
The characteristic equation is r 2 + 4r + 128 = 0, which has roots −4± 216−512 = −2 ± 2 31i, so the general
solution is
√
√
x(t) = Ae−2t cos(2 31t) + Be−2t sin(2 31t).
0
The initial
√ conditions are x(0) = 0 and x (0) = 3/12 = 1/4. Solving for A and B gives A = 0 and
B = 1/8 31: the solution is
√
1
x(t) = √ e−2t sin(2 31t) .
8 31
Here’s a graph:
0.2
0.15
0.1
0.05
0
-0.05
-0.1
0
0.5
1
1.5
2
1
2.5
3
3.5
4
√
The mass first returns to its equilibrium position when x(t) = 0; this happens when sin(2 31t) = 0, which
√
√
is when 2 31t = π, which is when t = π/2 31 = 0.282 seconds. Finally,
√
1
1
|x(t)| = | √ e−2t sin(2 31t)| ≤ | √ e−2t |
8 31
8 31
1
= √ e−2t .
8 31
If I want this to be less than 0.01 inches, i.e., less than 0.01/12 feet, I solve this inequality for t:
1
√ e−2t < 0.01/12.
8 31
So e−2t <
√
(0.01)(8) 31
,
12
so −2t < ln( (0.01)(8)
12
√
31
), so
√
1
(0.01)(8) 31
t > − ln(
) = 1.647 .
2
12
If t satisfies this inequality, then |x(t)| will be less than 0.01 inches, so it does answer the question. It
turns out that if you work at it, you can find a slightly smaller value for τ : if t > 1.59 seconds, then
|x(t)| < 0.01/12. Let’s not worry about how to get this better number right now.
17. The weight is 8/32 = 1/4; the spring constant is given by the equation 8 = k(1.5/12), so k = 64.
Thus the spring equation is
1/4x00 + γx0 + 64x = 0, or x00 + 4γx0 + 256x = 0.
The characteristic equation is r 2 + 4γr + 256 = 0. This has roots
p
−4γ ± 16γ 2 − 4 · 256
.
2
The system is critically damped if there is just one real root, which happens when the term under the
square root is zero: 16γ 2 − 4 · 256 = 0. Solving for γ gives γ = 8 . (Actually, it gives γ = ±8, but γ is
always assumed to be positive.) The units are lb · sec/ft.
√
√
28. (a) u00 + 2u = 0, u(0) = 0, u0 (0) = 2. √
The general solution is u(t) = c 1 sin( 2t) + c2 cos( 2t); the
initial conditions imply that c2 = 0 and c1 = 2, so the solution is
√
√
u(t) = 2 sin( 2t) .
√
(b) u0 (t) = 2 cos( 2t), so here’s the plot:
2
u(t)
u’(t)
1.5
1
0.5
0
-0.5
-1
-1.5
-2
0
2
4
6
2
8
10
(c) If you can plot parametrically
(with a calculator
√
√ or computer), that’s what you do. Otherwise, notice
that (u(t))2 = 2 sin2 ( 2t), and (u0 (t))2 = 4 cos2 ( 2t). So
√
√
2(u(t))2 + (u0 (t))2 = 4 sin2 ( 2t) + 4 cos2 ( 2t) = 4.
More succinctly:
2u2 + (u0 )2 = 4.
The graph of this is an ellipse.
2
u vs. u’
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
The initial conditions are u(0) = 0, u 0 (0) = 2, so the motion starts (when t = 0) at the top of the ellipse.
As time moves forward, u(t) increases (it’s a sine curve) and u 0 (t) decreases (it’s a cosine), so as t increases,
you go around the phase plot clockwise.
Section 3.9: 1, 5, 7, 17, 18
1. The relevant trig identities are
cos(α + β) = cos α cos β − sin α sin β,
cos(α − β) = cos α cos β + sin α sin β,
so
cos(α + β) − cos(α − β) = −2 sin α sin β.
