Question

PeoB.# 3(25%) 5k Ws 36 2c lokg Kadooo fun c-400 N-s/ F(t)= 300 Sin (200) r=o.im EVALUATE THE STEADY Bu FCE STATE RESPONSE Oss(t) w 32k

Answer 1

Answer:

-> Free body diagaram of mass :- 1 FEU) m T ake Equation of motion, mãe + Cóé +2km 7 T1 = F ( t) me tere taketti = 300 sin(205) Here, se is the Srorall displacemend, » Freebody diagram of disk. .5k22 F 202, K R = 2e = displacement by mags ů =r sé = ro 12 3 3 re and law Id =) Apply Newton's That Tir -(2005è, xr + SKX, *3r] > IL Tire (2cxró*r+Bkx 310 x 3r] - P& here, I = + MR? = 2 x 3m x Brj²

I = 27 mr Tir-[2Cr20+45 Kro] – 27 mrö Tir= 2crotu5kr 207+ 27 mr²8 T = [2$ me + 2c6+HS ke] substitute T, value in ean@ mé+cé +2kX+[27 mö +260 +m5kO] = 3oosin(205) mný tero+2kY&+(27 mö+2có ths koj= 300 sin(207) Y[(m+8+30+47KO] - 300 sinkot) 1 +47)x10kö + 3X4006 +47 XM100 00] =300 sinmot) 14.50 +120 0 +18,8000 = 300 sin (20+) By referring this ean steady state responsecan be found, mo+Co+ko ad@sinot) Steady state response is, xCt) = xo sin Cwtto) oct) = sin(wt+) here, Fo=300 k = tk 1 18,800 -Wn m 18,800 14.5 wn = 36 radis