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 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 F(t) 0 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 0
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 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 F(t) 0 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 0
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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)
5- Find the steady state solution (only particular solution) for the forced vibration system.
Take number of terms in your Fourier series terms from this range [ 30 – 55].
6- Plot the force in the table, and the Fourier series expansion of the force (same graph).
7- Plot the steady state solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F56a72e11-b103-4489-abab-5a781cca5d51%2F00549e9a-9945-41ef-b611-c4de61b5ed2e%2Fq4nnci7_processed.gif&w=3840&q=75)
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)
5- Find the steady state solution (only particular solution) for the forced vibration system.
Take number of terms in your Fourier series terms from this range [ 30 – 55].
6- Plot the force in the table, and the Fourier series expansion of the force (same graph).
7- Plot the steady state solution.
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