In this problem, we want to rewrite cos 9t − cos 7t, so I want to find α and β so that α + β = 9t and
α−β = 7t. It’s easy to solve these two equations to get α = 8t and β = t, so cos 9t−cos 7t = −2 sin 8t sin t .
5. The mass is 4/32 = 1/8 slugs. The spring constant is given by (weight) = k(distance), or 4 =
k(1.5/12), so k = 32. There is no damping, so the equation is
1 00
x + 32x = 2 cos 3t, or x00 + 256x = 16 cos 3t .
8
The initial conditions are x(0) = 1/6 and x 0 (0) = 0 .
7. The characteristic equation is r 2 + 256 = 0, which has roots ±16i, so the homogeneous equation has
general solution
xh = c1 cos 16t + c2 sin 16t.
To find a particular solution, the method of undetermined coefficients tells me to try
xp = A cos 3t + B sin 3t.
3
When I plug this in, I find that B = 0 and A = 16/247, so the general solution to the nonhomogeneous
equation is
16
x = c1 cos 16t + c2 sin 16t +
cos 3t.
247
16
151
= 1482
, so the solution to the initial value
Plugging in the initial conditions gives c 2 = 0 and c1 = 61 − 247
problem is
151
16
x=
cos 16t +
cos 3t .
1482
247
Here’s a graph:
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
-10
-5
0
5
10
If the external force is replaced by 4 sin ωt, then resonance will occur when ω is the natural frequence: it
will occur when ω = 16.
17. u00 + 41 u0 + 2u = 2 cos ωt, u(0) = 0, u0 (0) = 2. It’s probably not a bad idea to multiply by 4, so
that I don’t have to deal with fractions: 4u 00 + u0 + 8u = 8 cos ωt.
(a) The “steady-state part” is the particular solution, which is given in equationp(11) on √
page 203.
Here is what that equation becomes, in this case (m = 4, γ = 1, k = 8, F 0 = 8, ω0 = k/m = 2):
8
up = p
cos(ωt − δ) .
16(2 − ω 2 )2 + ω 2
δ is the inverse tangent of γω/m(ω02 − ω 2 ) = ω/4(2 − ω 2 ).
(b) The amplitude is the coefficient in the previous equation:
8
A= p
.
16(2 − ω 2 )2 + ω 2
(c) Here is a plot of A versus ω:
6
5
4
3
2
1
0
0
0.5
1
1.5
2
4
2.5
3
3.5
4
p
(d) A is largest when its denominator 16(2 − ω 2 )2 + ω 2 is smallest. This is smallest when its square
16(2 − ω 2 )2 + ω 2 is smallest. This is an easy function to maximize using basic calculus. I found that
√
p
ω = 126/64 = 3 14/8 ≈ 1.403 . Note that this is close to, but a bit smaller than, the natural frequency
√
ω0 = 2 ≈ 1.414. The correspond maximum value for A is approximately 5.679 .
18. u00 + u = 3 cos ωt, u(0) = 0, u0 (0) = 0.
(a) uh (t) = c1 cos t + c2 sin t. Since ω 6= 1, I’ll try up = A cos ωt + B sin ωt for a particular solution;
plugging this in and solving for A and B gives B = 0 and A = 3/(1 − ω 2 ). So the general solution to the
nonhomogeneous equation is
u(t) = c1 cos t + c2 sin t +
3
cos ωt.
1 − ω2
The initial conditions say that c2 = 0 and c1 = −3/(1 − ω 2 ), so the solution is
u(t) =
3
(cos ωt − cos t) .
1 − ω2
(b) Here’s a graph:
40
w=0.7
w=0.8
w=0.9
30
20
10
0
-10
-20
-30
-40
0
10
20
30
40
50
60
70
80
When t is near zero, there is not much difference between the solutions; however, for larger values of t, as
ω gets closer to 1, the amplitude of the response u(t) increases. This fits with what we expect: the closer
the frequency ω of the driving force is to the natural frequency (1 in this case), the larger the amplitude
of the response.
Also, by the way, the closer ω is to 1, the more pronounced the “beat” phenomenon is.
5