Question

value of W=11.1

omega value is 11.1

1. A mass-spring-dashpot system is described by my" + cy' + ky = Focoswt, 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. For m = 2.53 kg, c = 0.502 N/(m/s), k = 97.2 N/m, Fo 97.2 x 0.5 N = 48.6 N, and w will be given by your instructor, the equation becomes 2.534" +0.502y +97.2y = 48.6 cos(wt). (1) (a) Given the initial value y(0) = 2.20 (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 Yp = Acoswt + B sin wt, calculate the amplitude, which is A + 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 = 2f where ſ is the frequency in the simulation.

Answer 1

Gelven: my" +CY'tky = To coor Clot) For m=2.53 kg , C=0.502 N/mls • K=97.2 Nim Fo= 48.6 N and W=11.1 Then 2.53 y" +0.502y' +97-2y = 48.6 Cos (11.17) 9) Ganeral solution : y= Ict Yp whord, Yc= complimentary function Ypz particular integral For gci calculate for homogeneous D.E. 2.53y" to 502 y' +97.24 =0 (2.538² +0.502D+972)y=0 put D= m and g=1 2.53 m2 +0.502m+ 97.2=0 roots are: m=-0.99 £6.198 i Then yc for complex roote i.e. Carbis Ycz et (acoubt tez sim bt) Gc= ė 0.99€ (a casco-198t) + Ca sin (6.198t))

General sodution: yct) = 4ct Yp y ct)= @040tca cosC61984) +Ca sainC6-1987) +0.00613 sm (11.lt) -0.23597 CodCli.it) -o-qat yst)=ē "Lacos6-198+) +2 Sim (6.198+))70-00b Sincll.lt)-0.236 casc11iH) Yht) =-0.9920914 Cacob(6-198+) +G2 Sin (6.1984). "(-6-1984 Sim (6-1984) +6-1982 Cas (6-1987)) +€0.99€ +0006x11.) cos (11-1+) + 6-236x11.1 sin (11.17) use boundary conditions at ycol = 2.2 2.2 = G-0.236 G= 2.436 at yl (o)=0 0 = -0.99G76-1982 to-006x11-1 (2= Co.qgG-0.006 X11.1) 6.198 (2= (0.99x2.436 -0.006 X 11.1) 66198 = 0.373 Gc= 68.99€ (2.436 Cos (6-1984) +0.378 Sim (6-1984) 464) =ē°992.436 cos (6.1984)46.378 Sin (6-1987) +0.006 Sincul.it) -0.236 Coorcll-H)

(6) Yo= 0.006 Sim Cililt) - 0.236 Coos Cli.it) compare by Yp= A Coscut) + B Sim (wt) A= -0.236 and B=0.006 Trun Amplitude VARIB? √(-0.236)? (0.006]} √ A2B? = 0.236