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The parameter of a single-degree-of-freedom system are given by m=1 kg, c=5 N-sec/m, and k=16 N/m. Find the response of the system for the following initial conditions:

The parameter of a single-degree-of-freedom system are given by m=1 kg, c=5 N-sec/m, and k=16 N/m. Find the response of the system for the following initial conditions:

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
Problem 1.1MA
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ANSWER problem 2.149& 2.150.Use x(0)=.1m and x(dot)(0)=2 as a template for answering the problem stated problem. And note theta tells you which quadrant the tan inverse value will fall in so for example theta with two positive values so the angle inputted will be in quadrant 1 (in the unit circle).Negative value from tan inverse means subtract the inverse tan value from larger value and positive value means add to the smaller value. EX: quad 2 (in the unit circle) smaller value is 90 and larger value is 180

The parameter of a single-degree-of-freedom system are given by m=1 kg, c=5 N-sec/m, and k=16 N/m. Find the response of the system for the following initial conditions: (a) x(0) = 0.1 m, x(0)=2 On = k m mx + cx+kx=0→x+5x+16x=0→x= -5%, sin(wat + d) x² w²² + x² +2{@_xx @d Xo = tan -1 = X₁ x(1) = X₂e = X₁ 16 = = m x(0)=0.1m, x(0) = 2- 4 Xo Od x + 5@₂xo rad √√x² w²/² + x² +25@₂xX @d %= x(t) = X₂e tan = tan sec m sec c = 2√/km = 2√16(1) = 8 sec @, = 0,₂√1–5² = 4√/1—(0.625)² =3.1225 |x(0)=−0.1m, x(0)=−2- √x²w² + x³² +25@„xx @d (b) x(0) = -0.1 m, x(0)=-2 (0.1)(3.1225) 2+0.625(4)(0.1) m sec % = tan √√(0.1)²(4)² + (2)² + 2(0.625)(4)(0.1)(2) 3.1225 -5± √√25-4(16) −5±i√39 2 2 Xo Od x + {@₂xo N• sec tan -2.5t m -500nt sin (@t+)=0.7275e sin (3.1225t+0.1379) 5 m |(−0.1)²(4)² + (2)² + 2(0.625)(4)(−0.1)(2) 3.1225 XoOd (-0.1)(3.1225) 2+0.625(4)(-0.1) x + 5@₂xo →=-10.1167°, −0.1379 (rad), or 349.8833°, 6.1066 (rad) x(t)= X₂e −5°¹ sin(@₁t+) = 0.5693e sin (3.1225t+6.1066) -2.5t = sec tan ¹(0.1387) → = 7.8997°, 0.1379 (rad) = ·=0.7274 (m) 5 8 tan ¹(-0.1784) = = 0.625 = 0.5693 (m)
Transcribed Image Text:The parameter of a single-degree-of-freedom system are given by m=1 kg, c=5 N-sec/m, and k=16 N/m. Find the response of the system for the following initial conditions: (a) x(0) = 0.1 m, x(0)=2 On = k m mx + cx+kx=0→x+5x+16x=0→x= -5%, sin(wat + d) x² w²² + x² +2{@_xx @d Xo = tan -1 = X₁ x(1) = X₂e = X₁ 16 = = m x(0)=0.1m, x(0) = 2- 4 Xo Od x + 5@₂xo rad √√x² w²/² + x² +25@₂xX @d %= x(t) = X₂e tan = tan sec m sec c = 2√/km = 2√16(1) = 8 sec @, = 0,₂√1–5² = 4√/1—(0.625)² =3.1225 |x(0)=−0.1m, x(0)=−2- √x²w² + x³² +25@„xx @d (b) x(0) = -0.1 m, x(0)=-2 (0.1)(3.1225) 2+0.625(4)(0.1) m sec % = tan √√(0.1)²(4)² + (2)² + 2(0.625)(4)(0.1)(2) 3.1225 -5± √√25-4(16) −5±i√39 2 2 Xo Od x + {@₂xo N• sec tan -2.5t m -500nt sin (@t+)=0.7275e sin (3.1225t+0.1379) 5 m |(−0.1)²(4)² + (2)² + 2(0.625)(4)(−0.1)(2) 3.1225 XoOd (-0.1)(3.1225) 2+0.625(4)(-0.1) x + 5@₂xo →=-10.1167°, −0.1379 (rad), or 349.8833°, 6.1066 (rad) x(t)= X₂e −5°¹ sin(@₁t+) = 0.5693e sin (3.1225t+6.1066) -2.5t = sec tan ¹(0.1387) → = 7.8997°, 0.1379 (rad) = ·=0.7274 (m) 5 8 tan ¹(-0.1784) = = 0.625 = 0.5693 (m)
Problem 2.149 & 2.150: Response of Damped System Determine the values of 3, wa, and free-vibration response of the following viscously damped systems when x₁ = 0.1 (m), x = 10 (a) m=10 kg, c=150 (b) m=10 kg, c = 200 (c) m=10 kg, c = 250 N• sec m N sec m N• sec m k = 1,000 k = 1,000 N m N sec m N k = 1,000 ( m :
Transcribed Image Text:Problem 2.149 & 2.150: Response of Damped System Determine the values of 3, wa, and free-vibration response of the following viscously damped systems when x₁ = 0.1 (m), x = 10 (a) m=10 kg, c=150 (b) m=10 kg, c = 200 (c) m=10 kg, c = 250 N• sec m N sec m N• sec m k = 1,000 k = 1,000 N m N sec m N k = 1,000 ( m :
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