Underdamped Oscillator
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But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
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That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
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Exercise:
A linearly damped oscillator can be described by the following system of differential s: fracddtleftmatrixy v_ymatrixright leftmatrix & -omega_^ & -deltamatrixright leftmatrixy v_ymatrixright abcliste abc Find the eigenvalues and eigenvectors. Show that the system corresponds to a stable spiral for deltaomega_ underdamped. abc Derive the solutions for the initial conditions yy_ v_y start from rest and y v_yv_ initial push. abcliste
Solution:
abcliste abc The trace tau and the determinant Delta for the matrix describing the system are tau -delta -delta Delta -delta--omega_^omega_^ Using the general expression for eigenvalues of a times matrix we find lambda_ fractaupmsqrttau^-Delta frac-deltapmsqrtdelta^-omega_^ result-deltapmsqrtdelta^-omega_^ For the underdamped case we have delta^-omega_^ and therefore the square root is an imaginary number: lambda_ -deltapm isqrtomega_^-delta^ -deltapm iomega where omegasqrtomega_^-delta^ is a real value. The eigenvectors are found to be bf v_ resultleftmatrix-delta-iomega omega_^ matrixright bf v_ resultleftmatrix-delta+iomega omega_^ matrixright Both eigenvalues are complex and have a negative real part so the system corresponds to a stable spiral. abc The fundamental solutions are bf xi_t bf v_ e^lambda_ t bf xi_t bf v_ e^lambda_ t The general solutions are superpositions of the fundamental solutions: leftmatrixyt v_ytmatrixright a_ bf v_ e^lambda_ t+a_ bf v_ e^lambda_ t The coefficients a_ and a_ are given by the initial conditions. In the first case we have leftmatrixy_ matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright The second tells us that a_-a_ so the first can be written as y_ a_left-delta-iomega+delta-iomegaright -a_ i omega with the solution a_ fracy_-iomegafraci y_omega-a_ The displacement for the case where the pulum is started from rest with an initial displacement y_ is then yt fraciy_omegaleft-delta-iomega e^-delta te^iomega t--delta+iomega e^-delta te^-iomega tright fraci y_omega e^-delta tleftdelta lefte^-iomega t-e^iomega tright-iomegalefte^-iomega t+e^iomega trightright The terms with the complex exponentials can be simplified as follows: e^-iomega t-e^iomega t cos-omega t+isin-omega t-cosomega t-isinomega t cosomega t-cosomega t-isinomega t-isinomega t -isinomega t e^-iomega t+e^iomega t cos-omega t+isin-omega t+cosomega t+isinomega t cosomega t+cosomega t-isinomega t+isinomega t cosomega t It follows for the displacement yt y_ e^-delta tleftfrac-delta i^omegasinomega t-fracomega i^omegacosomega tright resulty_ e^-delta tleftcosomega t+fracdeltaomegasinomega tright For the velocity we find v_yt fraci y_omega omega_^ e^-delta tlefte^iomega t-e^-iomega tright fraci y_omega omega_^ e^-delta t isinomega t result-y_ fracomega_^omega e^-delta tsinomega t For the pulum starting from the equilibrium position with an initial velocity v_ the coefficients are defined as follows: leftmatrix v_matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright From the first we get a_ -delta-iomega+a_-delta + iomega Longrightarrow a_ -a_fracdelta+iomegadelta-iomega The second is then v_ omega_^ a_+a_ omega_^ a_ left-fracdelta + iomegadelta-iomegaright omega_^ a_frac-iomegadelta-iomega Longrightarrow a_ -v_fracdelta-iomegaiomegaomega_^ i v_fracdelta-iomegaomegaomega_^ a_ -a_fracdelta+iomegadelta-iomega -i v_fracdelta-iomegaomegaomega_^fracdelta+iomegadelta-iomega -iv_fracdelta+iomegaomegaomega_^ The displacement is given by yt i v_fracdelta-iomegaomegaomega_^-delta-iomegae^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^-delta+iomegae^-delta te^-iomega t fraci v_omegaomega_^e^-delta tleft-delta-iomegadelta+iomegae^iomega t & quadquadquadquad-delta+iomega-delta+iomegae^-iomega tright fraci v_omegaomega_^e^-delta tleft-omega^+delta^ e^iomega t+omega^+delta^e^-iomega tright fraci v_omegaomega_^e^-delta tleftomega_^ e^-iomega t-omega_^ e^iomega tright fraci v_omega e^-delta t-isinomega t resultfracv_omegae^-delta tsinomega t The velocity is as follows: v_yt i v_fracdelta-iomegaomegaomega_^omega_^ e^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^omega_^ e^-delta te^-iomega t fraci v_omega e^-delta tleftdelta-iomegae^iomega t-delta+iomegae^-iomega tright fraci v_omega e^-delta tleftdelta lefte^iomega t-e^-iomega tright-iomegalefte^iomega t+e^-iomega trightright fraci v_omega e^-delta tleftdelta i sinomega t-iomega cosomega tright resultfracv_omega e^-delta tleftomegacosomega t-deltasinomega tright abcliste
