In a block-spring system, m=0.27kg and k=3.2N/m. At t=0.12s, the velocity is vx=-0.164m/s, and the acceleration, ax=0.875m/s^2. Write the expression for the position as a function of time, x(t).

\fW K 23.21/m m = 0. 27 kg . . FC = A Cos ( wt + x ) Vac = - Awsin(wt + 8 ) ax= - Act Cos ( wt + $ ) & at t = 0.125 . , Vy = - 0. 164 m/s = >-0-164 = - A(/K ) Sin JK x0 . 12 +0 = > - 0. 164 = - Awsin (whatp) (': 02 = 317 an = - Aw? Cos ( woto to) - (11) but w = 3.2 10.27 2 3.44 rad/s dividing ean (1) by ( 2 ), = > - Awsin ( wtoto ) Vx an - A wzcos ( wto + of ) = > ton ( wto + $ ) = Vx x co ax = > (wto + g ) = tan' lax -> 8 = tan' ( wvn ) - wto = > o = tan' /3. 44 x (- 0. 164) - 3. 44 x 0 . 12 0-875 2 - 0.9854 rad Also, A = - Vx Cos ( wt + $ ) = ) A = + (- 0. 164) 3-44 x Sin (3. 44 x0-12 - 0.9853) = > A = 0 . 0879m So, X = 0.0879 Sin ( 3.44+ -0.985)