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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 1 2 ( ) ( ) ( ) 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 2 2 1 2 1 ( ) 1,2 1 2 2 ( ) tan 1,2 1 i i n i n i n i i n H i 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 2 2 2 2 1 1 2 2 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 d n 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 t t 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 F e 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 1 1 2 2 2 2 1 1 2 2 2 2 1 1 ( ) 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 2 1 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
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HW3Solutions_Winter2012 - ENME361 Vibrations Controls and...

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