Question

10. Consider the forced but undamped system described by the initial value prob- lem u" + 25u = sin(wt), (0) = 0, ta (0) =1. a. Find solution for u(t) when w #5. b. Plot the solution u(t) versus t for w = : 4, 4.5, and 4.9. Describe how the response u(t) changes as w varies. What happens when w gets close to 5? Note that the natural frequency of the system is wo = 5. (Hint: Make the plots using technology.)

Answer 1

U²+25 u = sinot Le col=0 L'(0,) a the auxilary Equation.be m m+25=0 m=tsi General > Solun Ug = G Gast +Csinst Particular Intigral Up = D725 sincet sincet I 25-0 - w125 Sincol Thein Requared slunde u (t.) = G Const į lasst tc, sinst + I Sinar 25-col G = 0 460)=0 / (0) 20 sw Coub tow 56 + 25-0 W 5lt 5G 25-02 ا 25-wrw (25.com) . 25.com w (25-646) -
1
+ U() = (es way Requared solun (25-ww] Sinst thinieot 5 b 7 when w-4, 4.5, 4.9. o u(t) 25-16-4 1 sinst thin4. sinst tllingt thfinkt (sl-se) { (25-16-4) sinst thin st u(t) ☺ Ino graph. A We't) 2 with Sinst & Sinkost 25-459-415 4.75 5 25 Sinsit + Singist 87475 4.75 sinst t sin 4.5t 95 Ult) 2 t- ht) (25-49-49) sinst t! Singt 5125-49 5 25-4.90 -391 Sinst 99 + Singt 495 100 3rd Graphe.
2
when le (+) W = =5 undefined & Snii 25-co a But limeting Graph not exist ult) : is - w-w) finst + sinof? (25-60) here Wo is be the natural frequens w ke clependent on arebitary Constant r W = 4.5 W = 4.9 ultos WEn fos sino 0.5 ty Es ty are The Graph of this function delt] changed when a change @ when w=4 pie greater then when ishen we 4g if has hight pie. W = 4.5
3
III 0:5 -0.5 0.5 -0.5 + 1 Х sin(52)+* sin(4x) 9 2
1 no graph
4:20 PM ...1.0KB/S @MO 33 -1 0 -1- + 1 Х 2 Х -391 495 sin(52)+ 99 100 sin(4 3 powered by desmos
2no graph
= 1 0 -1 + 1 Х 2 Х -391 495 sin(5x)+ 99 100 sin(4 3.
3no graph