# SolSec1_6 - Problems and Solutions Section 1.6(1.75 through...

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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 0 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 0 = x (0).

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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 ! r ! " l r ! l h Write the kinetic and potential energy to obtain the frequency: Kinetic energy: T max = 1 2 I o ˙ ! 2 + 1 2 I ˙ ! 2 From geometry, ! r x = and ˙ x = r ˙ ! T max = 1 2 I o + I ( ) ˙ x 2 r 2 Potential Energy: U max = m o + m ( ) g l ! l cos " ( ) Two term Taylor Series Expansion of cos φ ! 1 " # 2 2 : U max = m o + m ( ) gl ! 2 2 " # \$ % & For geometry, sin l r !
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