2) The speeds of oscillation, x, of a loaded beam is given by the equation: x³ 3.25 x² + x -0.063 = 0 Determine the positive value (root) of .x which is between (2) and (3), by use bisection method. a) x₂ = 3.75 b) x₂ = 1.75 c)x₂ = 2.75
2) The speeds of oscillation, x, of a loaded beam is given by the equation: x³ 3.25 x² + x -0.063 = 0 Determine the positive value (root) of .x which is between (2) and (3), by use bisection method. a) x₂ = 3.75 b) x₂ = 1.75 c)x₂ = 2.75
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Q3// choose the correct answer :
1)compute the temperature distribution in a rod that heated at both ends ,as in fig. below,
use gauss-sidle method T₁+12T₁-T₁-1=0 where T, represent the temperature
To
200°C
the solution of system of linear equations using Jacobi method
TY
(T₁
19.75
19.75
48.25
151 75A
b) 7(2)
29-75
48-25
a) T(2)
(7.(2)
48-25
\T₂(²)
150-75/
151-75/
The speeds of oscillation, x, of a loaded beam is given by the equation:
³3.25x² + x -0.063 = 0
Determine the positive value (root) of .x which is between (2) and (3), by use bisection method.
a) x₂ = 3.75
b) x₂ = 1.75
c)x₂ = 2.75
2)
c) 7(²)
T(2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda09cea8-12c9-4878-b0cd-df3849ecce3c%2F11a8857f-0f9f-47a4-bdff-0c4060e38225%2F2ens4k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q3// choose the correct answer :
1)compute the temperature distribution in a rod that heated at both ends ,as in fig. below,
use gauss-sidle method T₁+12T₁-T₁-1=0 where T, represent the temperature
To
200°C
the solution of system of linear equations using Jacobi method
TY
(T₁
19.75
19.75
48.25
151 75A
b) 7(2)
29-75
48-25
a) T(2)
(7.(2)
48-25
\T₂(²)
150-75/
151-75/
The speeds of oscillation, x, of a loaded beam is given by the equation:
³3.25x² + x -0.063 = 0
Determine the positive value (root) of .x which is between (2) and (3), by use bisection method.
a) x₂ = 3.75
b) x₂ = 1.75
c)x₂ = 2.75
2)
c) 7(²)
T(2)
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