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Unformatted text preview: Problems and Solutions Section 1.6 (1.75 through 1.81) 1.75 Show that the logarithmic decrement is equal to ! = 1 n ln x x n where x n is the amplitude of vibration after n cycles have elapsed. Solution: ln x t ( ) x t + nT ( ) ! " # # $ % & & = ln Ae ’ () n t sin ) d t + * ( ) Ae ’ () n t + nt ( ) sin ) d t + ) d nT + * ( ) ! " # # $ % & & (1) Since n ! d T = n 2 " ( ) , sin ! d t + n ! d T + # ( ) = sin ! d t + # ( ) Hence, Eq. (1) becomes ln Ae ! "# n t sin # d t + $ ( ) Ae ! "# n t + nT ( ) e ! "# n nt sin # d t + # d nt + $ ( ) % & ’ ’ ( ) * * = ln e "# n nT ( ) = n "# n T Since ln x t ( ) x t + T ( ) ! " # # $ % & & = ’( n T = ) , Then ln x t ( ) x t + nT ( ) ! " # # $ % & & = n ’ Therefore, ! = 1 n ln x o x n " original amplitude " amplitude n cycles later Here x = x (0). 1.76 Derive the equation (1.70) for the trifalar suspension system. Solution: Using the notation given for Figure 1.29, and the following geometry: r !...
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 J.G.Lee

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