The frequency equation of a 3 Degree of Freedom spring mass system (Figure Q1) is given as: m3w6 – 4km2w* + 3k?mw² = 0 where the value of the mass, m = 0.1 kg and spring stiffness coefficient, k = 10 N/m. By performing your calculation (final answer in 3 decimal points): Determine the most positive roots of the system, wusing Bisection Method. Use initial value of [4+(0.1*A), 12+(B/2)]. Iterate until 5th iteration. (a)
The frequency equation of a 3 Degree of Freedom spring mass system (Figure Q1) is given as: m3w6 – 4km2w* + 3k?mw² = 0 where the value of the mass, m = 0.1 kg and spring stiffness coefficient, k = 10 N/m. By performing your calculation (final answer in 3 decimal points): Determine the most positive roots of the system, wusing Bisection Method. Use initial value of [4+(0.1*A), 12+(B/2)]. Iterate until 5th iteration. (a)
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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Question
A= 1 B=6
![Q1
The frequency equation of a 3 Degree of Freedom spring mass system (Figure Q1) is given
as:
m3w6 –
4km?w* + 3k²mw² = 0
where the value of the mass, m = 0.1 kg and spring stiffness coefficient, k = 10 N/m.
By performing your calculation (final answer in 3 decimal points):
(a)
Determine the most positive roots of the system, w using Bisection Method. Use initial
value of [4+(0.1*A), 12+(B/2)]. Iterate until 5th iteration.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcbf6f585-4c53-46a7-aebb-a2e033dc9c74%2F56bac4f5-7e50-41b4-be5b-94c1d565a0c1%2Fjheb4k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1
The frequency equation of a 3 Degree of Freedom spring mass system (Figure Q1) is given
as:
m3w6 –
4km?w* + 3k²mw² = 0
where the value of the mass, m = 0.1 kg and spring stiffness coefficient, k = 10 N/m.
By performing your calculation (final answer in 3 decimal points):
(a)
Determine the most positive roots of the system, w using Bisection Method. Use initial
value of [4+(0.1*A), 12+(B/2)]. Iterate until 5th iteration.

Transcribed Image Text:X3
X2
k2
ki
m3
m2
m1
Figure Q1
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