homework 08 solutions - Section 3.3 12 30 38 42 48 r43r3...

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Unformatted text preview: Section 3.3 12. 30. 38. 42. 48. r4—3r3+3r2—r = r(r—1)3 = 0; r = 0,1,1,1; y(x)= 61+C2ex+03xex+C4x2ex First we spot the root r = — 1. Then long division of the polynomial r4—r3+r2—3r—6 by r + 1 yields the cubic factor r3 - 2r2 + 3r - 6. Next we spot the root r = 2, and another long division yields the quadratic factor r2 + 3 with roots r = ii J3 . Hence the general solution is y(x) = me“ + czez" + C3008 x J3 + C4sin x J3 . Given that r = 5 is one characteristic root, we divide (r — 5) into the characteristic polynomial r3 — 5r2 +100r — 500 and get the remaining factor r2 +100 . Thus the general solution is y(x) = Aes" +Bcolex+Csin10x_ Imposition of the given initial conditions yields the equations A+B = 0, 5A+10C =10, 25A—lOOB = 250 with solution A = 2, B = — 2, C = 0. Hence the desired particular solution is y(x) = 2e5" —200510x. (r2 + 4)3 r6 +12r4 + 48r2 + 64 , so the differential equation is y<6>+12y‘4’+48y"+64y = 0_ The general solution is y(x) = Ae" + Be” + Cefl" where a = (-l + iJ§ )l2 and ,8 = (—1 - 1' J3 )/ 2. Imposition of the given initial conditions yields the equations A+ B+ C =1 A+aB+ ,BC = 0 A+azB+fl2C = 0 that we solve for A = B = C = 1/3. Thus the desired particular solution is given by y(x) = JB-(e‘ + eHH‘EW2 + eH‘mm), which (using Euler's relation) reduces to the given real-valued solution. Section 3.4 3. The spring constant is k = 15 N/0.20 m = 75 N/m. The solution of 3x" + 75x = O with x(0) = 0 and x'(0) = —10 is x(t) = —2 sin 5t. Thus the amplitude is 2 m; the frequency is mo = \lk/m = \/75/3 = 5 rad/sec = 2.5/71' Hz ; and the period is 272/5 sec. (a) With m = 1/4 kg and k = (9 N)/(O.25 m) = 36 N/m we find that we = 12 rad/sec. The solution of x"+ 144x = 0 with x(0) = 1 and x’(0) = -5 is x(t) = cos 12t — (5/12)sin 12t = (13/12)[(12/13)cos 12: - (5/13)sin 12:] x(t) = (13/12)cos(12t— a) where a = 27r—tan'1(5/12) z 5.8884. (b) C = 13/12 z 1.0833m and T = 271/12 z 0.5236 sec. The gravitational acceleration at distance R from the center of the earth is g = GM/RZ. According to Equation (6) in the text the (circular) frequency a) of a pendulum is given by wz = g/L = GM/RZL, so its period is p = 272/0) = 272'RJL/GM. The period equation p = 3960 #10010 = (3960 + x) #100 yields x z 1.9795 mi z 10,450 ft for the altitude of the mountain. We must differentiate betWeen “true” time and “recorded” time. Let r be the ratio of true time to recorded time. This will depend on L and 9. Let Trec denote the amount of recorded time per cycle. This is a constant of the clock, independent of L and g. For instance, perhaps the clock is set up so that it records 3 seconds per cycle. Now note that T 2 Trecr. That is, the true time per cycle equals the recorded time per cycle multiplied by the ratio of true time to recorded time. Therefore, Trec'r = 271' g So7 if L1 and L2 are two pendulum lengths7 corresponding to true-to—recorded-time ratios T] and T2, then, using the fact that Trec and g are constants, we obtain 31_/£l 72‘— L2- In our problem here, let L2=30 in. Since a clock with this pendulum length “loses 10 minutes per day,” it has only recorded 23—;— h in a true 24 h period, so T _ 24 _ 86400 2 ' 233 _ 85800' If the clock keeps perfect time with pendulum length L1, then 7'1 = 1. Therefore, g “ 85800 V L2 ’ 86400’ so L1 = (80)(85800/86400)2 e 29.58 in. ...
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