1) The document discusses the spring-mass problem and vibration of springs based on Hooke's law. It provides the equations of motion for an undamped and damped spring-mass system with and without an imposed force. 2) Examples are presented of an oscillating mass on an undamped and damped spring to illustrate the equations of motion. 3) Harmonic motion and resonance are introduced, with equations provided for undamped vibrations using Laplace transforms. The document discusses the cases when the natural and damping frequencies are equal or not equal.

1. CALENG3
Differential
Equations
Luis F. Razon, Ph.D.
Chemical Engineering Department
De La Salle University
Lecture 7
Spring-Mass Problem
1

2. Hooke’s Law
2

3. Vibration of Springs
• Force due to spring:
k = spring constant(lbs/ft)
x = displacement
(downwards positive)
• Force due to acceleration
• Force due to retarding medium
 if b = 0 then motion is undamped
 if b  0 then motion is damped
+
kx
F 
)
(
' t
bx
F 
)
(
" t
x
g
w
ma
F 

3

4. Video courtesy of Mahdi Eynian
(http://www.youtube.com/user/MahdiEynian)
4

5. Video courtesy of Mahdi Eynian
(http://www.youtube.com/user/MahdiEynian)
5

6. Summation of Forces
• If no imposed force
• If a force of magnitude H(t)
is imposed
0
)
t
(
kx
)
t
(
'
bx
)
t
(
"
x
g
w



)
(
)
(
)
(
'
)
(
" t
H
t
kx
t
bx
t
x
g
w



H(t)
6

7. Video courtesy of Mahdi Eynian
(http://www.youtube.com/user/MahdiEynian)
7

8. Example
A spring is such that it would be
stretched 6 inches by a 12-lb weight.
Let the weight be attached to the spring
and pulled down 4 in. below the
equilibrium point. If the weight is
started with an upward velocity of 2
ft/sec, describe the motion. No damping
or impressed force is present.
8

9. 𝑤
𝑔
𝑥"(𝑡) + 𝑏𝑥′(𝑡) + 𝑘𝑥(𝑡) = 𝐻(𝑡)
A spring is such that it would be stretched 6 inches by a 12-lb weight. Let the weight be attached
to the spring and pulled down 4 in. below the equilibrium point. If the weight is started with an
upward velocity of 2 ft/sec, describe the motion. No damping or impressed force is present.
Given: x = 0.5 ft when F = 12 lb, w = 12 lb,
x(0) = 4 in. (“pulled down”) , x’(0) = - 2 ft/s (“upward velocity”), b=0, H(t) = 0
Find: x = f(t)
Solution:
12
32
𝑥"(𝑡) + (0)𝑥′(𝑡) + 24𝑥(𝑡) = 0
𝑘 =
𝐹
𝑥
=
12 𝑙𝑏
6 𝑖𝑛
=
12 𝑙𝑏
0.5 𝑓𝑡
= 24
𝑙𝑏
𝑓𝑡
𝑔 = 32
𝑓𝑡
𝑠2 (by convention)
𝑥"(𝑡) + 64𝑥(𝑡) = 0 𝑥 0 = 4 𝑖𝑛 =
1
3
𝑓𝑡
𝐿{𝑥"(𝑡)} + 64𝐿{𝑥(𝑡)} = 0
𝑥′
0 = −2
𝑓𝑡
𝑠
𝑠2
𝐿 𝑥 − 𝑠𝑥 0 − 𝑥′
0 + 64 𝐿 𝑥 = 0
𝑠2
𝐿 𝑥 − 𝑠
1
3
− (−2) + 64 𝐿 𝑥 = 0
(𝑠2
+64) 𝐿 𝑥 = 𝑠
1
3
− 2
𝐿 𝑥 =
𝑠
(𝑠2+64)
1
3
−
2
(𝑠2+64)
𝑥 =
1
3
cos 8𝑡 −
1
4
sin(8𝑡)
The weight will bounce up and down indefinitely.
9

10. 10

11. Example
A spring is such that it would be
stretched 6 inches by a 12-lb
weight. Let the weight be attached
to the spring and pulled down 4 in.
below the equilibrium point. If the
weight is started with an upward
velocity of 2 ft/sec, describe the
motion. There is a damping force of
magnitude 3|v|.
11

