Question

result given in part c of this exercise. 21.6. Consider a damped mass/spring system given by m dy gdy tr dt + ky = Fo cos(nt) where m. y. K and Fo are all positive constants. (This is the same as equation (214) a. Using the method of educated guess, derive the particular solution given by equation ser (21.10) on page 409. genelaidi b. Then show that the solution in the previous part can be rewritten as described by equation set (21.11) on page 409; that is, verify that the solution can be written as Dengan ypt) C cos(nt - 0) bios where the amplitude of these forced vibrations is S. Fo C = vík – mn2] + n2y2 1 DO Uus Ilustrate the

21.6 A,B,C,D

Answer 1

o 2116 + 8 dy mdy at2 tky = to cos (nt) dt 9) Y p (t) = A cosnt + B sinnt Yp' (t) = - An sinnt + Bn cosnt Yp" (t) = -An² cosnt - B h² sinnt L-Anscount - Bn sinnt) +8 [Ansinnt + Bh Cosnt +K[A cosnt +B sinnt +] = Fo Cosnt On comparing coefficient of sinnt Bram m - An8 +KB = 0 B (k-1 min) Comparing coefficient of cont - Amn + Bn8 + AK= Fo A [K-mn?] + Bn8=Fo B {(k-me?) + (nr3²] = Fo no

B = Fo hr (K-mn ²)2 + n²g2 A = Fo (K-mn²) (k-mn 2)2 + 2282 Cosnt + Fong sinht Yp (t) = Fo (k-m2²) (k-mn]2+ n²82 (k-mn ²)2 +2282 b) (k-min²) cosnt Yp (t) = Fo Fo V(k-mn ² +2782 + nr sinnt v (km2) 2222 Vik-mn 2)2 +2222 cos o 2 ut K-mn² (k-mn 2 2 2 2 2 2 than = sind hr v (kmn2 + 272 Yp (t) J (k-mn 2+2 222 cosnt cosp + sinnt sind [ con lnt-d)] Fo Yp (t) v (k-mn 27 1282 where c = Fo J (K-mn ²)²+ nr ² et

for (max with no respect to c this should be so X = (k-mn ² 2 + n² myn = 0 dX dn 2 (K-mn ²) (2mn) +2n8² n(82 - 2 m (k-mn²)) =0 82 - 2m (k-mn² = 0 2 m (k-mn²) Y? K-mn? 82 2 m mn² = k-82 2 m Y? 2 m2 m n=J&K 2 2 m2 Cmax Fo v (x - (K-21 ) )2 + ( + 1 - 2 2 2 (max - 84 4m² + kr²_84 m 2 m2

Cmax = 2 fom учKm — 82 re 12km then Charactestic eqn mo 2 + 87 tk=0 Ta -8 + 582_4km 2 m = – ri i 54 km - 82 2 m sinut loswt te e at Ya(t) = ci è at with a= 8 w = 34 km-82 2 m 2 m 80 (n.) + 1 k у rm2 +22 4m2 2 2 + 18 r 4m2 =w2 m 2m Since w= 4km-82 2 m Ell rece