Download Problem set 11

MATH 225 / MTHE 225
Problem Set 11
1. A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to
a viscous damper with a damping constant of 400 g/sec. If the mass is pulled down
an additional 2 cm and then released, find its position at any time. What are the
amplitude, time-varying amplitude, natural frequency, and phase shift?
2. A block of mass of 100 g stretches a spring 5 cm. Assume there is no damping and
gravitational constant g = 980 cm/sec2 .
(a) If the mass is set in motion from its equilibrium position with a downward velocity
of 10 cm/sec, determine the position of the mass at any time.
(b) What are the amplitude, natural frequency, and phase shift?
(c) When does the mass first return to its equilibrium position?
3. Consider a mass-spring system where mass M = 1, damping constant c = 4, spring
constant k = 4 and external force FE = 10 cos (3t). Determine the position of the mass
at any time.
4. A body of mass 4 kg will stretch a spring 80 centimeters. This same body is attached
to such a spring with an accompanying dashpot. Suppose the damping constant is
49 N. At t = 0, the mass is given an initial upward velocity of 87.5 centimeters per
second from its equilibrium position. Find a mathematical model that represents the
motion of the body and determine the resulting motion. After release, does the mass
ever cross equilibrium? If so, when does it first cross equilibrium? Assume the value
of gravitational acceleration as 9.8 m/sec2 and 1 meter = 100 centimeter.
5. A body of mass 6 kilogram is attached to to an undamped spring and stretches it by
2.94 meter. At t = 0, the mass is pulled down from its equilibrium position a distance
of 10 centimeters and released. Find a mathematical model that represents the motion
of the body and determine the resulting motion. What is the amplitude, frequency,
and phase shift? Assume the value of gravitational acceleration as 9.8 m/sec2 .
6. A body weighing 16 pounds will stretch a spring 6 inches. This same body is attached
to such a spring with an accompanying dashpot. Suppose the damping force is 2
pounds for every ft/sec velocity. At t = 0, the mass is pulled down from its equilibrium
position a distance of 1 foot and released with a downward velocity of 1 foot per second.
Find a mathematical model that represents the motion of the body and determine the
resulting motion. What is the amplitude, frequency, and phase shift? Assume the
value of gravitational acceleration as 32 ft/sec2 .
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7. A spring is stretched 1 m by a force of 5 N. A body of mass 2 kg is attached to the
spring with, accompanying dashpot. Suppose the damping force of the dashpot is 6
N when the velocity of the body is 1 m/s. At t = 0, the mass is pulled down from
its equilibrium position a distance of 10 centimeters and given an initial downward
velocity of 10 centimeters per second. Find a mathematical model that represents the
motion of the body and solve. Determine the resulting motion. After release, does the
mass every cross equilibrium?
8. A body weighing 32 pounds will stretch a spring 2 feet. This same body is attached
to such a spring with an accompanying dashpot. Suppose the damping constant is 8
lbs s/ft. At t = 0, the mass is pulled up from its equilibrium position a distance of 1
foot and released. Find a mathematical model that represents the motion of the body
and solve. Determine the resulting motion. After release, does the mass every cross
equilibrium
2
Solution
1.
Given mass M = 20, damping constant c = 400, and external force FE = 0 (since nothing is
mentioned). Now notice that at equilibrium condition (that is no oscillation), weight of the
block is equal to the restoring force of the spring, that is M g = kL where M is the mass,
k is the spring constant and L is the stretched length of the spring at equilibrium. That
means k = M g/L = (20 · 980)/5 = 3920 g/sec2 .
Let y(t) be the position (or displacement) of the mass from the equilibrium position at time
t. The differential equation for a mass-spring-dashpot system is given by
M y 00 + cy 0 + ky = FE .
Plugging the known values the differential equation becomes
20y 00 + 400y 0 + 3920y = 0 ⇒ y 00 + 20y 0 + 196 = 0.
−20 ±
2
So the auxiliary equation is: m + 20m + 196 = 0 ⇒ m1,2 =
√
−10 ± 4 6i. Hence the general solution is
√
√
y(t) = e−10t [C1 cos (4 6t) + C2 sin (4 6t)].
√
202 − 4 · 1 · 196
=
2
Now from the question we see that y(0) = 2 and y 0 (0) = 0 (since nothing is mentioned).
We see that,
√
√
√
√
√
y 0 (t) = −10e−10t [C1 cos (4 6t) + C2 sin (4 6t)] + 4 6e−10t [−C1 sin (4 6t) + C2 cos (4 6t)]
√
√
√
√
= e−10t [(−10C1 + 4 6C2 ) cos (4 6t) + (−4 6C1 − 10C2 ) sin (4 6t)]
Since y(0) = 2 ⇒ C1 cos 0 + C2 sin 0 = 2 ⇒ C1 = 2.
