Question

Mechanical vibration subject

4. Develop equation Xk M. 482 + p2 482(r2 + ur2 - 1)2 + (r2 - 1)2r2 = f(t,5) from Equation Trial 2019/2020 X2 (ka - m.w2)2 + wacả F3" [(k - mw2)(ka - mqw2) - makaw2]2 + [k - (m + ma)w2]2cw2 and develop equation Sop 1 2(4 + 1)(u + 2) for the optimal damping ratio.

Answer 1

Xk (207) +(r2 - ) (257) (-1+4+*)+[pr*f-(2-1)(-2-1)] 0 To derive Equation (5.-, which is the same as Equation note that c=k, = 0, = 0, which also means ß = 0. Since this is a moment equation, Fo is replaced by Mo. Therefore,

Xk 462 +r (257) +r? F. V(255)” (p2 – 1+ per?)* +(x2-1) 145*(n+her? - 1)* +(92-1)ºr? which is Equation (' ') after canceling r?. To derive Equation( '9), first let Equation ( ) be fir,5). Since ſ = 41/2 where B1/2' A= 452 +r? and B= 45*(x2 + por? - 1) +(n2-11 72, the + of ac 0 becomes BdA = AdB where dA 85 да as aB as dB = +85[r? + - 1) So, ***(*+pp+ -1)+(2-1) 485)={^4 +r"}(83Xrº +40° -1° (-2-1) = (y2 + for? - 1) (-2-1)= +(r? +4r? - 1) Taking the minus sign (the plus sign yields r = 0) (2+4)r? -2=0 2 V(2+) af Now take the other partial derivative = 0, which becomes де

BdA= AdB A dA = = 2r dr aB dB = 1653r(1+)(-+ + por? – 1)+4r°(92-1)+2r(-2-1) ar So, 2 {45+(x2+ -1) +(x2 - 1)*r*}{2r) ={«5+r"}(163°r(1+x)(x2 +40-1)+4r”(? - 1)+2r(p2–1] 2 Substituting r= yields, after rearranging (2+1) 2) + 2+ H ac[zintemente I[22--|(27 1)-(4x2160) 2 2μ 85*(1-4) 27 H+ 2+ 462 + 2+ Expanding and canceling terms yields 2 45*(1+ +x)(2 (2+1)+25°, FO 2+ u The physical solution for Ğ is S = V2(1+)(2+4)