HW5#2 - disp'The case that has the smallest natural frequency of vibration is when k = 4X10^6 N/m and it is 2.003 cycles per second disp(b Which

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Sheet1 Page 1 %Engineering 6, Spring 2004, Problem 5.2 %Ryan Uy, 2509 %Section B05, Tues. 3-4 % (f^6) - (5 * a * f^4) + (6 * a^2 * f^2) - (a^3) = 0 % a = k / (4*m*pi^2) %Set values for given variables m = 5000 k = [ 4e6 5e6 6e6] a = k / (4 * m * pi^2) %Set coefficient vectors for every a f1 = [1 0 -5*a(1) 0 6*a(1)^2 0 -a(1)^3] f2 = [1 0 -5*a(2) 0 6*a(2)^2 0 -a(2)^3] f3 = [1 0 -5*a(3) 0 6*a(3)^2 0 -a(3)^3] %Calculate and display roots for every a fprintf('Roots when k = 4X10^6 N/m: %.3f %.3f %.3f %.3f %.3f %.3f\n',roots(f1)) fprintf('Roots when k = 5X10^6 N/m: %.3f %.3f %.3f %.3f %.3f %.3f\n',roots(f2)) fprintf('Roots when k = 6X10^6 N/m: %.3f %.3f %.3f %.3f %.3f %.3f\n',roots(f3)) disp(' ') % (a) Which case has the smallest natural frequency of vibration and what is it?
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Unformatted text preview: disp('The case that has the smallest natural frequency of vibration is when k = 4X10^6 N/m, and it is 2.003 cycles per second' ) disp(' ') % (b) Which case has the largest natural frequency of vibration and what is it? disp('The case that has the largest natural frequency of vibration is when k = 6X10^6 N/m, and it is 9.935 cycles per second') disp(' ') % (c) Which case has the smallest spread between the natural frequencies and what is it? fprintf('The case that has the smallest spread between the natural frequencies is when k = 4X10^6 N/m, and it is %.3f cycles p e Sheet1 Page 2 %calculate a for every k %new line %new line %new line...
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This note was uploaded on 01/22/2011 for the course ENG 006 taught by Professor Lagerstrom during the Spring '06 term at UC Davis.

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HW5#2 - disp'The case that has the smallest natural frequency of vibration is when k = 4X10^6 N/m and it is 2.003 cycles per second disp(b Which

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