MT-1620 al.2002 Now use the homogeneous solution with this to get the total solution qt)=C snot C2cosot k The Initial Conditions are q 0)=0 F6 q(0)=0→C1=0 So the final solution is F6 q coS ot k with Plotting this Paul A Lagace @2001 Unit 20-7qt MIT - 16.20 Fall, 2002 Now use the homogeneous solution with this to get the total solution: F () = C1 sinω t + C2 cosω t + 0 k The Initial Conditions are: @ t = 0 q = 0 q˙ = 0 q () 0 = 0 ⇒ C = − F0 2 k q ˙() 0 = 0 ⇒ C1 = 0 So the final solution is: F q = 0 (1 − cosω t) k with ω = k m Plotting this: Paul A. Lagace © 2001 Unit 20 - 7