HW3Solutions_Winter2012

# HW3Solutions_Winter2012 - ENME361: Vibrations, Controls and...

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ENME361: Vibrations, Controls and Optimization I January 20, 2012 Winter 2012 Solutions for ENME361 HW 3 5.22 Determine an expression for the output of an accelerometer with the damping factor and natural frequency n , when it is mounted on a system executing periodic displacement motions of the form t A t A y 2 2 1 1 sin sin Solution 5.22 We use the material from Sections 5.6 and 5.9 to determine that the accelerometer output is given by 12 ( ) ( ) ( ) z t z t z t  where, from Eq. (5.98),     2 1 1 1 1 1 1 2 2 2 2 2 2 2 ( ) ( )sin ( ) ( ) ( )sin ( ) z t A H t z t A H t    and         2 22 1 2 1 ( ) 1,2 2 ( ) tan 1,2 1 i i n i n in i Hi i             6.1 Determine the response of a vibratory system governed by the following equation. ( ) 0.2 ( ) 3.5 ( ) 1.5sin( ) ( ) 4 ( 3) x t x t x t t u t t  Assume that m = 1 kg, the initial conditions are x (0) = 0.1 m and (0) 0 x m/s, and the excitation frequency = 1.4 rad/s. Plot the response. Solution 6.1 From the coefficients of the terms in the equation, we note that c = 0.2 N s/m, and k = 3.5 N/m. Then 3.5 1.87 rad/s 1 0.2 0.0535 2 2 1 1.87 n n c m 

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We shall break the solution into three parts by solving the responses to the initial conditions, the harmonic force input, and the delta function input separately and use the principle of superposition to determine the combined response of the system. The first part is the solution to an initial value problem with the initial velocity equal to zero. Then from Eq. (4.33) 2 ( ) sin( ) 1 n t o ic d X x t e t    where 22 2 2 11 0.1 0.100 1 1 0.0535 1 1 0.0535 tan tan 1.517 rad 0.0535 1 1.87 1 0.0535 1.868 rad/s 1.868 0.0535 0.1 o o dn n X A   Then the displacement due to the initial conditions is 0.1 0.1 2 0.1 ( ) sin(1.867 1.517) ( ) 0.1 sin(1.868 1.517) ( ) m 1 0.535 tt ic x t e t u t e t u t The second portion of the solution is obtained from the suddenly applied sinusoidal excitation. For this solution, we use Eqs. (5.7) and (5.8). Thus,     2 2 ( ) ( ) sin 1.4 ( ) sin 1 ( ) ( ) 1 n t o har n t Fe x t H t t u t k      where = / n = 1.4/1.87 = 0.749, F o / k = 1.5/3.5 = 0.429 m and,         2 2 2 2 2 2 ( ) 2.236 1 2 1 0.749 2 0.0535 0.749 2 2 0.0535 0.749 ( ) tan tan 0.1799 rad 1 1 0.749 21 2 0.0535 1 0.0535 ( ) tan tan 2.901 rad 2 (1 ) 2 0.0535 (1 0.749 ) t H               Then         0.1 2 0.1 0.749 ( ) 0.426 2.236 sin 1.4 0.1799 sin 1.868 2.901 ( ) 1 0.0535 0.953sin 1.4 0.1799 0.713 sin 1.868 2.901 ( ) m t har t e x t t t u t t e t u t
The third part of the solution is due to the delta function of magnitude 4 N s. For the delta function input, we use Eq. (6.1a) to determine the response and find that () 0 ( 3) 0.1( 3) 4 ( ) sin( [ ]) ( 3) 4 sin( [ 3]) ( 3) 2.141 sin(1.868[ 3]) ( 3) m n n t t delta d d t d d t x t e t d m e t u t m e t u t      The response of mass is ( ) ( ) ( ) ( ) ic har delta x t x t x t x t

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## This note was uploaded on 04/07/2012 for the course ENME 361 taught by Professor Yoo during the Winter '11 term at Maryland.

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HW3Solutions_Winter2012 - ENME361: Vibrations, Controls and...

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