317_pdfsam_math 54 differential equation solutions odd

# 317_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.5 where b is the damping constant. Notice that velocity of the arm of the dashpot is the diFerence between the velocities of mass m 2 and mass m 1 .Thema s s m 2 has a force F 3 acting on its left side due to the dashpot and a force F 4 acting on its right side due to the right spring. Using similar arguments, we ±nd F 3 = b ( y 0 x 0 )a n d F 4 = k 2 y. Applying Newton’s second law to each mass gives m 1 x 0 ( t )= F 1 + F 2 = k 1 x ( t )+ b [ y 0 ( t ) x 0 ( t )] , m 2 y 0 ( t )= F 3 + F 4 = b [ y 0 ( t ) x 0 ( t )] k 2 y. Plugging in the constants m 1 = m 2 =1, k 1 = k 2 =1 ,and b = 1, and simplifying yields x 0 ( t )+ x 0 ( t )+ x ( t ) y 0 ( t )=0 , x 0 ( t )+ y 0 ( t )+ y 0 ( t )+ y ( t )=0 . (5.34) The initial conditions for the system will be y (0) = 0 ( m 2 is held in its equilibrium position), x (0) = 2( m 1 is pushed to the left 2 ft), and x 0 (0) = y 0 (0) = 0 (the masses are simply released at time t = 0 with no additional velocity). In operator notation this system becomes ( D 2 + D +1)[ x ] D [
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Unformatted text preview: , − D [ x ] + y ( t ) + ( D 2 + D + 1) [ y ] = 0 . By multiplying the ±rst equation above by D and the second by ( D 2 + D + 1) and adding the resulting equations, we can eliminate the function y ( t ). Thus, we have n ( D 2 + D + 1 ) 2 − D 2 o [ x ] = 0 ⇒ ±²( D 2 + D + 1 ) − D ³ · ²( D 2 + D + 1 ) + D ³´ [ x ] = 0 ⇒ ±( D 2 + 1 ) ( D + 1) 2 ´ [ x ] = 0 . This last equation is a fourth order linear diFerential equation with constant coeﬃcients whose associated auxiliary equation has roots r = − 1, − 1, i , and − i . Therefore, the solution to this diFerential equation is x ( t ) = c 1 e − t + c 2 te − t + c 3 cos t + c 4 sin t 313...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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