Bc this include x(t) it will not be exactly as template proved but more complex so please include anyhting extra needed please, but please follow the strucutre in the template proved. Thank you. ANSWER problem 2.151. Use x(0)=.1m and x(dot)(0)=2 as a template for answering the problem stated problem 2.151. 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 thefollowing initial conditions:(a) x(0) = 0.1 m, x(0)=2On=kmmx + cx+kx=0→x+5x+16x=0→x=-5%, sin(wat + d)x² w²² + x² +2{@_xx@dXo= tan-1=X₁x(1) = X₂e=X₁16==mx(0)=0.1m, x(0) = 2-4Xo Odx + 5@₂xorad√√x² w²/² + x² +25@₂xX@d%=x(t) = X₂etan= tansecmsecc = 2√/km = 2√16(1) = 8sec@, = 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)msec% = tan√√(0.1)²(4)² + (2)² + 2(0.625)(4)(0.1)(2)3.1225-5± √√25-4(16) −5±i√3922Xo Odx + {@₂xoN• sectan-2.5tm-500nt sin (@t+)=0.7275e sin (3.1225t+0.1379)5m|(−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)-2.5tsectan ¹(0.1387) → = 7.8997°, 0.1379 (rad)=·=0.7274 (m)58tan ¹(-0.1784)== 0.625= 0.5693 (m) Problem 2.151: Energy Dissipated in Viscous DampingFind the energy dissipated during a cycle of simple harmonic motion givenby x(t)=0.2sinwot (m) by a viscously damped single-degree-of-freedomsystem with the following parameters.(a) m=10 kg, c=50(b) m = 10 kg, c=150N• secmN secmk = 1,000k = 1,000NmNm

Answered: Image /qna-images/answer/e4b0387d-cfb0-427d-9e77-9e0087cfa22d.jpg

Answered: Problem 2.151: Energy Dissipated in…

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

See similar textbooks

Related questions

Concept explainers

Question

Bc this include x(t) it will not be exactly as template proved but more complex so please include anyhting extra needed please, but please follow the strucutre in the template proved. Thank you. ANSWER problem 2.151. Use x(0)=.1m and x(dot)(0)=2 as a template for answering the problem stated problem 2.151. 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)

Problem 2.151: Energy Dissipated in Viscous Damping
Find the energy dissipated during a cycle of simple harmonic motion given
by x(t)=0.2sinwot (m) by a viscously damped single-degree-of-freedom
system with the following parameters.
(a) m=10 kg, c=50
(b) m = 10 kg, c=150
N• sec
m
N sec
m
k = 1,000
k = 1,000
N
m
N
m

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions