Mat 303 Calculus Iv Homework 7 Worksheet With Answers - State University Of New York At Stony Brook - 2013 Page 6
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6MAT 303 Spring 2013
Calculus IV with Applications
At t
0, x 0
c
0. Then, using c
0, v 0
4c
8, so c
2. Hence,
1
1
2
2
3t
pt
x t
2e
sin 4t, which we convert to Ce
cos
t
. We have that C
2, and
1
1
3
A
2
2 sin
, so
(taking an angle between 0 and 2 ), so
1
1
2
3
3t
x t
2e
cos 4t
.
2
With c
0, the solution is u t
A cos 5t
B sin 5t, with u t
5A sin 5t
5B cos 5t;
8
8
applying the initial conditions, A
0 and 5B
8, so u t
sin 5t
cos 5t
5
5
3
.
2
3.4.22. A 12-lb weight (mass m
0.375 slugs in fps units) is attached both to a verti-
cally suspended spring that it stretches 6 inches and to a dashpot that provides 3 lb of
resistance for every foot-per-second of velocity.
(a) If the weight is pulled down 1 foot below its static equilibrium position and then
released from rest at time t
0, find its position function x t .
(b) Find the frequency, time-varying amplitude, and phase angle of the motion.
12 lb
Solution (a): The mass in fps units is m
0.375, and the spring constant is k
0.5 ft
k
2
24 lb/ft. The circular frequency is given by
64, so
8 rad/s. The damping
0
0
m
c
constant is c
3 lb-s/ft, so p
4, and the system is therefore underdamped. Then
2m
2
2
p
48
4 3, so the general solution is
1
0
4t
4t
x t
Ae
cos 4 3t
Be
sin 4 3t,
4t
4t
v t
Ae
4 cos 4 3t
4 3 sin 4 3t
Be
4 sin 4 3t
4 3 cos 4 3t .
We measure the displacement vertically, considering a displacement downwards as being
positive, since it corresponds to stretching the string further. Thus, at time t
0 we have
4A
1
that x 0
1 and v 0
0. Then A
1 and
4A
4 3B
0, so B
. Thus,
4 3
3
the solution is
1
4t
4t
x t
e
cos 4 3t
e
sin 4 3.
3
pt
Solution (b): We reformulate our answer to part (a) in the form Ce
cos
t
. Then
1
2
2
2
1
4
2
2
4t
C
A
B
1
, so C
, and the time-varying amplitude is
e
. From
3
3
3
3
B
1
above, the pseudofrequency is
4 3 rad/s. Finally, tan
, with
in
1
A
3
Quadrant I so that both A and B are positive. Thus, the phase angle is
.
6
6
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