Accelerating the Pace of Engineering and Science - Speech or Presentation Example

I confirm that I have not received help from, or given help to, anyone else in constructing the solution to this assignment. here and signature above) No. 530542 Problem: A block of mass m g is released at a distance h m above one end of a certain vertical non-linear spring. The other end of the spring is fixed. 1. Given the constants k1, k2, a and b, we can solve for the force F(x) in the compressed spring using the equation F(x) = k1xa + k2xb. Question 1 requires a plot of the variation of force F(x) for values of x from 0 to 6h using MATLAB. The constants k1, k2, h, a and b can be computed using our student number: k1 = 45000 (1 + 5) = 270000 k2 = 45000 (1 + 0) = 8100 a = 1 + 0.01(0) = 1 b = 1.5 + 0.01(5) = 1.55 h = 0.43 [1 + (0.1)(6)] = 0.688 2. The next question asks us to find the compression d of the spring using bisection method from the formula for conservation of energy, which is . The program should compute for d such that the absolute error is less than 1 x 10-6. The additional constants needed to solve this are: m = 960 [1.2 – (0.1)(4)] = 768 g = 9.8 [1 + (0.01)(0)] = 9.8 3. The next value we need to solve using MATLAB is the energy E stored in the spring using the formula . For this problem, we are asked to use composite trapezoidal method to solve for the integral, with a prescribed relative error of less than 0.001 percent. Solution: 1. Plotting the force as a function of x from 0 to 6*0.688 can be easily done in MATLAB without having to write a new m-file. The MathWorks website provides a comprehensive guide to use this feature. The first step is to define the values that will be used to generate a 2-d plot (“plot,” par. 14). Thus, we input the following codes on MATLAB’s command window: >> x = 0 : 0.01: 6*0.688; >> y = 270000*x + 8100*x.^1.55; The first line defines the values of x. That is, from 0 to 6*0.688 with an increment of 0.01. We are therefore defining a one dimensional array. The increment can be any number but a smaller value will produce a smoother graph as MATLAB simply connects the dots. The dots correspond to each x and y coordinates defined. The next line defines y as a function of x. Next, we highlight the two variables we have just defined in the workspace then click the button. The resulting graph is shown below: 2. Problem two is easier as we already have a bisection method program from our previous exercises. We only have to make the necessary changes on that program to answer question 2. The equation to be used is . Thus, we replace the equation in the subfunction with f_value = (270000/2)*x.^2 + (8100/2.55)*x.^2.55 - 768*9.8*(x + 0.688); . Notice that all terms on the right side of the original equation was moved to the left so that the equation is now equal to zero (since we are looking for a point that lies in the x-axis). To give us an idea about the location of this point, we can plot it the same way we did in question 1 then choose an interval in the x-axis that encloses this point. Thus we input the following codes on the command window and click the plot button: >> x = 0 : 0.01: 1; >> f_value = (270000/2)*x.^2 + (8100/2.55)*x.^2.55 - 768*9.8*(x + 0.688); Looking at the graph above, we see that the point intersecting the x-axis is somewhere in the interval (0.1, 0.3). We will use this as our initial interval in the following bisection program: function bisection %BISECTION solve a given equation with the bisection method clc % specify number of bisection steps n = 17; % define the initial interval a = 0.1; b = 0.3; fprintf(\n initial interval [%g, %g] \n total bisection steps %d\n, a,b,n); %initialise and check that there is a root in the prescribed interval x_left = a; x_right = b; f_left = f(x_left); f_right = f(x_right); if f_left*f_right > 0 error(ERROR: no root in the specified interval); end %the bisection method for i=1:n if f_left == 0 %exact root is reached by the left bound of the interval fprintf(\n stage %g root %g with zero absolute error \n,i,x_left); return; end if f_right==0 %exact root is reached by the right bound of the interval fprintf(\n stage %g root %g with zero absolute error \n,i,x_right); return end %the bisection process x_mid = (x_left+x_right)/2.0; f_mid = f(x_mid); if f_left*f_mid 0 % there is a root in [x_mid,x_right] x_left = x_mid; f_left = f_mid; end %compute the approximate rood for the current bisection step root = (x_left+x_right)/2.0; %compute the absolute error for the current bisection step abs_err=(x_right-x_left)/2.0; fprintf(\n stage %g root %g absolute error < %g \n,i,root,abs_err); end %check satisfaction of equation at end of process residual = f(root); fprintf(\n final residual = %g \n,residual); end %Subfunction defines the equation f(x) = 0 function f_value = f(x) f_value = (270000/2)*x.^2 + (8100/2.55)*x.^2.55 - 768*9.8*(x + 0.688); end The output is: initial interval [0.1, 0.3] total bisection steps 17 stage 1 root 0.25 absolute error < 0.05 stage 2 root 0.225 absolute error < 0.025 stage 3 root 0.2125 absolute error < 0.0125 stage 4 root 0.21875 absolute error < 0.00625 stage 5 root 0.221875 absolute error < 0.003125 stage 6 root 0.223438 absolute error < 0.0015625 stage 7 root 0.224219 absolute error < 0.00078125 stage 8 root 0.224609 absolute error < 0.000390625 stage 9 root 0.224414 absolute error < 0.000195312 stage 10 root 0.224316 absolute error < 9.76562e-005 stage 11 root 0.224365 absolute error < 4.88281e-005 stage 12 root 0.22439 absolute error < 2.44141e-005 stage 13 root 0.224377 absolute error < 1.2207e-005 stage 14 root 0.224384 absolute error < 6.10352e-006 stage 15 root 0.22438 absolute error < 3.05176e-006 stage 16 root 0.224379 absolute error < 1.52588e-006 stage 17 root 0.224378 absolute error < 7.62939e-007 final residual = 0.0282083 3. As in question 2, we can use the composite trapezoidal program from our previous exercises. First, we replace the equation defined in the subfunction with f_value = 270000*x + 8100*x.^(1.55); as this is the function that we need to integrate, then use the value of d computed from question 2. Since we need to compute for the percent relative error, we include a code in the program that calculates the exact answer to be able to compare it with the approximated answer. function trapezoidal(N) %INTTRAPEZOIDAL integration using the composite trapezoidal rule a = 0; b = 0.224378; n = 1; if nargin ~= 1 N = quadl(@f, a, b); %computes for the correct answer if there is no input from user end for n = 1:1:inf; %An infinite loop. n begins at 1 with an increment of 1 with each iteration h = (b-a)/n; integral = 0.0; for m = 1:n x_left = a+(m-1)*h; x_right = a+m*h; f_left = f(x_left); f_right = f(x_right); integral = integral+h/2*(f_left+f_right); end r_err = abs((integral - N)/N)*100; if r_err < 0.001 %percent relative error should be less than 0.001 break; %breaks if inequality is satisfied end end fprintf(\n Trapezoidal approximation to integral using %g subintervals is %g \n,n,integral); fprintf(\n Relative error is %g percent.\n, r_err); end function f_value = f(x) f_value = 270000*x + 8100*x.^(1.55); end The output is as follows: Trapezoidal approximation to integral using 19 subintervals is 6867 Relative error is 0.000902803 percent. Works Cited “plot.” The MathWorks: Accelerating the pace of engineering and science. 2009. The MathWorks, Inc. 09 Dec. 2009 Read More