Question

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Suppose two undamped masses are connected to each other, and to the opposite walls, by springs as shown. 20 kg/s 4 kg/s2 30 kg/sa www.1 kg wm 2 kg Find the angular frequencies w1 and W2 of the solutions. (You do not need to find the solution itself.)

Answer 1

eqь position ki - 20klѕ ka = 443 /s? — m₂ = м, K, °304/ 8 arral excited state at timet due to longitudinal → a oscillation ki +ъ. most у му , 2, mestoving force equm of motion -kis miä E - 4ч – К., (4-x) — k, C, - 24 ) motion of maasky k ekoras a К слотаby («-») motion of me? 12. – 2а к, сфоло 3, clare» 4 (14) - L, X, m, 2 – К., - , (-*) - , (а, -ч)

for my block, kz ng miä + (k tkz) au 482 => 4 + 2404 put, z = 045m, = x Ze=22min2Z => ZI +2 444 = 252 Zz ci) for of M₂ block Mz a + (k₂ + kg) X2 = kg a - 2 3 + 34 x2 = 42

4 Zi => da ŽJE + 17 Je za 2 2 + 17 Za = 252 Z ci) & vibrating with equa ci) and ui) are coupled eguro assume, periodic sola with both oscillators & same fregc. So, trial sol ? Z = A eswt = Beint Za from (i) > = 212 Beuwt Ców) 2 A cook + 24 A piwt .. Ciri) ỳ (24- co*) 4 = 22 0 from Cii) >> sast = 252 A ejwt (ów) ² B ejut +17 Be sy (17-02) m = 217 A ... .(iv) by Cri) xliv), we get (24-002) ( 14-02) AB = (20212 48. W9.4102 + 408 = 8. (N) 4 41w 2 + 400 = 0 sol of quadratic equo. (v) Ź (41 1 1 641)2- 4.4.400 (4119). → 2 مودم = 250 co2 = 32 => W. 5- 3 } 452s co- normal frer, of oscillation So, frez. of oscillalni au w=55 W = 4.125