Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u” + 1 4 u’ + 2u = 2 cos ωt

Question

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u” + 1 4 u’ + 2u = 2 cos ωt, u(0) = 0, u'(0) = 2 (a) Determine the steady state part of the solution of this problem.

in progress 0
Phúc Điền 5 months 2021-08-21T07:30:00+00:00 1 Answers 16 views 0

Answers ( )

  1. Thunguyet
    0
    2021-08-21T07:31:40+00:00
    Reply

    Answer:

    Therefore the required solution is

    U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

    Explanation:

    Given vibrating system is

    u''+\frac{1}{4}u'+2u= 2cos \omega t

    Consider U(t) = A cosωt + B sinωt

    Differentiating with respect to t

    U'(t)= – A ω sinωt +B ω cos ωt

    Again differentiating with respect to t

    U”(t) =  – A ω² cosωt -B ω² sin ωt

    Putting this in given equation

    -A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

    \Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

    Equating the coefficient of sinωt and cos ωt

    \Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

    \Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0………(1)

    and

    \Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

    \Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0……..(2)

    Solving equation (1) and (2) by cross multiplication method

    \frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

    \Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

    \therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

    Therefore the required solution is

    U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )