SUMMARY: FOURIER SERIES ALETHED FOR DAMPED + DRIVEN HARMONIC OSCILLATOR SOLVE: MX = - kx-bx +F(Z) * + 2 3x + wix = in F ( t ) B = b 2 m W. = ( 4/M F ( + + E ) = F (2 ) W = 21/ 2 EXPAND F (Z) IN A FOURIER SERIES TO CONSTRUCT THE PARTICULAR SOLUTION XP (2) F ( 2 ) = za. + S (ancos not + busin nut) a = 7 4/2 F ( 2 ) dt 2 / 2 4/ 2 an = F (E ) cos ( not ) di -e / 2 by = F ( t ) sin (natalt Do X p (2 ) = in S . ao + Es On ( w) Lan cos ( nut- Sn ( w) + n = 1 busin ( nwz - 8 x( w))] WITH gn ( w ) = ( ( w 2 - n ? w = ) 2 + 4n 2 82 (2) 1/2 Cos Sn ( w) = ( w2 - new2 ) gn (w) Sindn ( w ) = 2 Bnwan (w) FULL SOLUTION IS X ( t ) = X, (8) + Xp(2)Problem 4. Consider a driven, but undamped (6 = 0) harmonic oscillator with driving force FOE) = F0c082(wt) Hence, the equation of motion is 533 + 0033: = (%)F06032(wt) where F0 is a positive constant, (do is the natural angular frequency of the undriven oscillator, and w > wo. Determine an expression for the position as a function of time, 33(15), using the Fourier Series method with initial conditions, 117(0) 2 "0(0) 2 0, by proceeding through the following steps: (i) Determine the Fourier coefficients of F(t) using the integral method described in class. Your results can be easily checked by noting the trigonometric identity, 6052(wt) = 1/2 + 1/2603(2wt) (i.e. you should nd that a0 2 F0, a2 2 F0 / 2, and the others are all zero). Integrals: ff\" 6052($)cos(n$)d:v = g 71,2 (n = 1,2, ..) f; 6052($)5in(n$)d$ = 0 (ii) Determine the appropriate amplitude and phase functions 9,,(w) and 6,,(w). iii Construct the particular solution, a: t usin the results from parts i and ii . p 8 (iv) Add the homogeneous solution to :L'p) so that the full solution is 33(t) = Acos(w0t) + Bsin(w0t) + x1905) (v) Apply the initial conditions to determine A and B