Question

4. (30%) Consider the following system that consists of a mass m-10kg, coil spring of stiffness k-1000N/m, and damping c-200Ns/m. 1) Suppose that the mass is initially at rest and is given an initial velocity of 3 m/s Find the free vibration response of the mass. 2) Suppose that at a later time, a harmonic force F (t)- sin15t is acted on the mass. Determine the amplitude of the forced vibration response. F, sin

Answer 1

1) Here consider As per given | r = 46 {cx = .| N w։ 185) | c - \oo A-5 – c = (ooo 9 / «... k Հ ԷԿ- 100 - 3/> "ԱՍՆԱՆՆՆՆՆՆՆԷ Here at t=0 , x=0 Ջ x = 3 *) s wna Natural treatency WH : Յ.2 Jeesec, Now — 6. 16 case is շVk.5 2 (ope. . As & <1, system is inderdamped So Necessary equation of Motion for this xa me whst sin (wat + 4) Wo=(Taqa Jon | - 3.2 (Հ. — (8.1.3 ) : 5.4 6 217 (se , - x= Me sin (2.764 + 4) | —o, so ž= -0.5 me - , < c .*Օ.Տ - si) (24 6 : e ces (2-3 c+ + +)12.6

so at t=0 x = 3mis I to X=0 o m sinep : 4 = 0 t=0 x = 3mis & 3 = o+ M os u 2076 - So 3 = H (1) 2.76 = 3 = 1.09 2.76 a free vibration Response will be Sar1.09 -0.5t sin (2.967 * 18o jº ansot 1, 12] Here function Frt) = sin 15t acred Here Harmonic the mass on so Solution will be for forced vibration) x= c + xp ving with the function. moc +coct ko = Fosin net = 15 radl sec. te= Me - whit sin (wait to) Xo=xsin (wero) x = 1.09 e os sin (2.768 x 180) (from above solution). so

For xp ve neauine x Vcrumuger course MULLI V (1000-98.1 (158 72 + (100x15) V 2,24x108 + 0.023 x 106 4.463 x 108 2.11 x 104 x = 4,74810-3 Xp = 4,24x103 sin (154 x 180) So Total forced vibration response will be x= 1.00 e os sin (2.96 dx120go + 2,748103 sin (154 x 148 Ans. 2