A linearly damped oscillator can be described by the following system of differential s: fracddtleftmatrixy v_ymatrixright leftmatrix & -omega_^ & -deltamatrixright leftmatrixy v_ymatrixright abcliste abc Find the eigenvalues and eigenvectors. Show that the system corresponds to a stable spiral for deltaomega_ underdamped. abc Derive the solutions for the initial conditions yy_ v_y start from rest and y v_yv_ initial push. abcliste
Solution:
abcliste abc The trace tau and the determinant Delta for the matrix describing the system are tau -delta -delta Delta -delta--omega_^omega_^ Using the general expression for eigenvalues of a times matrix we find lambda_ fractaupmsqrttau^-Delta frac-deltapmsqrtdelta^-omega_^ result-deltapmsqrtdelta^-omega_^ For the underdamped case we have delta^-omega_^ and therefore the square root is an imaginary number: lambda_ -deltapm isqrtomega_^-delta^ -deltapm iomega where omegasqrtomega_^-delta^ is a real value. The eigenvectors are found to be bf v_ resultleftmatrix-delta-iomega omega_^ matrixright bf v_ resultleftmatrix-delta+iomega omega_^ matrixright Both eigenvalues are complex and have a negative real part so the system corresponds to a stable spiral. abc The fundamental solutions are bf xi_t bf v_ e^lambda_ t bf xi_t bf v_ e^lambda_ t The general solutions are superpositions of the fundamental solutions: leftmatrixyt v_ytmatrixright a_ bf v_ e^lambda_ t+a_ bf v_ e^lambda_ t The coefficients a_ and a_ are given by the initial conditions. In the first case we have leftmatrixy_ matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright The second tells us that a_-a_ so the first can be written as y_ a_left-delta-iomega+delta-iomegaright -a_ i omega with the solution a_ fracy_-iomegafraci y_omega-a_ The displacement for the case where the pulum is started from rest with an initial displacement y_ is then yt fraciy_omegaleft-delta-iomega e^-delta te^iomega t--delta+iomega e^-delta te^-iomega tright fraci y_omega e^-delta tleftdelta lefte^-iomega t-e^iomega tright-iomegalefte^-iomega t+e^iomega trightright The terms with the complex exponentials can be simplified as follows: e^-iomega t-e^iomega t cos-omega t+isin-omega t-cosomega t-isinomega t cosomega t-cosomega t-isinomega t-isinomega t -isinomega t e^-iomega t+e^iomega t cos-omega t+isin-omega t+cosomega t+isinomega t cosomega t+cosomega t-isinomega t+isinomega t cosomega t It follows for the displacement yt y_ e^-delta tleftfrac-delta i^omegasinomega t-fracomega i^omegacosomega tright resulty_ e^-delta tleftcosomega t+fracdeltaomegasinomega tright For the velocity we find v_yt fraci y_omega omega_^ e^-delta tlefte^iomega t-e^-iomega tright fraci y_omega omega_^ e^-delta t isinomega t result-y_ fracomega_^omega e^-delta tsinomega t For the pulum starting from the equilibrium position with an initial velocity v_ the coefficients are defined as follows: leftmatrix v_matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright From the first we get a_ -delta-iomega+a_-delta + iomega Longrightarrow a_ -a_fracdelta+iomegadelta-iomega The second is then v_ omega_^ a_+a_ omega_^ a_ left-fracdelta + iomegadelta-iomegaright omega_^ a_frac-iomegadelta-iomega Longrightarrow a_ -v_fracdelta-iomegaiomegaomega_^ i v_fracdelta-iomegaomegaomega_^ a_ -a_fracdelta+iomegadelta-iomega -i v_fracdelta-iomegaomegaomega_^fracdelta+iomegadelta-iomega -iv_fracdelta+iomegaomegaomega_^ The displacement is given by yt i v_fracdelta-iomegaomegaomega_^-delta-iomegae^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^-delta+iomegae^-delta te^-iomega t fraci v_omegaomega_^e^-delta tleft-delta-iomegadelta+iomegae^iomega t & quadquadquadquad-delta+iomega-delta+iomegae^-iomega tright fraci v_omegaomega_^e^-delta tleft-omega^+delta^ e^iomega t+omega^+delta^e^-iomega tright fraci v_omegaomega_^e^-delta tleftomega_^ e^-iomega t-omega_^ e^iomega tright fraci v_omega e^-delta t-isinomega t resultfracv_omegae^-delta tsinomega t The velocity is as follows: v_yt i v_fracdelta-iomegaomegaomega_^omega_^ e^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^omega_^ e^-delta te^-iomega t fraci v_omega e^-delta tleftdelta-iomegae^iomega t-delta+iomegae^-iomega tright fraci v_omega e^-delta tleftdelta lefte^iomega t-e^-iomega tright-iomegalefte^iomega t+e^-iomega trightright fraci v_omega e^-delta tleftdelta i sinomega t-iomega cosomega tright resultfracv_omega e^-delta tleftomegacosomega t-deltasinomega tright abcliste
