Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 3.7 Free Mechanical Vibrations due 03/25/2021 at 11:59pm MST
1. (1 point)
The following differential equations represent oscillating
springs.
(i) s00 +36s=0, s(0) = 3, s0(0) = 0.
(ii) 36s00 +s=0, s(0) = 16, s0(0) = 0.
(iii) s00 +16s=0, s(0) = 4, s0(0) = 0.
(iv) 16s00 +s=0, s(0) = 8, s0(0) = 0.
Which differential equation represents:
(a) The spring oscillating most quickly (with the shortest pe-
riod)? [?/i/ii/iii/iv]
(b) The spring oscillating with the largest amplitude?
[?/i/ii/iii/iv]
(c) The spring oscillating most slowly (with the longest period)?
[?/i/ii/iii/iv]
(a) The spring oscillating with the largest maximum velocity?
[?/i/ii/iii/iv]
Solution:
SOLUTION
All the differential equations have solutions of the form
s(t) = Acos ωt+Bsinωt. Since for all of them, s0(0) = 0, we
have s0(0) = 0=C1ωsin0 −Bωcos 0 =0, giving Bω=0. Thus,
either B=0 or ω=0. If ω=0, then s(t)is a constant function,
and since the equations represent oscillating springs, we don’t
want s(t)to be a constant function. Thus, B=0, so all four
equations have solutions of the form s(t) = Acosωt.
(i) s00 +36s=0 and s(0) = Acos0 =A=3. Thus, s(t) =
3cos(6t).
(ii) 36s00 +s=0 and s(0) = Acos0 =A=16. Thus, s(t) =
16cost
6.
(iii) s00 +16s=0 and s(0) = Acos0 =A=4. Thus, s(t) =
4cos(4t).
(iv) 16s00 +s=0 and s(0) = Acos 0 =A=8. Thus, s(t) =
8cost
4.
(a) Spring (i) has the shortest period.
(b) Spring (ii) has the largest amplitude.
(c) Spring (ii) has the longest period.
(a) Spring (i) has the largest maximum velocity (because the
velocity is s0(t)).
Answer(s) submitted:
•i
•ii
•ii
•i
(correct)
Correct Answers:
•i
•ii
•ii
•i
2. (1 point)
For the differential equation
s00 +bs0+9s=0,
find all the values of bthat make the general solution over-
damped, those that make it underdamped, and those that make
it critically damped.
(For each, give an interval or intervals for b for which the
equation is as indicated. Thus if the the equation is overdamped
for all b in the range 2<b≤3and 4≤b<∞, enter (2,3],
[4,infinity); if it is overdamped only for b =3, enter [3,3].)
If the equation is overdamped, b∈
If the equation is underdamped, b∈
If the equation is critically damped, b∈
Solution:
SOLUTION
Recall that s00+bs0+cs =0 is overdamped if the discriminant
b2−4c>0, critically damped if b2−4c=0, and underdamped
if b2−4c<0. In all cases, this is true if the motion is damped:
that is, the real part of the exponential in the solutions is neg-
ative. Since b2−4c=b2−36, the solution is overdamped if
b>6. If b<−6, the condition b2−4c>0 is satisfied, but the
exponentials in the solution will be positive and the system is
therefore not damped. Similarly, the system will be critically
damped if b=6 (but not b=−6, because that will give expo-
nential growth), and underdamped if 0 <b<6.
Answer(s) submitted:
•(-inf,0)
•(0,9)
•
(incorrect)
Correct Answers:
•(6,infinity)
•(0,6)
•[6]
3. (1 point)
A mass m=4 kg is attached to both a spring with spring
constant k=197 N/m and a dash-pot with damping constant
c=4 N ·s/m .
The mass is started in motion with initial position x0=3 m
and initial velocity v0=8 m/s .
Determine the position function x(t)in meters.
x(t) =
Note that, in this problem, the motion of the spring is un-
derdamped, therefore the solution can be written in the form
1
x(t) = C1e−pt cos(ω1t−α1). Determine C1,ω1,α1and p.
C1=
ω1=
α1=
(assume 0 ≤α1<2π)
p=
Graph the function x(t)together with the ”amplitude enve-
lope” curves x=−C1e−pt and x=C1e−pt .
Now assume the mass is set in motion with the same initial
position and velocity, but with the dashpot disconnected ( so
c=0). Solve the resulting differential equation to find the posi-
tion function u(t).
