EML 4225 Spring 2018 HW 5 PDF

Title EML 4225 Spring 2018 HW 5
Author JamieLeigh Wilkins
Course  Introduction to Vibrations and Controls
Institution University of Central Florida
Pages 3
File Size 160.9 KB
File Type PDF
Total Downloads 62
Total Views 139

Summary

Homework for this class...


Description

Homework #5

1. Find the natural frequencies and mode shapes: m 1 = 50 kg, m 2 = 200 kg, k = 1000 N/m, and l = 1 m. (Start with free-body diagram and equations of motion) Reference: Homework 1

2. A vibration system given by

2 0  x1   300 − 200  x1  0 0 1  x  + − 200 400   x  = 0   2    2    a) Find the natural frequencies. b) Find the mode shapes associated with the natural frequencies.

Homework #5

1. Find the natural frequencies: From Homework 1-3 solution,

ω 1 = 10 rad/s or ω2 = 40 rad/s For ω1 = 10 rad/s 1200  x   4000 − (250)(10)  =  1200 1360 − (40)(10) θ  

0   0

 x (1)   1  1500 x +1200 θ = 0 θ = -1.25x =>  (1)  =     1200x + 960θ = 0 θ  − 1.25 For ω2 = 40 rad/s 1200  4000 − (250)(40)   x  0   =   1200 1360 − (40)(40) θ  0   − 6000 x + 1200θ = 0 θ = 5x =>   1200 x − 240θ = 0

 x( 2 )  1  ( 2)  =   θ  5

2. A vibration system given by

2 0 x1   300 − 200  x 1  0   =      +  0 1 x2   − 200 400   x 2  0 a) Find the natural frequencies. b) Find the mode shapes associated with the natural frequencies. Let

2 s 2 + 300  x1   X1  st e =     =>   x2  X 2   − 200

− 200   X 1  0    =   s 2 + 400  X 2  0 

For non-trivial solution,

  2s 2 + 300 − 200   4 2 det    = 0 => 2s + 1100s + 120000 − 40000 = 0 => 2   − 200 s + 400   s 4 + 550 s 2 + 40000 = 0 => s 2 =

− 550 ± 550 2 − 4 * 40000 − 550 ± 377.49 = 2 2

s2 = -86.254 or -463.75 The natural frequencies are ω 1 = 9.287 and ω 2 = 21.535 rad/s For s2 = -86.254, ω1 = 9.287 rad/s (1)

 1  X 1   127.492X 1 − 200X 2 = 0 => X 2 = 0.637X 1 =>   =    0.637  X 2  − 200 X 1 + 313.746 X 2 = 0 For s2 = --463.75, ω2 = 21.535 rad/s

X 1  − 627.5 X 1 − 200 X 2 = 0 => X 2 = -3.14X 1 =>    X 2  − 200 X 1 − 63.75 X 2 = 0

( 2)

 1  =  − 3.14 ...


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