12. 𝑤
𝑔
𝑥"(𝑡) + 𝑏𝑥′(𝑡) + 𝑘𝑥(𝑡) = 𝐻(𝑡)
A spring is such that it would be stretched 6 inches by a 12-lb weight. Let the weight be attached
to the spring and pulled down 4 in. below the equilibrium point. If the weight is started with an
upward velocity of 2 ft/sec, describe the motion. There is a damping force of magnitude 3|v|
Given: x = 0.5 ft when F = 12 lb, w = 12 lb,
x(0) = 4 in. (“pulled down”) , x’(0) = - 2 ft/s (“upward velocity”), b=3, H(t) = 0
Find: x = f(t)
Solution:
12
32
𝑥"(𝑡) + (3)𝑥′(𝑡) + 24𝑥(𝑡) = 0
𝑘 =
𝐹
𝑥
=
12 𝑙𝑏
6 𝑖𝑛
=
12 𝑙𝑏
0.5 𝑓𝑡
= 24
𝑙𝑏
𝑓𝑡
𝑔 = 32
𝑓𝑡
𝑠2 (by convention)
𝑥"(𝑡) + 8𝑥′(𝑡) + 64𝑥(𝑡) = 0 𝑥 0 = 4 𝑖𝑛 =
1
3
𝑓𝑡
𝐿{𝑥"(𝑡)} + 8𝐿{𝑥′(𝑡)} + 64𝐿{𝑥(𝑡)} = 0
𝑥′
0 = −2
𝑓𝑡
𝑠
𝑠2
𝐿 𝑥 − 𝑠𝑥 0 − 𝑥′
0 + 8𝑠𝐿 𝑥 − 8𝑥 0 + 64 𝐿 𝑥 = 0
𝑠2
𝐿 𝑥 − 𝑠
1
3
− −2 + 8𝑠𝐿 𝑥 −
8
3
+ 64 𝐿 𝑥 = 0
(𝑠2
+8𝑠 + 64)𝐿 𝑥 − 𝑠
1
3
− −2 −
8
3
= 0
12

13. (𝑠2
+8𝑠 + 64)𝐿 𝑥 − 𝑠
1
3
− −2 −
8
3
= 0
𝐿 𝑥 =
1
3
𝑠
𝑠2 + 8𝑠 + 64
+
2
3
1
𝑠2 + 8𝑠 + 64
𝐿 𝑥 =
1
3
𝑠 + 2
𝑠2 + 8𝑠 + 16 + 48
=
1
3
𝑠 + 2
𝑠 + 4 2 + 48
2
completing the square
𝐿 𝑥 =
1
3
𝑠 + 4 + 2 − 4
𝑠 + 4 2 + 48
2
𝑥 = 𝐿−1
1
3
𝑠 + 4
𝑠 + 4 2 + 48
2 −
2
3 48
48
𝑠 + 4 2 + 48
2
𝑥 =
1
3
𝑒−4𝑡
cos 48𝑡 −
2
3 48
𝑒−4𝑡
sin 48𝑡
The weight will bounce up and down and settle down eventually at x=0.
13

14. x
t 14
𝑥 =
1
3
𝑒−4𝑡 cos 48𝑡 −
2
3 48
𝑒−4𝑡 sin 48𝑡

15. Example
A spring, with spring constant 0.75 lb/ft,
lies on a long smooth (frictionless) table.
A 6-lb weight is attached to the spring
and is at rest at the equilibrium
position. A 1.5-lb force is applied to the
support along the line of action of the
spring for 4 sec and is then removed.
Describe the motion.
15