In this case the position function u(t)can be written as
u(t) = C0cos(ω0t−α0). Determine C0,ω0and α0.
C0=
ω0=
α0=
(assume 0 ≤α0<2π)
Finally, graph both function x(t)and u(t)in the same window
to illustrate the effect of damping.
Solution: The differential equation is mx00 +cx0+kx =0, i.e.
4x00 +4x0+197x=0.
The characteristic equation, 4r2+4r+197 =0 has roots r=
−1
2±7i. Thus the general solution has the form
x(t) = e−t/2(c1cos(7t) + c2sin(7t)).
Substituting the initial conditions yields c1=3 and c2=19
14 .
Hence the position function is given by
x(t) = e−t/23cos(7t) + 19
14 sin(7t).
The solution can be written in the form x(t) = C1e−pt cos(ω1t−
α1)where
C1=qc2
1+c2
2=q2125
196
ω1=7
α1=tan−1c2
c1=tan−119
42
p=1
2
If we assume c=0, the differential equation reduces to
4u00 +197u=0.
The roots of the characteristic equation, 4r2+197 =0 are
r=±√197
2i. Thus the general solution has the form
u(t) = Acos √197
2t!+Bsin √197
2t!.
Substituting the initial conditions yields A=3 and B=16
√197 .
We have
C0=√A2+B2=q2029
197
ω0=√197
2
α0=tan−1B
A=tan−116
3√197 .
Answer(s) submitted:
•11.98
•
•
•
•
•
•
•
(incorrect)
Correct Answers:
•(3) *(exp((-1/2)*t))*(cos((7)*t)) + (19/14) *(exp((-1/2)*t))*(sin((7)*t))
•3.29269444903317
•7
•0.424832162919342
•0.5
•3.2092822228322
•7.0178344238091
•0.363133261711284
4. (1 point) This problem is an example of critically damped
harmonic motion.
A mass m=6 kg is attached to both a spring with spring
constant k=54 N/m and a dash-pot with damping constant
c=36 N ·s/m .
The ball is started in motion with initial position x0=4 m
and initial velocity v0=−13 m/s.
Determine the position function x(t)in meters.
x(t) =
Graph the function x(t).
Now assume the mass is set in motion with the same initial
position and velocity, but with the dashpot disconnected ( so
c=0). Solve the resulting differential equation to find the posi-
tion function u(t).
In this case the position function u(t)can be written as
u(t) = C0cos(ω0t−α0). Determine C0,ω0and α0.
C0=
2
ω0=
α0=
(assume 0 ≤α0<2π)
Finally, graph both function x(t)and u(t)in the same window
to illustrate the effect of damping.
Solution: The differential equation is mx00 +cx0+kx =0, i.e.
6x00 +36x0+54x=0.
The characteristic equation, 6r2+36r+54 =0 has the repeated
root r=−3. Thus the general solution has the form
x(t) = c1e−3t+c2te−3t.
Substituting the initial conditions yields c1=4 and c2=−1.
Hence the position function is given by
x(t) = 4e−3t−1te−3t.
If we assume c=0, the differential equation reduces to
6u00 +54u=0.
The roots of the characteristic equation, 6r2+54 =0 are r=
±3i. Thus the general solution has the form
u(t) = Acos (3t) + Bsin(3t).
Substituting the initial conditions yields A=4 and B=−13
3.
We have
C0=√A2+B2=q313
9
ω0=3
α0=tan−1B
A+2π=tan−1−13
12 +2π.
Answer(s) submitted:
•
•
•
•
(incorrect)
Correct Answers:
•(4) *exp((-3)*t) + (-1)*t *exp((-3)*t)
•5.89726867098471
•3
•5.45780845665885
5. (1 point) This problem is an example of over-damped har-
monic motion.
A mass m=4 kg is attached to both a spring with spring
constant k=168 N/m and a dash-pot with damping constant
c=52 N ·s/m .
The ball is started in motion with initial position x0=−5 m
and initial velocity v0=1 m/s.
Determine the position function x(t)in meters.
x(t) =
Graph the function x(t).
Solution: The differential equation is mx00 +cx0+kx =0, i.e.
4x00 +52x0+168x=0.
The characteristic equation, 4r2+52r+168 =0, has roots
r1=−6 and r2=−7. Thus the general solution has the form
x(t) = c1e−6t+c2e−7t.
