Question

ki 23,The position of a certain lightly damped oscillator satisfies the initial van 20 y" +0.1y' + y = (-1)*+18(1 - kn), y(0) = 0, y(0) = 0. k=1 Observe that, except for the damping term, this problem is the same as Problem 1o. (a) Try to predict the nature of the solution without solving the problem. (b) Test your prediction by finding the solution and drawing its graph. (c) Determine what happens after the sequence of impulses ends.

Answer 1

(a) Because of small, posithe damping Constant, we have to calculate Inverse Laplace transform of a function. It ward Crente exponen Wol tactos with negatie exponents in the Solution. That should cham the amplitude of the solution. tM) -> Amplitude

(b) Cy'+ory'+y) - 2(y)+11/9) +209) - SNYCS)+DHI SY($-sy(0) --D1960+y(5) = Sys) +0:15 yrs +Y) =C4D-IS+DYC9 eccy** -ku) - -** (814-km) KET (S4o.1s+DX) = (-**!-KTS YLS) = 20 kt & kis Soush 2-DK -S (S+ +394 AW 399 By Taking inverse laplace Transtom" com home very best (t-T honor 9N

(c) After the sequence which still behave of impulse ends, the amplitude of the solution as an oscillatory function decreases.