√
√
Since y 0 (0) = 0 ⇒ (−4 6C1 −√10C2 ) sin 0 + (−10C1 + 4 6C2 ) cos 0 = 0
⇒ (−10C1 + 4 √
6C2 ) = 0 √
√
⇒ C2 = 10C1 /4 6 = 20/4 6 = 5/ 6
√
√
5
−10t
So, the position function is: y(t) = e
2 cos (4 6t) + √ sin (4 6t) .
6
s
r
r
2
q
5
25
49
7
2
Let D = C12 + C22 = 22 + √
= 4+
=
= √ . Now let cos α =
=
6
6
D
6
6
√
√
2 6
5/ 6
5
5
−1
and sin α =
= . That means α = sin
(we can just leave it in this way
7
D
7
7
since we don’t have this value from the trigonometry formula sheet).
3
Therefore,
"
#
√
√
√
√
√
6
5
2
5/
y(t) = e−10t 2 cos (4 6t) + √ sin (4 6t) = De−10t
cos (4 6t) +
sin (4 6t)
D
D
6
h
√
√ i
√
= De−10t cos α · cos (4 6t) + sin α · sin (4 6t) = De−10t cos (4 6t − α)
√
5
7 −10t
−1
=√ e
cos 4 6t − sin
.
7
6
√
7
7
Here, amplitude is √ e−10t , time varying amplitude is √ , natural frequency is 4 6 and
6
6
5
phase shift is sin−1
.
7
2.
(a) Given mass M = 100, damping constant c = 0 (since there is damping), and external
force FE = 0 (since nothing is mentioned). Now notice that at equilibrium condition
(that is no oscillation), weight of the block is equal to the restoring force of the spring,
that is, M g = kL where M is the mass, k is the spring constant and L is the stretched
length of the spring at equilibrium. That means k = M g/L = (100 · 980)/5 = 19600
g/sec2 .
Let y(t) be the position (or displacement) of the mass from the equilibrium position at
time t. The differential equation for a mass-spring-dashpot system is given by
M y 00 + cy 0 + ky = FE .
Plugging the known values the differential equation becomes
100y 00 + 19600y = 0 ⇒ y 00 + 196y = 0.
√
So the auxiliary equation is: m2 + 196 = 0 ⇒ m2 = −196 ⇒ m1,2 = ± −196 = 0 ± 14i.
Hence the general solution is
y(t) = e0·t [C1 cos (14t) + C2 sin (14t)] ⇒ y(t) = C1 cos (14t) + C2 sin (14t)
Now from the question we see that y(0) = 0 (since nothing is mentioned) and y 0 (0) =
10 (we are treating velocity as positive since it is acting downwards according to the
question).
We see that, y 0 (t) = −14C1 sin (14t) + 14C2 cos (14t)
Since y(0) = 0 ⇒ C1 cos 0 + C2 sin 0 = 0 ⇒ C1 = 0.
Since y 0 (0) = 10 ⇒ −14C1 sin 0 + 14C2 cos 0 = 10 ⇒ C2 = 10/14 = 5/7
5
So, the position function is: y(t) = sin (14t).
7
4
s
2
5
5
0
02 +
= . Now let cos α =
= 0 and
7
7
D
5/7
π
5
sin α =
= 1. That means α = sin−1 1 = . Here, y(t) = sin (14t) = 0 cos (14t) +
D
7
2
5
0
5/7
sin (14t) = D
cos (14t) +
sin (14t) = D [cos α · cos (14t) + sin α · sin (14t)] =
7
D
D
5
π
5
D cos (14t − α) = cos 14t −
. So amplitude is , natural frequency is 14, phase
7
2
7
π
shift is .
2
q
(b) Let D = D =
C12 + C22 =
5
Remark: Here y(t) = sin (14t). Now if you just want to know the amplitude and
7
natural frequency and NOT the phase-shift, you do NOT need to go through the above
mentioned calculation; you can simply see them in the expression of yp ; that is amplitude
5
= and natural frequency = 14.
7
5
sin (14t) = 0 ⇒
7
sin (14t) = 0 ⇒ 14t = πn for some integer n. Since n = 0 means the starting point, we
π
sec.
should use n = 1 in order to get our required answer, that is t =
14
(c) The mass returns to equilibrium position means y(t) = 0. That means
3.
The differential equation for any mass-spring-dashpot system is
M y 00 + cy 0 + ky = FE .
Plugging the known values this differential equation becomes
y 00 + 4y 0 + 4y = 10 cos (3t)
• For yc : Set y 00 + 4y 0 + 4y = 0.
So the auxiliary equation is m2 + 4m + 4 = 0 ⇒ (m + 2)2 = 0 ⇒ m1,2 = −2, −2.
