Overdamped Oscillation
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Exercise:
An overdamped oscillation is characterised by the eigenvalues * lambda_ -delta quad textrmand quad lambda_ -delta and the corresponding eigenvectors * vec v_ pmatrix -delta delta^ pmatrix quad textrmand quad pmatrix -delta delta^ pmatrix Determine the solution for the initial conditions given by * y A v_y delta A
Solution:
The general solution can be written as pmatrix yt v_yt pmatrix a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t a_ pmatrix -delta delta^ pmatrix e^-delta t + a_ pmatrix -delta delta^ pmatrix e^-delta t The coefficients a_ and a_ can be found from the initial conditions: pmatrix A delta A pmatrix a_ pmatrix -delta delta^ pmatrix + a_ pmatrix -delta delta^ pmatrix The two components can be simplified as A -delta a_+a_ A delta a_+a_ Adding the two s yields A -delta a_+delta a_ delta a_ Longrightarrow a_ fracAdelta Adding three times the first to the second one yields A -delta a_+delta a_ - delta a_ Longrightarrow a_ -frac Adelta The solution is thus yt fracAdeltadelta e^-delta t-fracAdelta delta e^-delta t Aleft e^-delta t - e^-delta tright v_yt -fracAdeltadelta^ e^-delta t + fracAdelta delta^ e^-delta t Adelta left-e^-delta t+A e^-delta t right
An overdamped oscillation is characterised by the eigenvalues * lambda_ -delta quad textrmand quad lambda_ -delta and the corresponding eigenvectors * vec v_ pmatrix -delta delta^ pmatrix quad textrmand quad pmatrix -delta delta^ pmatrix Determine the solution for the initial conditions given by * y A v_y delta A
Solution:
The general solution can be written as pmatrix yt v_yt pmatrix a_ vec v_ e^lambda_ t + a_ vec v_ e^lambda_ t a_ pmatrix -delta delta^ pmatrix e^-delta t + a_ pmatrix -delta delta^ pmatrix e^-delta t The coefficients a_ and a_ can be found from the initial conditions: pmatrix A delta A pmatrix a_ pmatrix -delta delta^ pmatrix + a_ pmatrix -delta delta^ pmatrix The two components can be simplified as A -delta a_+a_ A delta a_+a_ Adding the two s yields A -delta a_+delta a_ delta a_ Longrightarrow a_ fracAdelta Adding three times the first to the second one yields A -delta a_+delta a_ - delta a_ Longrightarrow a_ -frac Adelta The solution is thus yt fracAdeltadelta e^-delta t-fracAdelta delta e^-delta t Aleft e^-delta t - e^-delta tright v_yt -fracAdeltadelta^ e^-delta t + fracAdelta delta^ e^-delta t Adelta left-e^-delta t+A e^-delta t right