Ch 2-1 to 2-3

Ch 2-1 to 2-3 - Survey Result: 1 Group 1 More examples in...

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Survey Result: , FRXOG OHDUQ EHWWHU LI «± 1
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2 More examples in class (similar to homework assignments) Website with course information and notes Slow down Practice exam problems before exams Notes are clearer Group 1
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3 Explain mathematical derivation/terms in equations more clearly Information better organized More demos of vibration Spend more time for example problems More applications of vibrations (real world) Speak clearly Snacks Explain better how equation used better More office hours 5 minutes break Too long (3X a week) Big class size Show more visual demonstration/computer models Text book is too dense/Slow read Review sessions before exams More feedback on homework Group 2
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4 , GRQ¶W OLNH VSULQJV Harder examples 'RQ¶W OHW PH SURFUDVWLQDWH New materials are not given at the end of class Better prepared homework solutions Solutions to homework problems are listed Move this class to Tues/Thurs Less people in the room/More sessions Hmw assigned too early Email solutions to homework problems Some derivations are too slow Better Greek letter formation Step by step lectures 'RQ¶W VWDQG EHIRUH WKH ERDUG Notes are not directly from the book More visuals Go over the homework problems Tests are not too difficult Class is interesting , GLGQ¶W WDNH RWKHU FRXUVHV ZLWK WKLV More concepts are explained Speak louder 2 min break at the 50 min. mark Assign reading materials Announce due dates in class Cleaner projector/ Projector in the way Convert overheads to PowerPoint Group 3
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Solutions: Homework 01 5
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Chapter 2 Response to Harmonic Excitation 14
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Harmonic Excitation of Undamped Systems M k x Displacement F=F 0 c o s Z t Consider the spring-mass damper system with the following applied (external) force: Ȧ is the driving frequency F 0 is the magnitude of the applied force We take c = 0 to start with. A Thought Experiment () cos( ) F 15
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Equations of Motion ! " ! # " $ #$%! Z # " 0 2 0 0 0 ( ) ( ) cos( ) ( ) ( ) cos( ) where , n m x t k F f ZZ ± ² ² Solution is the sum of homogenous and particular solution. The particular solution assumes form of the forcing function: F.B.D. 16
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Substitute Particular Solution into the Equation of Motion: 2 22 0 0 cos cos cos solving for yields: p n x X t f Z ZZ ±² ± Thus the particular solution has the following form: ! " ! # " % & & ' ±Z ' #$%! # " 17
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Add particular and homogeneous solutions to get general solution: 0 12 22 homogeneous ( ) sin cos cos par t i c ular n f x A ZZ Z ± ± ² and are constants of integration. These constants will be determined from the I.C. Remarks: 18
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