Therefore, yc (t) = C1 e−2t + C2 te−2t
• For yp (t) : Try yp (t) = A cos (3t) + B sin (3t). So we have
yp0 (t) = −3A sin (3t) + 3B cos (3t) and yp00 (t) = −9A cos (3t) − 9B sin (3t)
Now plugging these back in the main differential equation we get
−9A cos (3t) − 9B sin (3t) + 4(−3A sin (3t) + 3B cos (3t)) + 4(A cos (3t) + B sin (3t)) =
10 cos (3t)
⇒ (−9A + 12B + 4A) cos (3t) + (−9B − 12A + 4B) sin (3t) = 10 cos (3t)
⇒ (−5A + 12B) cos (3t) + (12A − 5B) sin (3t) = 10 cos (3t)
5
Now comparing the coefficient we get: −5A + 12B = 10 and −12A − 5B = 0,
that is: −60A + 144B = 120 and −60A − 25B = 0,
that is: (−60A + 144B) − (60A − 25B) = 120 − 0,
that is: 169B = 60 ⇒ B = 120/169.
120
5B
5 120
50
Now since −12A − 5B = 0 and B =
we have A = −
=− ·
=−
.
169
12
12 169
169
50
120
So yp (t) = −
cos (3t) +
sin (3t)
169
169
50
120
Therefore y(t) = yc (t) + yp (t) = C1 e−2t + C2 te−2t −
cos (3t) +
sin (3t).
169
169
4.
Given mass M = 4, damping constant c = 49, and external force FE = 0 (since nothing is
mentioned). Now notice that at equilibrium condition (that is no oscillation), weight of the
block is equal to the restoring force of the spring, that is, M g = kL where M is the mass,
k is the spring constant and L is the stretched length of the spring at equilibrium. That
means k = M g/L = (4 · 9.8)/(80/100) = 49 kg/sec2 .
The differential equation for any mass-spring-dashpot system is
M y 00 + cy 0 + ky = FE .
Plugging the known values this differential equation becomes
4y 00 + 49y 0 + 49y = 0
−49 ±
2
So the auxiliary equation is: 4m + 49m + 49 = 0 ⇒ m1,2 =
√
−49 ± 7 33
. Hence the general solution of the differential equation is
8
y(t) = C1 e
√
−49−7 33
t
8
+ C2 e
√
−49+7 33
t
8
√
492 − 4 · 4 · 49
=
2·4
.
Now from the question we see that y(0) = 0 (since from equilibrium position) and y 0 (0) =
87.5
14
−
=−
(we are treating velocity as negative since it is acting upwards according to
100
16
the question).
√
√
√
√
−49−7 33
−49+7 33
−49 − 7 33
−49 + 7 33
0
t
t
8
8
Now y (t) =
C1 e
+
C2 e
.
8
8
Since y(0) = 0 ⇒ C1 + C2 = 0 ⇒ C2 = −C1 .
√
√
14
−49 − 7 33
−49 + 7 33
14
0
Since y (0) = − ⇒
C1 +
C2 = −
16
8
8
16
√
√
√
−49 − 7 33
−49 + 7 33
14
14
14
1
⇒
C1 +
(−C1 ) = − ⇒ −
33C1 = − ⇒ C1 = √
8
8
16
8
16
2 33
6
1
⇒ C2 = − √
2 33
i
√
−49+7 33
1 h −49−7√33 t
t
8
8
√
−e
.
Therefore the position function of the resulting motion is: y(t) =
e
2 33
Now “crossing the equilibrium” means:
√
√
i
√
−49+7 33
1 h −49−7√33 t
−49 − 7 33
−49 + 7 33
t
y(t) = 0 ⇒ √
=0⇒
e 8
−e 8
t=
t ⇒ t = 0.
8
8
2 33
That means the mass only crosses the equilibrium when t = 0, that is at the starting point
and never crosses it afterwards.
5.
1
cos
Do yourself. Answer: 6y 00 + 20y = 0, y(0) = 0.1, y 0 (0) = 0. y(t) =
10
r
1
10
Amplitude = , natural frequency =
and phase-shift = 0.
10
3
r
!
10
t .
3
6.
√
1
Do yourself. Answer: y 00 + 2y 0 + 32y = 0, y(0) = 1, y 0 (0) = 1. y(t) = e−2t cos ( 60 t) +
r2
r
√
√
√
3 −2t
23 −2t
60
23 −2t
−1
√ e sin ( 60 t) =
e cos ( 60 t + α) where α = tan
. Amplitude =
e ,
20
20
20
60
√
√
60
.
natural frequency = 60 and phase-shift = tan−1
20
7.
Do yourself.
8.
Do yourself.
7