homework 08 solutions - Section 3.3 12 30 38 42 48 r4—3r3...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern