242_pdfsam_math 54 differential equation solutions odd

242_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 The maximum displacement of the mass is found by determining the Frst time the velocity of the mass becomes zero. Therefore, we have y 0 ( t )=0= ± 2+0 . 2 5 ² e 2 5 t 2 5 h 0 . 1+ ± . 2 5 ² t i e 2 5 t , which gives t = 2 2 5(2 + 0 . 2 5) = 1 5(2 + 0 . 2 5) . Thus the maximum displacement is y ³ 1 5(2 + 0 . 2 5) ´ = ³ 0 . ± . 2 5 ² µ 1 5(2 + 0 . 2 5) ¶´ e 2 5 / [ 5(2+0 . 2 5)] 0 . 242 (m) . 11. The equation of the motion of this mass-spring system is y 0 +0 . 2 y 0 + 100 y =0 ,y (0) = 0 0 (0) = 1 . Clearly, this is an underdamped motion because b 2 4 mk =(0 . 2) 2 4(1)(100) = 399 . 96 < 0 . So, we use use equation (16) on page 213 in the text for a general solution. With α = b 2 m = 0 . 2 2 = 0 . 1a n d β = 1 2 m 4 mk
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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 University of California, Berkeley.

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