Spring-Mass Systems

Consider the system of two masses

and two springs with no external force. Visualize a wall on the left and to the right a spring , a mass, a spring and another mass. Assume that the spring constants are

. See Figure 1 below.

[Graphics:Images/SpringMassMod_gr_3.gif]

Figure 1. Coupled masses with spring attached to the wall at the left.

Assume that the masses slide on a frictionless surface and that the functions denote the displacement from static equilibrium of the masses , respectively. It can be shown by using Newton's second law and Hooke's law that the system of D. E.'s for is

Remark:

The eigenfrequencies can be obtained by taking the square root of the eigenvalues of the matrix .

Consider the system of two masses and three springs with no external force. Visualize a wall on the left and to the right a spring , a mass, a spring, a mass, a spring and another wall. Assume that the spring constants are . See Figure 2 below.

[Graphics:Images/SpringMassMod_gr_195.gif]

Figure 2. Coupled masses with springs attached to walls at the left and right.

Assume that the masses slide on a frictionless surface and that the functions denote the displacement from static equilibrium of the masses , respectively. It can be shown by using Newton's second law and Hooke's law that the system of D. E.'s for is

Remark:

The eigenfrequencies can be obtained by taking the square root of the eigenvalues of the matrix .

Eigen Frequencies:

Consider three springs in parallel, with two of the springs having spring constant k and attached to two walls on either end, and the third spring of spring constant k placed between two equal masses m. To solve for the motion of the masses using the normal formalism, equate forces


			(1)

			(2)

Writing (1) and (2) in matrix form gives

(3)

To find the equations of motion from energy considerations, note that the kinetic energy is defined by

(4)

so the rate of change of K is

(5)

Similarly, the potential energy is defined by


			(6)

so the rate of change of U is



			(7)

But energy is conserved, so

(8)

and

(9)

Now, this equation must hold for arbitrary and , so each piece must vanish separately ("separation of variables "), yielding the coupled equations (3). The eigenmodes of the system follow from (3). Looking for a harmonic solution using the trial solution ,


			(10)

(11)

This has solutions when the determinant is 0, so

(12)

(13)

(14)

To find the eigenvectors, plug back in. For ,

(15)

so and the first eigenvalue and its associated eigenvector are

(16)

This corresponds to masses moving in opposite directions. For ,

(17)

(18)

This corresponds to masses moving in the same direction.

If the middle spring has the same spring constant as those on either side, then , and the eigenfrequencies are

			(19)
			(20)

Comments

Runge-Kutta-Fehlberg Method

One way to guarantee accuracy in the solution of an I.V.P. is to solve the problem twice using step sizes h and and compare answers at the mesh points corresponding to the larger step size. But this requires a significant amount of computation for the smaller step size and must be repeated if it is determined that the agreement is not good enough. The Runge-Kutta-Fehlberg method (denoted RKF45) is one way to try to resolve this problem. It has a procedure to determine if the proper step size h is being used. At each step, two different approximations for the solution are made and compared. If the two answers are in close agreement, the approximation is accepted. If the two answers do not agree to a specified accuracy, the step size is reduced. If the answers agree to more significant digits than required, the step size is increased. Each Runge-Kutta-Fehlberg step requires the use of the following six values: Then an approximation to the solution of the I.V.P....

Van Der Pol System

The van der Pol equation is an ordinary differential equation that can be derived from the Rayleigh differential equation by differentiating and setting . It is an equation describing self-sustaining oscillations in which energy is fed into small oscillations and removed from large oscillations. This equation arises in the study of circuits containing vacuum tubes and is given by If , the equation reduces to the equation of simple harmonic motion The van der Pol equation is , where is a constant. When the equation reduces to , and has the familiar solution . Usually the term in equation (1) should be regarded as friction or resistance, and this is the case when the coefficient is positive. However, if the coefficient is negative then we have the case of "negative resistance." In the age of "vacuum tube" radios, the " tetrode vacuum tube " (cathode, grid, plate), was used for a power amplifie...

Powell's Method

The essence of Powell's method is to add two steps to the process described in the preceding paragraph. The vector represents, in some sense, the average direction moved over the n intermediate steps in an iteration. Thus the point is determined to be the point at which the minimum of the function f occurs along the vector . As before, f is a function of one variable along this vector and the minimization could be accomplished with an application of the golden ratio or Fibonacci searches. Finally, since the vector was such a good direction, it replaces one of the direction vectors for the next iteration. The iteration is then repeated using the new set of direction vectors to generate a sequence of points . In one step of the iteration instead of a zig-zag path the iteration follows a "dog-leg" path. The process is outlined below. Let be an initial guess at the location of the minimum of the function . Let for be the ...

Newton-Raphson Method

This method is generally used to improve the result obtained by one of the previous methods. Let x o be an approximate root of f(x) = 0 and let x 1 = x o + h be the correct root so that f(x 1 ) = 0. Expanding f(x o + h) by Taylor ’s series, we obtain f(x o ) + hf’(x o ) + h 2 f’’(x o )/2! +-----= 0 Neglecting the second and higher order derivatives, we have f(x o ) + hf’(x o ) = 0 Which gives h = - f(x o )/f’(x o ) A better approximation than x o is therefore given be x 1 , where x 1 = x o – f(x o )/f’(x o ) Successive approximations are given by x 2 , x 3 ,----,x n+1 , where x n+1 = x n – f(x n )/f’(x n ) Which is the Newton-Raphson formula. Example: - Find a real root of the equation x 3 -5x + 3 = 0. Sol n : - Let, f(x) = x 3 -5x + 3 = 0 f’(x) = 3x 2 - 5 Choosing x o = 1 Step-1: f(x o ) = -1 f(x o ) = -2 So, x 1 =1 – ½ = 0.5 Step-2: f(x 1 ) = 0.625 f’(x 1 ) = -4.25 x 2 = 0.5 + 0.625/4.25 = 0.647 S...

Numerical Analysis

Search This Blog