Physical examples of eigen values and eigen vector and application in engineering

1. A Physical Example of
Eigen values & Eigen
vector with Application
In Engineering
Mr. Roshan Bhagat
Assistant Professor
PhD. Pursuing : Mechanical Engineering
Master of Technology : Thermal Engineering
Bachelor of Technology : Mechanical Engineering
roshan.bhagat25@gmail.com

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
Engineering
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Let us consider two masses 𝑚1 and 𝑚2 attached to the spring
𝑘1and 𝑘2 and 𝑥1 and 𝑥2 vary as function of time
Let 𝑥1 > 0 and 𝑥2 > 𝑥1 , 𝑎𝑛𝑑 𝑘1 = 𝑘2
𝑓 = 𝑚1 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
−𝑘 𝑥1 + 𝑘 𝑥2 − 𝑥1 = 𝑚1 ×
𝑑𝑣
𝑑𝑡
−𝑘 𝑥1 + 𝑘 𝑥2 − 𝑥1 = 𝑚1 ×
𝑑2
𝑥1
𝑑𝑡2
−𝑘 𝑥1 + 𝑘 𝑥2 − 𝑥1 − 𝑚1 ×
𝑑2 𝑥1
𝑑𝑡2 = 0
𝑚1 ×
𝑑2
𝑥
𝑑𝑡2 + 𝑘 𝑥1 − 𝑘 𝑥2 − 𝑥1 = 0 … … … (1)

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
Engineering
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Now consider mass 𝑚2 and using the below equation
𝑓 = 𝑚2 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
−𝑘 𝑥2 − 𝑥1 = 𝑚2 ×
𝑑2
𝑥2
𝑑𝑡2
−𝑘 𝑥2 − 𝑥1 − 𝑚2 ×
𝑑2 𝑥2
𝑑𝑡2 = 0
𝑚2 ×
𝑑2
𝑥2
𝑑𝑡2 + 𝑘 𝑥2 − 𝑥1 = 0 … … … (2)

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
Engineering
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Now let us assume the value of 𝑚1 = 10 , 𝑚2 = 20 𝑎𝑛𝑑 𝑘 =
15 𝑎𝑛𝑑 𝑢𝑠𝑖𝑛𝑔 𝑖𝑛 1 & (2)
10 ×
𝑑2 𝑥1
𝑑𝑡2 + 15 𝑥1 − 15 𝑥2 − 𝑥1 = 0 … … … (3)
20 ×
𝑑2
𝑥2
𝑑𝑡2
+ 15 𝑥2 − 𝑥1 = 0 … … … (4)
From vibration theory
𝑥𝑖 = 𝐴𝑖 sin(𝜔𝑡 − ∅)
𝑤ℎ𝑒𝑟𝑒, 𝑖 = 1,2,3 … … .
𝐴𝑖 = 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑚𝑎𝑠𝑠
𝜔 = 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛
∅ = 𝑃ℎ𝑎𝑠𝑒 𝑠ℎ𝑖𝑓𝑡

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
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𝑥1 = 𝐴1 sin(𝜔𝑡 − ∅)
𝑥2 = 𝐴2 sin(𝜔𝑡 − ∅)
Using,
𝑥𝑖 = 𝐴𝑖 sin(𝜔𝑡 − ∅)
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑑𝑥𝑖
𝑑𝑡
= 𝐴𝑖 cos(𝜔𝑡 − ∅) × 𝜔
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
𝑑2
𝑥𝑖
𝑑𝑡2 = −𝐴𝑖 sin 𝜔𝑡 − ∅ × 𝜔2 … … . (5)

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
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𝑓𝑜𝑟, 𝑖 = 1,2
𝑑2 𝑥1
𝑑𝑡2 = −𝐴1 sin 𝜔𝑡 − ∅ × 𝜔2 … … . (6)
𝑑2
𝑥2
𝑑𝑡2
= −𝐴2 sin 𝜔𝑡 − ∅ × 𝜔2 … … . (7)
𝑝𝑢𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
𝑑2
𝑥1
𝑑𝑡2 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (3)
10 ×
𝑑2
𝑥1
𝑑𝑡2 + 15 𝑥1 − 15 𝑥2 − 𝑥1 = 0 … … … (3)
10 × −𝐴1 sin 𝜔𝑡 − ∅ × 𝜔2
+ 15 𝐴1 sin(𝜔𝑡 − ∅) − 15 𝐴2 sin 𝜔𝑡 − ∅ − 𝐴1 sin 𝜔𝑡 − ∅ = 0 . . (8)

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
Engineering
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𝑝𝑢𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
𝑑2
𝑥2
𝑑𝑡2
𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (4)
20 ×
𝑑2
𝑥2
𝑑𝑡2 + 15 𝑥2 − 𝑥1 = 0 … … … (4)
20 × −𝐴2 sin 𝜔𝑡 − ∅ × 𝜔2 + 15 𝐴2 sin(𝜔𝑡 − ∅) − 𝐴1 sin(𝜔𝑡 − ∅) = 0 … … . (9)
𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 8 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
−10𝐴1 𝜔2
+ 15 𝐴1 − 15 𝐴2 − 𝐴1 = 0
−10𝐴1 𝜔2
+ 15 𝐴1 − 15𝐴2 + 15𝐴1 = 0
−10𝐴1 𝜔2 + 30 𝐴1 − 15𝐴2 = 0 … … … . (10)
𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 9 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
20 × −𝐴2 𝜔2
+ 15 𝐴2 − 𝐴1 = 0
−20𝐴2 𝜔2 + 15 𝐴2 − 15𝐴1 = 0 … … … (11)

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Roshan D Bhagat ; Applied Mathematics ; Physical Example of Application of Eigen Values & Eigen Vector in
Engineering
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𝑓𝑟𝑜𝑚 10 𝑎𝑛𝑑 11 𝑤𝑒 𝑔𝑒𝑡
(−10𝜔2
+ 30)𝐴1−15𝐴2 = 0 … … … . (12)
−15𝐴1 + (−20𝜔2 + 15) 𝐴2 = 0 … … … (13)
𝑈𝑠𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 12 𝑎𝑛𝑑 13 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 𝑚𝑎𝑡𝑟𝑖𝑥 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡
−10𝜔2
+ 30 −15
−15 −20𝜔2 + 15
𝐴1
𝐴2
=
0
0
𝑜𝑟
−𝜔2
+ 3 −1.5
−0.75 −𝜔2
+ 0.75
𝐴1
𝐴2
=
0
0
𝑜𝑟
3 −1.5
−0.75 0.75
𝐴1
𝐴2
− 𝜔2 𝐴1
𝐴2
=
0
0
𝐴 𝑋 − 𝜆 𝑋 = 0
𝐴 𝑋 = 𝜆 𝑋
𝑤ℎ𝑒𝑟𝑒 𝜆 𝑖𝑠 𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒 𝑎𝑛𝑑 𝑋 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