Question

10 Ich *74) . CONSIDER THE FOLLOWING SYSTEM: 100 W/m 500 N/m. t>XH) mom 89 кл Н į X16)= om} 20 kg Is the 2* (o)= omiss (a) if f1t): = : 25 sin (2017) + 0.1 sin (20015+) N, . FIND THE STEADY STATE PART OF THE RESPONSE. (b) IF FCt) = 110N, FIND THE TRANSIENT PART OF THE RESPONSE. IF F(H) = 25 sin (w+), FIND THE VALUE OF W THAT RESULTS IN THE MAXIMUM STEADY STATE AMPLITUDE OF VIBRATION

Answer 1

62 L00N/m m woon im m SxLt). gave F (4) Clogs keg= connection) (Parallel K Kz KtK2 = 500xico 500 HD keq = 500 Nm wn = key = 500 g = 0.967 radise Given, F(X) = 25 sin twt). Steady State Amplitude of vibration is given by Asteady = to ]-> Constant entire V1-2)2723+)? for Asteady to be maximum, Denominator, (1-1324238) should be minised Differentiating wots and equate it to zero d (6-834+2389=0 26-12-28+2 (238) 723=0. -48(1-12) + 838 0 48(+1+827232) 20. Solms, L) 8.0 (Not possible). Lo st=1-232 » [= 11-233 Marming en de steen 5) condition for

We know c, damping coefficient is given by C=23 wnm 20-28 370.967 x 89. 12 = 0.11619) Substituting 3 in 8 8-11-232 n = 1- 2(0-11619) W = 0.9864 wn TW = 0.9538 rad /sec). at Steady stabe with Maximum Amplitude