ASE 365 - Lecture 16

# ASE 365 - Lecture 16 - Recap x x = K M vibration free...

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Recap x x = + K M : vibration free undamped for EOMs ). ( ) ( t f t u x = solution s" synchronou " for Looking other. each with consistent be must ODEs These function scalar the for ODEs of a to Leads ). ( t f one set . let if only energy in ve conservati be can solution whose , form the has one Each 2 ϖ = → ≥ = + C C Cf f . allow to need we which in , equations the give EOMs Then u u ≠ =- =- ) det( ) ( 2 2 M K M K ϖ ϖ Example k m m k x2 x 1 k = -- + --- =- +- = 2 2 ) ( ), ( 2 1 2 1 2 1 2 2 1 2 1 1 x x k k k k x x m m kx x x k x m x x k kx x m : EOMs = +- + =- + ⇒ = ) 2 ( ) 2 ( ) ( ) ( ) ( 2 1 2 2 1 1 2 1 2 1 f ku ku f mu f ku ku f mu t f u u t x t x , If . ) ( 2 2 . 2 2 ) 2 ( ) 2 ( 2 1 2 2 2 2 1 2 1 2 1 2 1 2 2 1 1 = -- + -- = +- = =- = +- + =- + f u u k k k k m m f f mu ku ku mu ku ku f ku ku f mu f ku ku f mu ϖ ϖ ϖ give EOMs and Then need we ODEs two the For ) is which , or , written be can t requiremen cy (Consisten . 2 2 2 2 2 2 1 2 2 1 2 2 2 1 1 2 2 1 u u M K u u m m u u k k k k mu ku ku mu ku ku ϖ ϖ ϖ ϖ = = -- = +- =- ) 3 )( ( 3 4 ) ( ) 2 ( 2 2 2 2 : ) det( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 =-- = +- =-- =---- = - -- =- ≠ k m k m k km m k m k m k k k m k m m k k k k M K ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ need we , For u and roots two the has and in quadratic is equation The . 3 ) det( 2 2 2 1 2 2 m k m k M K = = =- ϖ ϖ ϖ ϖ . ) ( 2 u =- M K ϖ of solution nontrivial a is there these of each For . 1 1 2 2 1 2 1 2 1 2 1 2 1 = = -- = - -- = u u u u k k k k u u m m m k k k k k m k solution nontrivial the has equation the , For ϖ . 1 1 3 2 2 3 2 2 1 2 1 2 1 2 2 - = = ---- = - -- = u u u u k k k k u u m m m k k k k k m k solution nontrivial the has equation the , For ϖ . ing correspond the describing nts displaceme of vector a is there one, each For occur. can vibration free us) (synchrono which at 2 are there system, 2DOF this For : tion interpreta Physical vibration of mode natural s frequencie natural ....
View Full Document

## This note was uploaded on 09/18/2011 for the course ASE 365 taught by Professor Staff during the Spring '10 term at University of Texas.

### Page1 / 30

ASE 365 - Lecture 16 - Recap x x = K M vibration free...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online