Question

MAT 204 Elementary Differential Equations 1. A mass-spring-dashpot system is described by my" + cy' + ky = Fo coswt, see $3.6 Eq. (17). This second-order differential equation has been used in simulations, such as this one at the PhET site: https://phet.colorado.edu/en/simulation/legacy/resonance. = 48.6 N, and w will be For m = 2.53 kg, c = 0.502 N/(m/s), k 97.2 N/m, Fo = 97.2 x 0.5 N given by your instruetes, the equation becomes 2.53y" + 0.502y +97.2y = 48.6 cos(wt). 8.1 (1) & omega (a) Given the initial value y(0) = 2.20, 7(0) = 0, solve Eq. (1), Round to three significant figures. Show all your work, and clearly highlight your conclusion—the combination of complementary function and particular solution. (b) From your particular solution, which is of the form = A coswt + B sin wt, calculate the amplitude, which is A2+ B2. Clearly highlight your conclusion the amplitude. You are encouraged to use the PhET simulation to verify your amplitude. Note that the angular frequency w = 2nf where f is the frequency in the simulation.

This is Differential equations, please help me, solve and show step by step.

Answer 1

- Foloswt m mylly'tky =) y" + y! + y 2 to lasat D² + ) ។ ។ y = for laut C m m m The characteristic equation is 1²+ (net 2 C سلم ti m 2m 6 2 =-ati dal tests a C am lt + Belt C. f => Ae y = A atiwelt & BE Ga-illalt = é at (Pelalett -1042) Ве T y = ė at il -at е Fologist +isan wat) + B ( comedit @ i sónuut) [(012) Cowdet ti (0-3) Sém edit] sin wat y = ē at C, Coswelt taba

Particular Integral solution P.I4 B, sinut 18,604 Wt d² ditt m Flogit -B, w sinut - B Walopant (-B,w+sinut - B. J* Copelt) + c(3, 0 copiebt - B Wsenet) + KCB, sinat + Blowot) = (-m B, w ² – ( W. Be + kB, ) senet + (-mw ²B + CWB, tkB ) Codit = Floent to sendt - ww² B, - CWB2 = 0 CWB + (-mw) B₂ = Fo (k-mn ²) B - CW B2 B2 komot ca Substituiting By value in @ we get a CWB + XB, - F B ~ Kamwt) (komu?) لینے B. CW Fo (We # (k-mwa)2 K-mwt. ca Fo cw B = (²W² + (k-mw2/2 Amplitude = √3,²+ B² -at C, con udt the sin udt C am 0.502 2 (2.53) -0.09902 Wd = k m = 6.1975 am

-0.01902t y = e cos 6,20€ +C, Sin6, Lot P.I I = B, sin 8.it + B₂ Log 8.it ·0.09902t [ low 6.20t +4, sin 6.16] B. Sin&. it + B₂ Cop 8.lt At t=0 y = 2.2, 4-0 =) 2.2 = c + B₂ B = folk-mw?) c²w²+ (kamw?) B₂ = -0,0703 → = 2.27037 After differentiating y 2 Putly'=0 -0.099026 +2+B=0 (2= 0.099024-B, = 0.2951 ✓ B 2+3² fo=48.6 Be - 48.59

-0.09902t. y=e 0,2951,6mm 6,27 +2.2703661 6,2t + f. 48,59 sin seit -0.0703 Caps.lt ob) Comparing particular Integral obtained above with the particular Integral given in question B,= A; B, = B 2 ALAB2 B 24 B2 - Fo=48.6