16. A spring, with spring constant 0.75 lb/ft, lies on a long smooth (frictionless) table. A 6-lb
weight is attached to the spring and is at rest at the equilibrium position. A 1.5-lb force is
applied to the support along the line of action of the spring for 4 sec and is then removed.
Describe the motion.
𝑤
𝑔
𝑥"(𝑡) + 𝑏𝑥′(𝑡) + 𝑘𝑥(𝑡) = 𝐻(𝑡)
Given: k=0.75 lb/ft, b=0 (“frictionless”), w = 6 lb, x’(0)=0 (“at rest”), x(0)=0 (“at the equilibrium position”)
H(t) = 1.5 lb, 0 ≤ t < 4; H(t) = 0, t ≥ 4
Find: x = f(t)
Solution:
𝐻 𝑡 =
1.5 0 ≤ 𝑡 < 4
0 𝑡 ≥ 4
= 1.5 − 1.5u t − 4
6
32
𝑥"(𝑡) + (0)𝑥′(𝑡) + 0.75𝑥(𝑡) = 1.5 − 1.5u t − 4
𝑥"(𝑡) + 4𝑥(𝑡) = 8 − 8u t − 4
𝐿{𝑥"(𝑡)} + 4𝐿{𝑥(𝑡)} = 𝐿{8 − 8u t − 4 }
𝑠2𝐿 𝑥 − 𝑠𝑥 0 − 𝑥′ 0 + 4𝐿 𝑥 =
8
𝑠
−
8𝑒−4𝑠
s
(𝑠2+4)𝐿 𝑥 − 0 − 0 =
8
𝑠
−
8𝑒−4𝑠
s
𝐿 𝑥 =
8
𝑠(𝑠2+4)
−
8𝑒−4𝑠
s(𝑠2+4)
16

17. 𝐿 𝑥 =
8
𝑠(𝑠2+4)
−
8𝑒−4𝑠
s(𝑠2+4)
8
𝑠(𝑠2+4)
=
𝐴
𝑠
+
𝐵𝑠 + 𝐶
𝑠2 + 4
A=2 B=-2
𝑥 = 𝐿−1
2
𝑠
−
2𝑠
𝑠2 + 4
−
2𝑒−4𝑠
𝑠
+
2𝑠𝑒−4𝑠
𝑠2 + 4
𝑥 = 2 − 2 cos 2𝑡 − 2𝑢 𝑡 − 4 + 2 cos 2 𝑡 − 4 𝑢(𝑡 − 4)
𝑥 =
2 − 2cos(2𝑡) 0 ≤ 𝑡 < 4
−2cos(2𝑡) + 2 cos 2 𝑡 − 4 𝑡 ≥ 4
The weight will oscillate about x=2
then shift its center of oscillation when t=4
and oscillate about x=0 indefinitely
17

18. 18

19. Harmonic Motion and Resonance
w
bg


2
w
kg

2

w
g
t
H
t
K
)
(
)
( 
)
(
)
(
)
(
'
2
)
(
" 2
t
K
t
x
t
x
t
x 

 

)
(
)
(
)
(
" 2
t
K
t
x
t
x 
 
Let
Undamped Vibrations, b=0
)
(
)
(
)
(
'
)
(
" t
H
t
kx
t
bx
t
x
g
w



19

20. Harmonic Motion and Resonance
t
A
t
K 
sin
)
( 
𝑥(0) = 0 𝑥′(0) = 0
2
2
2
2
)
(
)
0
(
'
)
0
(
)
(








s
A
s
u
x
sx
s
u
s
𝑠2𝑢(𝑠) + 𝛽2𝑢(𝑠) =
𝐴𝜔
𝑠2 + 𝜔2
𝑢(𝑠) =
𝐴𝜔
(𝑠2 + 𝜔2)(𝑠2 + 𝛽2)
suppose
by Laplace Transforms
)
(
)
(
)
(
" 2
t
K
t
x
t
x 
 
20

21. Harmonic Motion and Resonance
𝑢(𝑠) =
𝐴𝜔
(𝑠2 + 𝜔2)(𝑠2 + 𝛽2)

 
𝑢(𝑠) =
𝐴𝜔
𝜔2 − 𝛽2
1
(𝑠2 + 𝛽2)
−
1
(𝑠2 + 𝜔2)
𝑥(𝑡) =
𝐴𝜔
𝜔2 − 𝛽2
1
𝛽
sin 𝛽 𝑡 −
1
𝜔
sin 𝜔 𝑡
if then
x
time
21

22. Harmonic Motion and Resonance

 
𝑢(𝑠) =
𝐴𝜔
(𝑠2 + 𝛽2)2
𝑥(𝑡) =
2𝐴
𝛽2
sin 𝛽 𝑡 − 𝛽𝑡 cos 𝛽 𝑡
if then
𝑢(𝑠) =
𝐴𝜔
(𝑠2 + 𝜔2)(𝑠2 + 𝛽2)
x
time 22

23. 23

24. 24

25. 25

26. 26

27. 27

28. 28