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Unformatted text preview: Assigned on 19.10.2009 Monday Due date 26.10.2009 Monday, beginning of the lecture ME429 Fall 2009 HW3 - SOLUTION 1) According to given initial conditions, the system starts with zero velocity and and a nonzero displacement x o . Since the initial displacement is positive, the motion of the mass starts from left to right, which is in negative x direction. Response of the mass can be readily written as: ( ) 1 2 cos sin n n N x t C t C t k μ ω ω = + + where n k m ω = and N mg = With the initial conditions applied, it can be found that 1 o N C x k μ =- and 2 C = . Then the response equation becomes ( ) cos o n N N x t x t k k μ μ ω =- + However this solution is valid only for the first half cycle of the system only, that is; n t π ω ≤ ≤ . When n t π ω = , the mass will be at its extreme left position, which can be found as: 2 cos o o n N N N x x x k k k π μ μ μ π ω =- + = -- Since the motion started with a displacement of x o and in the end of a half cycle time the displacement value became ( ) 2 o x N k μ-- , the reduction in magnitude of x in time n π ω is 2 N k μ . In the second half cycle, the mass moves from left to right, so the response equation to be used is ( ) 1 2 cos sin n n N x t C t C t k μ ω ω = +- which is valid for 2 n n t π ω π ω ≤ ≤ . For this equation, initial conditions to be applied are the ....
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This note was uploaded on 03/28/2010 for the course AEROSPACE aero520 taught by Professor Mok during the Spring '10 term at Istanbul Technical University.
- Spring '10