On Free Mechanical Vibrations: On Free Mechanical Vibrations As derived in section 4.1( following Newton’s 2nd law of motion and the Hooke’s 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:: 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.: 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 I: Underdamped Motion (b2 < 4mk)
Case II: Overdamped 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 min One local max One local min
Case III: Critically Damped Motion (b2 = 4mk): Case III: Critically Damped Motion (b2 = 4mk) We have repeated root -b/2m. Thus the a general solution is:
Example: Example The 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.
Solution.: 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,: When b = 16, b2 - 4mk = 0, we have repeated root -8, thus the solution to the I.V.P is 1 t y
When b = 20, b2 - 4mk = 64, thus two distinct real roots are: r1 = - 4 and r2 = -16, the solution to the I.V.P. is: When b = 20, b2 - 4mk = 64, thus two distinct real roots are 1 1 t y
Next we consider forced vibrations: with the following D. E.
Next we consider forced vibrations
We know a solution to the above equation has the form: We know a solution to the above equation has the form where:
In fact, we have
Thus in the case 0 < b 2 < 4mk (underdamped), a general solution has the form:: Thus in the case 0 < b 2 < 4mk (underdamped), a general solution has the form:
Remark on Transient and Steady-State solutions.: Remark on Transient and Steady-State solutions.
Introduction: Consider the following interconnected fluid tanks Introduction A B 8 L/min X(t) Y(t) 24 L 24 L X(0)= a Y(0)= b 6 L/min 2 L/min 6 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. : 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 equations: Set up the differential equations For tank A, we have:
and for tank B, we have
This gives us a system of First Order Equations: This gives us a system of First Order Equations
On the other hand, suppose: We have the following 2nd order Initial Value Problem:
Let us make substitutions:
Then the equation becomes:
On the other hand, suppose
A system of first order equations: Thus a 2nd order equation is equivalent to a system of 1st order equations in two unknowns. A system of first order equations
General Method of Solving System of equations: is the Elimination Method.: Let us consider an example: solve the system
General Method of Solving System of equations: is the Elimination Method.
Slide22: We want to solve these two equations simultaneously, i.e.
find two functions x(t) and y(t) which will satisfy the given equations simultaneously
There 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 .…(**): (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 the
general 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), : To find x(t), we solve the 2nd eq. Y(t) = 4x(t) - 7y(t)+ 10t for x(t), We obtain:
Generalization: Generalization Let 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:
Example:: 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:
The auxiliary equation for the corresponding homogeneous eq. is (r - 1)(r2 - 1) = 0: Which implies r = 1, 1, -1. Hence the general solution to the homogeneous equation is
xh = 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
To find y, note that (3) - (4) yields : (D2 - D)[x] + y = -1 - t2.: Which implies y = (D - D2)[x] -1 - t2. To find y, note that (3) - (4) yields : (D2 - D)[x] + y = -1 - t2.
Chapter 7: Laplace Transforms: This is simply a mapping of functions to functions
This is an integral operator. Chapter 7: Laplace Transforms f F L f F
More precisely: 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
Example: Example 1. Consider f(t) = 1, for all t > 0. We have
Other examples,: Other examples, 2. Exponential function f(t) = e t .
3. Sine and Cosine functions say: f(t) = sin ßt,
4. Piecewise continuous (these are functions with finite number of jump discontinuities).
Example 4, P.375: Example 4, P.375 A function is piecewise continuous on [0, ), if it is piecewise continuous on [0,N] for any N > 0.
Function of Exponential Order : 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.: 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. 380: A table of Laplace Transforms can be found on P. 380 Remarks:
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 Transform: Properties of Laplace Transform Recall :
Proof.
How about the derivative of f(t)?: How about the derivative of f(t)?
Generalization to Higher order derivatives.: Generalization to Higher order derivatives.
Derivatives of the Laplace Transform: Derivatives of the Laplace Transform
Some Examples.: 1. e -2t sin 2t + e 3t t2.
2. t n.
3. t sin (bt). Some Examples.