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B-9-15 Hint: Transform into modal coordinates to find the solution in modal coordinates then transform the modal coordinates
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Solution

* ky-x) kily-x) Kqx Using Newtons law, kily-1)-ky1=Mü Rearranging Më t(kitka)x - kıy=0

my tky - k,x=0 In matrix form, 13:8) O Solution will be of the form, x= X sinut y=Ysinot

To kitka -k, X sin wt + sin wt = 1 0:33- [J-1 po -1-- (kitka) - Mart ki i-mw2

110-10 w -10 -10 10 - 2 (110-100²) (10-0²) - 100=0 low-210w² + 2010=0 Take w²X 102? -2101 +1000 = 0 1,37-2984 do=13.7016 Cor

w;F 2.7016 radls Wa= 3.7016 rads From 0, U2 TID - low ŷ (kitky)-Mw? = 110-1002 For w=wi - 10 = 0.2702 Yh10-1067-2986) For w=

-2702 Therefore mode shapes will be, (3): (Dame) and (3). = (10570) General solution is, { poss} = A(7), son curt + 6) +B(3).

Initial conditions x(0) = 0.05m 90 = om are: ż(0) = 0 m/s (o)= 0 m/s 0.05 = A(0.2702) sin 0,- BC0.3702) sin da – o. Asin , +

Acos 0,00 :A0 cosp=0 p= 7/2 @ +(0-3702)3 golves 0.05= (0-2702 +0.3762)A sner A=0.0781 m : 0-602702) *© gives,

О= -(3-700x0.3764 +3-7016 xО-272)&cos ф, всю фаво : Bto Costa = 0 2- (0-2762) 3 gies D. DS= - (o:3762 +0-2њ2) 8 sn pa В2 -0-

.. Ut) = 0.0211 sin(d.7016 t+1/2) + 0.0289 sin (3.70166 +T%) y(t) = 0.0781 sin(2.7016 + +1/2) - 0.0781 sin (3-70166 +Tya) me

This is the solution of the system for given initial conditions.

Plot of the solution

x,y (m) -0.15 3 3 4 6 7 8 5 Time (s) 9 10

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