Question

The figure shows the mass m at the end of a bar of length / is restrained by a spring and dashpot. The mass is initially at rest and vibrates in the vertical plane under the action of the force F(1). Determine the equation of motion, natural frequency, and damping ratio of the system when m = 45 kg, k = 9700 N/m, c = 950 N.s/m, a - 0.8 m, and I = 2 m. Neglect the mass of the bar.

Answer 1

page- loves diam [Solution → vivem that A bar of l Length hinged from one end with support and commect with spring and dashput and under gies vibration. TT lomass =m 45kg K= 97vo Nim damping ca-efficient=950 N-5lm ja=0.8m с 2 Negle eat the mass of the bar wee TO Jom F(A) F(t) final position initial position

page-① gf o angle them we displace by final positiom at 10 X f € l → FIA) mg Since, it i case of single degree of freedom with forced vibration. for very small, angelar displacement Restoring force spring force Fs = 147

Fd - Page ① (11) Damping force c Coi = velocity! distursting force F(t) Sy from Dralemobert's principal Restoring Torque = -IÖ [ Ô angular acceleration) sa [kn + ck - F(4)] -I I å tro + KR Flat) x1 U Ia moment of Imerdia of mass (m) about co (map) and for small ot x = loja e = (61a and che ha mig ö + c.loja + K (0)9 = F(H) x1 differential eavation of single degree of freedom with forced vibration

Page 4 So, equation in op motion in fmely ö + Calo + 119) = FULDA > O + lca + 119 mi? 0= [F(A)) m12 Sul compare with standard differenti al equation of single degree of freedom with damped forced Nibration ' Standard equation 18- 2 + [2&wm).6 +Wmo Fo gimwt) Sig for natural frequency (woon) w.ma = Rea 970xx org = 97088 45 X 212 m12 4524 : - Wm 970489 45XA

page-6 Wra 1940 45 6.565 → wms 6.56) sadis a G Angular natural frequency now frequency in (H2) PE >> Won 292 16:56) HZ 21 f = (1.04) H2 6 Liment natural frequency for damping Ratio (W from equation ① and ② > >> 2 & Won a mir ca m12 X1 2 Wm

Pag이 4 M g50x'o8x오 45tf x 2x 6.56 รุง 236106 00321 약 clamping ratro