1 answer

21.6 A,B,C,D result given in part c of this exercise. 21.6. Consider a damped mass/spring system...

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)
and satisfies cos(0) = and sin() = B y2 2m2 Next, by finding the maximum of C with respect to show that the angular resonant
21.6 A,B,C,D
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
and satisfies cos(0) = and sin() = B y2 2m2 Next, by finding the maximum of C with respect to show that the angular resonant frequency of the system is and that the corresponding maximum amplitude is provided y <2km. What happens if, instead, y 22km ? d. Assume that y < V2Km. Then the system is underdamped and, as noted in chapter 16. the general solution to the corresponding homogeneous equation is 415 no = y (1) = cje of cos(wt) + c2eaf sin(at) where a is the decay coefficient and w is the angular quasi-frequency. Verify that this w the resonant angular frequency no, and the natural angular frequency ory of the corresponding undamped system are related by 2m Fo Y4km - 2 (70)2 + 2) = w? = (es)? - () and, from this, conclude that no <o<w.

Answers

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 cosnB = 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 +22for (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)) =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 t

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