Question

2. A spring is stretched 10 cm by a force of 3 newtons. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 newtons when the velocity of the mass is 5 m/sec. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/sec, determine its position u at time t. Find the quasi frequency and the ratio of je to the natural frequency of the corresponding undamped motion.

Answer 1

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R$om? Giren data Spring stretched 10cm Force (F)- 3N Mass (m) = akg velocilj v= 5mls The car of mass & spring is mult du't bu flt) m man & camping const Kt Spring Const Now u' 0.6 N- 5 & Spring cont Fr=kL Le elongation of Spring L=10 cm tere Now k=fs L 31 olm 30kglam Now initially F(E)=0 car o becomes Now au" + 0-64' +30U 20

The auxiliary eam 2m² +0.6m +30 =0 -0.6 I Ju-4(30)(2) 212) -0-15 +3-87 i -0.15- 3.87 i is Now the Generation -oist -Đ/ct Stn (387) U= Ge Cos (3-87)t + Ce Now with initial conditious Ulo) = 50m (given) U10) = 0-05m) { u'(o) =10cmls ulo) = Olmad (gives) a = 0.05 G=1 -oist -0.15t Now Sm (3.87t) U = 0.05e Cos (3.87t) + ole 0.15t 7)] u=oosté Cos (3.847 -0.507 & Hore 'M= 3.87 rad/s. M 0.999 m 3.87 ✓ 3012

Answer 2

The value for C2 is incorrect.