On Free Mechanical VibrationsAs derived in section 4.1( following Newtons 2nd law of motion and the Hookes law), the D.E. for the mass-spring oscillator is given by:

In the simplest case, when b = 0, and Fe = 0, i.e. Undamped, free vibration, we can rewrite the D.E:As

When b 0, but Fe = 0, we have damping on free vibrations.The D. E. in this case is:

Case I: Underdamped Motion (b2 < 4mk)

Case II: Overdamped Motion (b2 > 4mk)In this case, we have two distinct real roots, r1 & r2. Clearly both are negative, hence a general solution:No local max or minOne local maxOne local min

Case III: Critically Damped Motion (b2 = 4mk)We have repeated root -b/2m. Thus the a general solution is:

ExampleThe motion of a mass-spring system with damping is governed by

This is exercise problem 4, p239.Find the equation of motion and sketch its graph for b = 10, 16, and 20.

1. b = 10: we have m = 1, k = 64, and b2 - 4mk = 100 - 4(64) = - 156, implies = (39)1/2 . Thus the solution to the I.V.P. is Solution.

When b = 16, b2 - 4mk = 0, we have repeated root -8,thus the solution to the I.V.P is1ty

r1 = - 4 and r2 = -16, the solution to the I.V.P. is: When b = 20, b2 - 4mk = 64, thus two distinct real roots are11ty

with the following D. E.

Next we consider forced vibrations

We know a solution to the above equation has the formwhere:

In fact, we have

Thus in the case 0 < b 2 < 4mk (underdamped), a general solution has the form:

Remark on Transient and Steady-State solutions.

Consider the following interconnected fluid tanksIntroductionAB8 L/minX(t)Y(t)24 L24 LX(0)= aY(0)= b6 L/min2 L/min6 L/min

Suppose both tanks, each holding 24 liters of a brine solution, are interconnected by pipes as shown . Fresh water flows into tank A at a rate of 6 L/min, and fluid is drained out of tank B at the same rate; also 8 L/min of fluid are pumped from tank A to tank B, and 2 L/min from tank B to tank A. The liquids inside each tank are kept well stirred, so that each mixture is homogeneous. If initially tank A contains a kg of salt and tank B contains b kg of salt, determine the mass of salt in each tanks at any time t > 0.

Set up the differential equationsFor tank A, we have:

and for tank B, we have

This gives us a system of First Order Equations

We have the following 2nd order Initial Value Problem:

Let us make substitutions:

Then the equation becomes:

On the other hand, suppose

Thus a 2nd order equation is equivalent to a system of 1st order equations in two unknowns.A system of first order equations

Let us consider an example: solve the system

General Method of Solving System of equations: is the Elimination Method.

We want to solve these two equations simultaneously, i.e.find two functions x(t) and y(t) which will satisfy the given equations simultaneouslyThere are many ways to solve such a system.One method is the following: let D = d/dt,then the system can be rewritten as:

(D - 3)[x] + 4y = 1, ..(*)-4x + (D + 7)[y]= 10t .(**)The expression 4(*) + (D - 3)(**) yields:{16 + (D - 3)(D + 7)}[y] = 4+(D - 3)(10t), or (D2 + 4D - 5)[y] =14 - 30t. This is just a 2nd order nonhomogeneous equation. The corresponding auxiliary equation is r2 + 4r - 5 = 0, which has two solution r = -5, and r = 1, thus yh = c1e -5t + c2 e t. And thegeneral solution is y = c1e -5t + c2 e t + 6t + 2.To find x(t), we can use (**).

To find x(t), we solve the 2nd eq.Y(t) = 4x(t) - 7y(t)+ 10t for x(t), We obtain:

GeneralizationLet L1, L2, L3, and L4 denote linear differential operators with constant coefficients, they are polynomials in D. We consider the 2x2 general system of equations:

Rewrite the system in operator form:(D2 - 1)[x] + (D + 1)[y] = -1, ...(3)(D - 1)[x] + D[y] = t2 ...(4) To eliminate y, we use D(3) - (D + 1)(4) ;which yields:{D(D2 - 1) - (D + 1)(D - 1)}[x] = -2t - t2. Or {(D(D2 - 1) - (D2 - 1)}[x] = -2t - t2. Or{(D - 1)(D2 - 1)}[x] = -2t - t2. Example:

Which implies r = 1, 1, -1. Hence the general solution to the homogeneous equation isxh = c1e t + c2te t + c3e -t.Since g(t) = -2t - t2, we shall try a particular solution of the form : xp = At2 + Bt + C, we find A = -1, B = -4, C = -6,The general solution is x = xh + xp.The auxiliary equation for the corresponding homogeneous eq. is (r - 1)(r2 - 1) = 0

Which implies y = (D - D2)[x] -1 - t2. To find y, note that (3) - (4) yields : (D2 - D)[x] + y = -1 - t2.

This is simply a mapping of functions to functions

This is an integral operator.Chapter 7: Laplace TransformsfFLfF

Definition: Let f(t) be a function on [0, ). The Laplace transform of f is the function F defined by the integral

The domain of F(s) is all values of s for which the integral (*) exists. F is also denoted by L{f}.More precisely

Example1. Consider f(t) = 1, for all t > 0. We have

Other examples,2. Exponential function f(t) = e t .3. Sine and Cosine functions say: f(t) = sin t,4. Piecewise continuous (these are functions with finite number of jump discontinuities).

Example 4, P.375A function is piecewise continuous on [0, ), if it is piecewise continuous on [0,N] for any N > 0.

Function of Exponential Order Definition. A function f(t) is said to be of exponential order if there exist positive constants M and T such that

That is the function f(t) grows no faster than a function of the form For example: f(t) = e 3t cos 2t, is of order = 3.

Existence Theorem of Laplace Transform.Theorem: If f(t) is piecewise continuous on [0, ) and of exponential order , then L{f}(s) exists for all s > .

Proof. We shall show that the improper integral converges for s > . This can be seen easily, because [0, ) = [0, T] [ T, ). We only need to show that integral exists on [ T, ).

A table of Laplace Transforms can be found on P. 380Remarks: 1. Laplace Transform is a linear operator. i.e. If the Laplace transforms of f1 and f2 both exist for s > , then we have L{c1 f1 + c2 f2} = c1 L{f1 } + c2 L{f2 } for any constants c1 and c2 . 2. Laplace Transform converts differentiation into multiplication by s.

Properties of Laplace TransformRecall :

Proof.

How about the derivative of f(t)?

Generalization to Higher order derivatives.

Derivatives of the Laplace Transform

1. e -2t sin 2t + e 3t t2.

2. t n.

3. t sin (bt).Some Examples.