Substituting the initial conditions yields the system
c1+c2=−5
−6c1−7c2=1
with solution c1=−34 and c2=29. Hence the position func-
tion is given by
x(t) = −34e−6t+29e−7t.
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•(-5/2 - -63/2) *exp((-13/2 - 1/2)*t) + (-5/2 + -63/2) *exp((-13/2 + 1/2)*t)
6. (1 point) A mass of 3kg stretches a spring 40cm. Suppose
the mass is displaced an additional 5cm in the positive (down-
ward) direction and then released.
Suppose that the damping constant is 3 N ·s/m and assume
g=9.8 m/s2is the gravitational acceleration.
(a) Set up a differential equation that describes this system.
Let xto denote the displacement, in meters, of the mass from its
equilibrium position, and give your answer in terms of x,x0,x00.
(b) Enter the initial conditions:
x(0) = m,
x0(0) = m/s
(c) Is this system under damped, over damped, or critically
damped?
• ?
• over damped
• critically damped
• under damped
Solution: (a)
The differential equation is given by mx00 +cx0+kx =0 where
mis the mass, cis the damping constant and kis the spring con-
stant.
To find the spring constant, k, we use the formula mg =kL
where m=3,g=9.8 and L=0.4m. Substituting the values
3
gives k=29.4
0.4.
The differential equation is the given by
3x00 +3x0+29.4
0.4x=0
(b)
Since the mass is displaced an additional 5cm and then release,
the initial condtions are
x(0) = 0.05 m
x0(0) = 0 m/s
(c)
Since the roots of the characteristic equation are complex, the
system is underdamped
Answer(s) submitted:
•
•
•
•over damped
(incorrect)
Correct Answers:
•3*x’’+3*x’+73.5*x = 0
•0.05
•0
•under damped
7. (1 point) A mass weighing 8 lb stretches a spring 15 in.
Suppose the mass is displaced an additional 11 in in the positive
(downward) direction and then released with an initial upward
velocity of 4 ft/s. The mass is in a medium, that exerts a viscu-
ouse resistance of 1 lb when the mass has a velocity of 10 ft/s.
Assume g=32 ft/s2is the gravitational acceleration.
(a) Find the mass m( in lb ·s2/ft).
m=
(b) Find the damping coefficient c(in lb ·s/ft).
c=
(c) Find the spring contant k(in lb/ft).
k=
(d) Set up a differential equation that describes this system.
Let xto denote the displacement, in meters, of the mass from its
equilibrium position, and give your answer in terms of x,x0,x00.
(e) Enter the initial conditions:
x(0) = ft,
x0(0) = ft/s
(f) Is this system under damped, over damped, or critically
damped?
• ?
• over damped
• critically damped
• under damped
Solution: (a) The weight is mg =8 so the mass is m=
8/32 =0.25.
(b) Since the viscuouse resistance is 1 lb when the mass has
a velocity of 10 ft/s, the damping coefficients is c=1/10 =0.1.
(c) To find the spring constant, k, we use the formula w=kL
where w=8 and L=15/12 =1.25 ft. Substituting the values
gives k=6.4.
(d) The differential equation is given by mx00 +cx0+kx =0
where mis the mass, cis the damping constant and kis the
spring constant:
0.25x00 +0.1x0+6.4x=0
(e) Since the mass is displaced an additional 11in with an initial
upward velocity of 4ft/s, the initial conditions are
x(0) = 11/12 ft
x0(0) = −4 ft/s
(f) Since the roots of the characteristic equation are complex,
the system is underdamped
Answer(s) submitted:
•
•
•
•
•
•
•
(incorrect)
Correct Answers:
•0.25
•0.1
•6.4
•0.25*x’’+0.1*x’+6.4*x = 0
•0.916667
•-4
•under damped
8. (1 point) A spring with an m-kg mass and a damping con-
stant 6 (kg/s) can be held stretched 1 meters beyond its natural
length by a force of 3 newtons. If the spring is stretched 2 meters
beyond its natural length and then released with zero velocity,
find the mass that would produce critical damping.
m=kg
Solution: Since the spring can be held stretched 1 meters
beyond its natural length by a force of 3 newtons, the spring
4
3
1
Answer(s) submitted:
•3
(correct)
Correct Answers:
•3
constant is given by k = = 3.
The motion is critically damped if c2 − 4mk = 0 where c = 6
and k = 3.
Solving for m gives
m = 3
5
