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For the mass spring damper system shown in the figure, assume that m = 0.25 kg, k= 2500 N/m, and c = 10 N.s/m. The values of force measured at 0.05-second intervals in one cycle are given below. time 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 F(t) 44 12 14 19 33 34 12 22 time 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 Force 32 11 18 25 30 49 40 35 21 time 0.90 0.95 F(t) 11 k m +x F(t) 3- If we excite the system with initial displacement and velocity as 5 mm and 0.2 m/s resnectivelv. plot the response of the free vibration system. 6- Plot the force in the table, and the Fourier series expansion of the force (same graph). 7- Plot the steady state solution. 8- Find the total solution for the forced system with the initial conditions given in part 3. 9- Plot the total solution found in part 8.

For the mass spring damper system shown in the figure, assume that m = 0.25 kg, k= 2500 N/m, and c = 10 N.s/m. The values of force measured at 0.05-second intervals in one cycle are given below. time 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 F(t) 44 12 14 19 33 34 12 22 time 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 Force 32 11 18 25 30 49 40 35 21 time 0.90 0.95 F(t) 11 k m +x F(t) 3- If we excite the system with initial displacement and velocity as 5 mm and 0.2 m/s resnectivelv. plot the response of the free vibration system. 6- Plot the force in the table, and the Fourier series expansion of the force (same graph). 7- Plot the steady state solution. 8- Find the total solution for the forced system with the initial conditions given in part 3. 9- Plot the total solution found in part 8.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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For the mass spring damper system shown in the figure, assume that m = 0.25 kg, k= 2500 N/m, and c = 10 N.s/m. The values of force measured at 0.05-second intervals in one cycle are given below. time 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 F(t) 12 14 44 19 33 34 12 22 time 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 Force 32 11 18 25 30 49 40 35 21 time 0.90 0.95 F(t) 11 k m +x F(t) 3- If we excite the system with initial displacement and velocity as 5 mm and 0.2 m/s resnectivelv. plot the response of the free vibration system. 6- Plot the force in the table, and the Fourier series expansion of the force (same graph). 7- Plot the steady state solution. 8- Find the total solution for the forced system with the initial conditions given in part 3. 9- Plot the total solution found in part 8.
Transcribed Image Text:For the mass spring damper system shown in the figure, assume that m = 0.25 kg, k= 2500 N/m, and c = 10 N.s/m. The values of force measured at 0.05-second intervals in one cycle are given below. time 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 F(t) 12 14 44 19 33 34 12 22 time 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 Force 32 11 18 25 30 49 40 35 21 time 0.90 0.95 F(t) 11 k m +x F(t) 3- If we excite the system with initial displacement and velocity as 5 mm and 0.2 m/s resnectivelv. plot the response of the free vibration system. 6- Plot the force in the table, and the Fourier series expansion of the force (same graph). 7- Plot the steady state solution. 8- Find the total solution for the forced system with the initial conditions given in part 3. 9- Plot the total solution found in part 8.
3). At t=0 sec., Xo =5 mm = 0.005 m and xo = 0.2 m/s %3D Put all the value in x(t), 0.005 = X * sin(Ø) .... . 1). And 0.2 = X * sin(Ø) * (- * wn) + X * cos(Ø) * wd. 2). ... Here, wd= (V1- §2) * wn = 97.979 rad/s From 1). And 2). Ø= 58.51 (in degree) = 1.021 rad. X = 0.00586 m So, solution will be, x(t) = 0.00586 * e-0.2+100-t + sin(97.979 * t + 1.021) m %3D
Transcribed Image Text:3). At t=0 sec., Xo =5 mm = 0.005 m and xo = 0.2 m/s %3D Put all the value in x(t), 0.005 = X * sin(Ø) .... . 1). And 0.2 = X * sin(Ø) * (- * wn) + X * cos(Ø) * wd. 2). ... Here, wd= (V1- §2) * wn = 97.979 rad/s From 1). And 2). Ø= 58.51 (in degree) = 1.021 rad. X = 0.00586 m So, solution will be, x(t) = 0.00586 * e-0.2+100-t + sin(97.979 * t + 1.021) m %3D
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