Question

The following system is composed by two masses . The first mass my - 20 kg, moving horizontally on positive rightwards) . The second mass m2 - 2.5 kg, moving horizontally xx. positive rightwards) The first mass is connected to the ground on the left) by two springs, each with stiffness k-232 Nm. The second mass is connected to the first mass by another spring, also with stiffness -232 N/m. A harmonic force is applied to the first mass FIO)Foconut), with frequency w = 7.6 radis and amplitude Fo. 48 N. w a) Find the steady-state amplitude of motion (particular solution) for mass 1 and mass 2 (no sign is required, only amplitude as absolute value) • Steady-state amplitude of xy in metres particular solution, amplitude with no sign metres • Steady-state amplitude of xz in metres particular solution, amplitude with no sign) metres Check b) Continuing from the previous question) Assume you can change the value of the mass me, keeping all other parameters in the system unmodified Find the new value of mass my in kg resulting in a null steady-state motion of the first mass (particular solution kg Check

Answer 1

The response for the green harmonic response is given by Xi : ACOS 7.66 + BSIn 7.66 (w) cw) į; : -Aw'ccruet - Bu? sincet 1, : c coswet + o sincet ta : -cw coscoot - DW'sincet substitute 1, 4 xy tn eq @ € 6 - Smco? ( Acosust + B Sincest) + 3 KCACOSWA +Bsincet) K cccoscoot tosincet) : 48 coswet Allow 5 8mw?+3K) A - CK]-coswet 86-8mw*+3K)B-KD) sincet co-efficient of sincet Ę coscer By comparing (-8 mw+3K) A -CK = 48 --- --lii) (-8 mus +3K) B - DK ♡ -wm [ecorwt + sinwet) -K(Acost + osincet) + KC Coswety + Dsm wer) -- (ii) (-mwit kolc -KA live) o (-meu?+ k) 0 - KB equal to zero are eq Sinic (iilę live B:0, 0:0

ma (-8mwt3K) A -CK (-mwa'tk) C-AK As (-mu'tk), - k substitute ③ in ① -smw'+3 k)(-mu'tk)_ *7c: 48 K C 48K (-$mw +3K)(-muk)- 12 A = fmw'tk) 48K Esmuo** 3k) Amerika... Read 48 (-mwita ) (-8 mw+3k ) (-mwi+k) * *

COO mo :3.5 kq, W: 7.6 radlice, k = 232 mm 401 -2.5 x 7.62+232) A: (-8X9.5 x 7.6243x232) (-9.5x7.62239)-9322 4204.8 .94049.92 A. -0.04491 :X,649:-0.0 447 ) Cos 1.67 48 * 232 (-8X9.5x 7.62+3x232) (-2.5x7.62+232) – 2322 C: -0.11841 X₂ Ctl. = -0.11841 COS 7.66 0.04471m steady state amplitude of mass1 steady state amplitude of mass 2 232 N/m 75 = 0 case cu m, W: 7.6 rad /sec, x, tto) A coswet .0 .: A = 0 o 46-mwak) (-8mw? 3 k) (-mw tk/k2 0 (-mwi+K) na K ميا ma : m: 4.0166 kg 4.0166 KO The new mass