Page 81 - Advanced engineering mathematics

2.4 Spring Motion 61

SECTION 2.3 PROBLEMS

In each of Problems 1 through 6, find the general solution, 11. y − 6y + 8y = 3e x
using the method of variation of parameters for a particular

12. y + 6y + 9y = 9cos(3x)
solution.

13. y − 3y + 2y = 10sin(x)

1. y + y = tan(x) 14. y − 4y = 8x + 2e 3x

2

2. y − 4y + 3y = 2cos(x + 3) 15. y − 4y + 13y = 3e − 5e 3x

2x

3. y + 9y = 12sec(3x) 16. y − 2y + y = 3x + 25sin(3x)

2

4. y − 2y − 3y = 2sin (x)
In each of Problems 17 through 24, solve the initial value
−x

5. y − 3y + 2y = cos(e ) problem.

2

6. y − 5y + 6y = 8sin (4x) 2x

17. y − 4y =−7e + x; y(0) = 1, y (0) = 3

In each of Problems 7 through 16, find the general solu- 18. y + 4y = 8 + 34cos(x); y(0) = 3, y (0) = 2
tion, using the method of undetermined coefficients for a 19. y + 8y + 12y = e −x + 7; y(0) = 1, y (0) = 0

particular solution.
2x

20. y − 3y = 2e sin(x); y(0) = 1, y (0) = 2
2
7. y − y − 2y = 2x + 5 21. y − 2y − 8y = 10e −x + 8e ; y(0) = 1, y (0) = 4
2x

8. y − y − 6y = 8e 2x 22. y − y + y = 1; y(1) = 4, y (1) =−2

2
2
9. y − 2y + 10y = 20x + 2x − 8 23. y − y = 5sin (x); y(0) = 2, y (0) =−4

10. y − 4y + 5y = 21e 2x 24. y + y = tan(x); y(0) = 4, y (0) = 3

2.4 Spring Motion
A spring suspended vertically and allowed to come to rest has a natural length L. An object
(bob) of mass m is attached at the lower end, pulling the spring d units past its natural length.
The bob comes to rest in its equilibrium position and is then displaced vertically a distance y 0
units (Figure 2.3) and released from rest or with some initial velocity. We want to construct a
model allowing us to analyze the motion of the bob.
Let y(t) be the displacement of the bob from the equilibrium position at time t, and take
this equilibrium position to be y = 0. Down is chosen as the positive direction. Now consider the
forces acting on the bob. Gravity pulls it downward with a force of magnitude mg. By Hooke’s
law, the spring exerts a force ky on the object. k is the spring constant, which is a number
quantifying the “stiffness" of the spring. At the equilibrium position, the force of the spring is
−kd, which is negative because it acts upward. If the object is pulled downward a distance y
from this position, an additional force −ky is exerted on it. The total force due to the spring
is therefore −kd − ky. The total force due to gravity and the spring is mg − kd − ky. Since at
the equilibrium point this force is zero, then mg = kd. The net force acting on the object due to
gravity and the spring is therefore just −ky.
There are forces tending to retard or damp out the motion. These include air resistance or
perhaps viscosity of a medium in which the object is suspended. A standard assumption (verified
by observation) is that the retarding forces have magnitude proportional to the velocity y . Then

for some constant c called the damping constant, the retarding forces equal cy . The total force

acting on the bob due to gravity, damping, and the spring itself is −ky − cy .

Finally, there may be a driving force f (t) acting on the bob. In this case, the total external
force is

F =−ky − cy + f (t).
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