An undamped single-degree-of-freedom system is subjected to dynamic excitation as shown in Figure 1.• System properties: m = 1, c = 0, k = (6π)2.• Force excitation: p(t) = posin(ωt) where po = 9 and ω = 2π.• Initial conditions: u(t = 0) = 0 and ̇u(t = 0) = 0.Please, complete Parts (a) through (d) using any computational tool of your preference. The preferred toolis MATLAB. Print and turn in a single pdf file that will include your code/calculations and your plots.(a) Generate the solution using a linear interpolation of the load over each time step (note that hereyou can use the undamped coefficients). Plot the displacement response for the first 4 seconds andcompare to the exact closed form solution. Repeat using the following time step sizes, ∆t = 0.01,0.05, 0.15, 0.20 seconds. Include the closed form solution and the solutions for different ∆t values in asingle plot. Please, provide your observations by comparing the closed form solution with the solutionsderived using the four different ∆t values. (b) Generate the solution using the Central Difference method. Plot the displacement response forthe first 4 seconds and compare to the exact closed form solution. Repeat using the following timestep sizes, ∆t = 0.01, 0.05, 0.15, 0.20 seconds. Include the closed form solution and the solutions fordifferent ∆t values in a single plot. Please, provide your observations by comparing the closed formsolution with the solutions derived using the four different ∆t values. (c) Generate the solution using the Newmark method for β = 1/6 (linear acceleration). Plot thedisplacement response for the first 4 seconds and compare to the exact closed form solution. Repeatusing the following time step sizes, ∆t = 0.01, 0.05, 0.15, 0.20 seconds. Include the closed form solutionand the solutions for different ∆t values in a single plot. Please, provide your observations by comparingthe closed form solution with the solutions derived using the four different ∆t values. (d) Generate the solution using the Newmark method for β = 1/4 (average acceleration). Plot thedisplacement response for the first 4 seconds and compare to the exact closed form solution. Repeatusing the following time step sizes, ∆t = 0.01, 0.05, 0.15, 0.20 seconds. Include the closed form solutionand the solutions for different ∆t values in a single plot. Please, provide your observations by comparingthe closed form solution with the solutions derived using the four different ∆t values.
An undamped single-degree-of-freedom system is subjected to dynamic excitation as shown in Figure 1.
• System properties: m = 1, c = 0, k = (6π)2.
• Force excitation: p(t) = posin(ωt) where po = 9 and ω = 2π.
• Initial conditions: u(t = 0) = 0 and ̇u(t = 0) = 0.
Please, complete Parts (a) through (d) using any computational tool of your preference. The preferred tool
is MATLAB. Print and turn in a single pdf file that will include your code/calculations and your plots.
(a) Generate the solution using a linear interpolation of the load over each time step (note that here
you can use the undamped coefficients). Plot the displacement response for the first 4 seconds and
compare to the exact closed form solution. Repeat using the following time step sizes, ∆t = 0.01,
0.05, 0.15, 0.20 seconds. Include the closed form solution and the solutions for different ∆t values in a
single plot. Please, provide your observations by comparing the closed form solution with the solutions
derived using the four different ∆t values.
(b) Generate the solution using the Central Difference method. Plot the displacement response for
the first 4 seconds and compare to the exact closed form solution. Repeat using the following time
step sizes, ∆t = 0.01, 0.05, 0.15, 0.20 seconds. Include the closed form solution and the solutions for
different ∆t values in a single plot. Please, provide your observations by comparing the closed form
solution with the solutions derived using the four different ∆t values.
(c) Generate the solution using the Newmark method for β = 1/6 (linear acceleration). Plot the
displacement response for the first 4 seconds and compare to the exact closed form solution. Repeat
using the following time step sizes, ∆t = 0.01, 0.05, 0.15, 0.20 seconds. Include the closed form solution
and the solutions for different ∆t values in a single plot. Please, provide your observations by comparing
the closed form solution with the solutions derived using the four different ∆t values.
(d) Generate the solution using the Newmark method for β = 1/4 (average acceleration). Plot the
displacement response for the first 4 seconds and compare to the exact closed form solution. Repeat
using the following time step sizes, ∆t = 0.01, 0.05, 0.15, 0.20 seconds. Include the closed form solution
and the solutions for different ∆t values in a single plot. Please, provide your observations by comparing
the closed form solution with the solutions derived using the four different ∆t values.
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