1. Math 331.5: Homework 17
Solutions
For the following problems, we use the differential equation
my + γy + ky = g(t)
where
y is the displacement of the mass from its equilibrium position
m is the mass
γ is the damping coefficient
k is the spring constant
g(t) is the external force
To find the constants above, we use the following:
Newton’s Law
w = mg
where
w is the weight
g is the acceleration due to gravity
Hooke’s Law
mg = kL
where
L is the elongation of the spring by the mass
If the solution is in the form
y = A cos(ω0t) + B sin(ω0t)
we can rewrite
y = R cos(ω0t − δ)
where
R =
√
A2 + B2 is the amplitude
ω0 is the frequency
T = 2π/ω0 is the period
δ is the phase
To find δ we use the reference angle
δR = arctan(B/A)
and then the relations
cos(δ) = A/R sin(δ) = B/R
to choose the correct quadrant
2. 2
1. A mass weighing 2 lb. stretches a spring 6 in. Suppose the mass is pulled down an additional 3
in. and then released, and that there is no damping.
(i) Determine the position y of the mass at any time t.
my + γy + ky = g(t)
w = 2 lb
g = 32 ft/s2
m = w/g = 2/32 = 1/16
γ = 0
g(t) = 0
L = 6 in = 1/2 ft
k = mg/L = 2
1/2 = 4
ODE:
1
16
y + 4y = 0
Initial conditions:
y(0) = 3 in = 1/4 ft
y (0) = 0
General solution:
y = A cos(8t) + B sin(8t)
Solution to IVP:
y =
1
4
cos(8t)
(ii) Plot y versus t.
(iii) Find the frequency, period, amplitude and phase of the motion.
ω0 = 8
T = 2π/8
R = 1/4
δ = 0
3. 3
2. A mass weighing 3 lb. stretches a spring 3 in. The mass is pushed upward, contracting the
spring a distance of 1 in., and then set in motion with a downward velocity of 2ft/sec. Suppose
there is no damping.
(i) Determine the position y of the mass at any time t.
my + γy + ky = g(t)
w = 3 lb
g = 32 ft/s2
m = w/g = 3/32
γ = 0
g(t) = 0
L = 3 in = 1/4 ft
k = mg/L = 3
1/4 = 12
ODE:
3
32
y + 12y = 0
Initial conditions:
y(0) = −1 in = −1/12 ft
y (0) = 2
General solution:
y = A cos(8
√
2t) + B sin(8
√
2t)
Solution to IVP:
y = −
1
2
cos(8
√
2t) +
1
4
√
2
sin(8
√
2t)
(ii) Plot y versus t.
(iii) Find the frequency, period, amplitude and phase of the motion.
ω0 = 8
√
2
T = π/4
√
2
R = 11/288
δR = arctan(−3/
√
2) ≈ −1.13
δ = π + arctan(−3/
√
2) ≈ 2.01 lies in quadrant II
4. 4
3. A mass of 100 grams stretches a spring 3 cm. Suppose the mass is set in motion from its
equilibrium position with a downwards velocity of 10 cm/sec and there is no damping.
(i) Find the position y of the mass at time t.
my + γy + ky = g(t)
m = 100 g = 0.1 kg
γ = 0
g(t) = 0
L = 3 cm = 0.03 m
g = 9.8 m/s2
k = mg/L = (0.1)(9.81)/0.03 = 32.7 kg/s2
ODE:
0.1y + 32.7y = 0
Initial conditions:
y(0) = 0
y (0) = 10 cm/s = 0.1 m/s
General solution:
y = k1 cos(
√
327t) + k2 sin(
√
327t)
Solution to IVP:
y(t) =
√
327
3270
sin(
√
327t)
(ii) Plot y versus t.
(iii) Determine when the mass first returns to its equilibrium position.
The mass returns to the equilibrium position when y(t) = 0.
√
327
3270
sin(
√
327t) = 0
sin(
√
327t) = 0
t =
nπ
√
327
for some integer n
Thus, the mass first returns to equilibrium after π√
327
≈ 0.17 seconds.
5. 5
4. A mass weighing 16 lb. stretches a spring 3 in. The mass is attached to a viscous damper with
a damping constant of 2 lb·s/ft. Suppose the mass is set in motion from its equilibrium position
with a downward velocity of 3 in/sec.
(i) Find the position y of the mass at time t.
my + γy + ky = g(t)
w = 16 lb
g = 32 ft/s2
m = w/g = 16/32 = 1/2
γ = 2
g(t) = 0
L = 3 in = 1/4 ft
k = mg/L = 16
1/4 = 64
ODE:
1
2
y + 2y + 64y = 0
Initial conditions:
y(0) = 3 in = 1/4 ft
y (0) = 0
General solution:
y = Ae−2t
cos(2
√
31t) + Be−2t
sin(2
√
31t)
Solution to IVP:
y =
1
8
√
31
e−2t
sin(2
√
31t)
(ii) Plot y versus t.
(iii) Find the quasi frequency and quasi period of the motion.
The quasi frequency is β = 2
√
31 and the quasi period is 2π/β = π/
√
31
(iv) Determine when the mass first returns to its equilibrium position.
The mass returns to the equilibrium position when y(t) = 0.
y =
1
8
√
31
e−2t
sin(2
√
31t) = 0
sin(2
√
31t) = 0
t =
nπ
2
√
31
for some integer n
Thus, the mass first returns to equilibrium after π
2
√
31
≈ 0.28 seconds.
6. 6
5. A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a damper with
coefficient γ. Determine the value of γ for which the system is critically damped. Be sure to give
the units for γ.
my + γy + ky = g(t)
w = 8 lb
g = 32 ft/s2
m = w/g = 8/32 = 1/4
γ = γ
g(t) = 0
L = 1.5 in = 1/8 ft
k = mg/L = 8
1/8 = 64
ODE:
1
4
y + γy + 64y = 0
The system is critically damped when γ2
− 4km = 0.
γ2
− 4km = γ2
− 64 = 0
γ = 8 lb · s/ft
7. 7
6. The position of a certain undamped spring-mass system satisfies the initial value problem
y + 2y = 0
y(0) = 0
y (0) = 2
(i) Find the solution of the IVP.
y =
√
2 sin(
√
2t)
y = 2 cos(
√
2t)
(ii) Plot y versus t and y versus t on the same axes.
Figure 1
(iii) Plot y versus y; that is, plot (y(t), y (t)) parametrically with t as the parameter. This
plot is known as a phase plot and the yy -plane is called the phase plane. Observe that a
closed curve in the phase plane corresponds to a periodic solution. What is the direction
of motion on the phase plot as t increases?
Figure 2
8. 8
7. The position of a certain undamped spring-mass system satisfies the initial value problem
y +
1
4
y + 2y = 0
y(0) = 0
y (0) = 2
(i) Find the solution of the IVP.
y =
16
√
127
e−t/8
sin
√
127
8
y = 2e−t/8
cos
√
127
8
−
2
√
127
e−t/8
sin
√
127
8
(ii) Is the system damped or overdamped? If it is damped, what is the quasi frequency?
The system is damped and the quasi frequency is
√
127
8 .
(iii) Plot y versus t and y versus t on the same axes.
-10 -7.5 -5 -2.5 0 2.5 5 7.5 10
-5
-2.5
2.5
5
Figure 3
(iv) Plot y versus y in the phase plane. What is the direction of motion on the phase plot as
t increases?
