# 285hw6 - HW 61 15 1. Sec.3.4:3. We have 15 = k 0:2 ) k =...

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HW 6 1 1. Sec.3.4:3. We have 15 = k & 0 : 2 ) k = 15 0 : 2 = 75 . x 0 = 0 and v 0 = ± 10 m= sec . Then we have the following problem: x 00 + ! 2 0 x = 0 ; ! 0 = r k m = r 75 3 = 5; x 0 = 0 ; v 0 = ± 10 . Then x ( t ) = A cos 5 t + B sin 5 t . x (0) = 0 = A . Hence x ( t ) = B sin 5 t . x 0 (0) = ± 10 ) ± 10 = 5 B ) B = ± 2 . Hence x ( t ) = ± 2 sin !t = 2 sin ( !t ± ) . Amplitude = 2 , period = 2 ! 0 = 2 5 sec; frequency = 5 rad= sec , or 1 perod = 5 2 = : 796 hz: 2. Sec.3.4:7. We are going to use the formula p 1 p 2 = R 1 p L 1 R 2 p L 2 : We have 1 = 3960 p 100 : 1 (3960 + r ) p 100 ; hence r = 1 : 9795 mi or r = 10 ; 450 ft. F = ma gives 2 hg ± 2 xg = 2 hx 00 ; where x = x ( t ) is the depth of the bottom of the buoy beneath the surface at time t . After simpli±cation we get x 00 + g ±h x = g . Besides we have initial conditions x (0) = x 0 (0) = 0 . Let us consider homogeneous 1 Note to the grader: please grade the following problems: Sec.3.4:7, Sec.3.4:10, Sec. 3.5: 17, Sec. 3.5: 49, Sec. 3.5: 55 (20pts for each problem). 1

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equation x 00 + g x = 0 ; then the characteristic equation is r 2 + g = 0 = ) r = r g i: We shall look for a particular solution of the non-homogeneous equation in the form x p = c since right-hand side is constant. Thus 0 + g c = g = ) c = The general solution of the given non-homogeneous equation will be x ( t ) = + c 1 cos r g t + c 2 sin r g t ; so we have x 0 ( t ) = ± c 1 r g sin r g t + c 2 r g cos r g t: Now we apply the initial conditions x (0) = + c 1 = 0 = ) c 1 = ± &h; x 0 (0) = c 2 r g = 0 = ) c 2 = 0 : Thus x ( t ) = ± cos q g t; that is x ( t ) = (1 ± cos ! 0 t ) ; where ! 0 = r g : Thus, the buoy undergoes simple harmonic motion about the equilib- rium position x e = with period p = 2 ± ! 0 = 2 ± s g : If = 0 : 5 g cm 3 ; h = 200 cm ; g = 980 cm s 3 then the amplitude of oscillation is = 100 cm and the period isgiven by p = 2 ± s g = 2 ± r 0 : 5 ² 200 980 ³ 2 : 01 s: 2
m = 1 2 ; c = 3 ; k = 4 ; x 0 = 2 ; v 0 = 0 . 1 2 x 00 + 3 x 0 + 4 x = 0 ; x (0) = 2 ; x 0 (0) = 0 : We have c 2 4 km = 9 8 > 0 , and this is an overdamped motion . 1 2 r 2 + 3 r + 4 = 0 ) r 2 + 6 r + 8 = 0 ) r = 2 ; r = 4 : General solution: x ( t ) = C 1 e 2 t + C 2 e 4 t . Using initial conditions, C 1 + C 2 = 2 2 C 1 4 C 2 = 0 ) C 1 = 4 ; C 2 = 2 : Answer: x ( t ) = 4 e 2 t 2 e 4 t 4 e 2 t 2 e 4 t 0 1 2 3 4 5 6 7 8 9 10 0.0 0.5 1.0 1.5 2.0 x y 5. Sec.3.4:17. Because c 2 = 4 km (8 2 = 4 ± 16 ± 1 = 64) we have critically damped case. x

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## This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285hw6 - HW 61 15 1. Sec.3.4:3. We have 15 = k 0:2 ) k =...

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