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Exercise:
A linearly damped oscillator can be described by the following system of differential s: fracddtleftmatrixy v_ymatrixright leftmatrix & -omega_^ & -deltamatrixright leftmatrixy v_ymatrixright abcliste abc Find the eigenvalues and eigenvectors. Show that the system corresponds to a stable spiral for deltaomega_ underdamped. abc Derive the solutions for the initial conditions yy_ v_y start from rest and y v_yv_ initial push. abcliste
Solution:
abcliste abc The trace tau and the determinant Delta for the matrix describing the system are tau -delta -delta Delta -delta--omega_^omega_^ Using the general expression for eigenvalues of a times matrix we find lambda_ fractaupmsqrttau^-Delta frac-deltapmsqrtdelta^-omega_^ result-deltapmsqrtdelta^-omega_^ For the underdamped case we have delta^-omega_^ and therefore the square root is an imaginary number: lambda_ -deltapm isqrtomega_^-delta^ -deltapm iomega where omegasqrtomega_^-delta^ is a real value. The eigenvectors are found to be bf v_ resultleftmatrix-delta-iomega omega_^ matrixright bf v_ resultleftmatrix-delta+iomega omega_^ matrixright Both eigenvalues are complex and have a negative real part so the system corresponds to a stable spiral. abc The fundamental solutions are bf xi_t bf v_ e^lambda_ t bf xi_t bf v_ e^lambda_ t The general solutions are superpositions of the fundamental solutions: leftmatrixyt v_ytmatrixright a_ bf v_ e^lambda_ t+a_ bf v_ e^lambda_ t The coefficients a_ and a_ are given by the initial conditions. In the first case we have leftmatrixy_ matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright The second tells us that a_-a_ so the first can be written as y_ a_left-delta-iomega+delta-iomegaright -a_ i omega with the solution a_ fracy_-iomegafraci y_omega-a_ The displacement for the case where the pulum is started from rest with an initial displacement y_ is then yt fraciy_omegaleft-delta-iomega e^-delta te^iomega t--delta+iomega e^-delta te^-iomega tright fraci y_omega e^-delta tleftdelta lefte^-iomega t-e^iomega tright-iomegalefte^-iomega t+e^iomega trightright The terms with the complex exponentials can be simplified as follows: e^-iomega t-e^iomega t cos-omega t+isin-omega t-cosomega t-isinomega t cosomega t-cosomega t-isinomega t-isinomega t -isinomega t e^-iomega t+e^iomega t cos-omega t+isin-omega t+cosomega t+isinomega t cosomega t+cosomega t-isinomega t+isinomega t cosomega t It follows for the displacement yt y_ e^-delta tleftfrac-delta i^omegasinomega t-fracomega i^omegacosomega tright resulty_ e^-delta tleftcosomega t+fracdeltaomegasinomega tright For the velocity we find v_yt fraci y_omega omega_^ e^-delta tlefte^iomega t-e^-iomega tright fraci y_omega omega_^ e^-delta t isinomega t result-y_ fracomega_^omega e^-delta tsinomega t For the pulum starting from the equilibrium position with an initial velocity v_ the coefficients are defined as follows: leftmatrix v_matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright From the first we get a_ -delta-iomega+a_-delta + iomega Longrightarrow a_ -a_fracdelta+iomegadelta-iomega The second is then v_ omega_^ a_+a_ omega_^ a_ left-fracdelta + iomegadelta-iomegaright omega_^ a_frac-iomegadelta-iomega Longrightarrow a_ -v_fracdelta-iomegaiomegaomega_^ i v_fracdelta-iomegaomegaomega_^ a_ -a_fracdelta+iomegadelta-iomega -i v_fracdelta-iomegaomegaomega_^fracdelta+iomegadelta-iomega -iv_fracdelta+iomegaomegaomega_^ The displacement is given by yt i v_fracdelta-iomegaomegaomega_^-delta-iomegae^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^-delta+iomegae^-delta te^-iomega t fraci v_omegaomega_^e^-delta tleft-delta-iomegadelta+iomegae^iomega t & quadquadquadquad-delta+iomega-delta+iomegae^-iomega tright fraci v_omegaomega_^e^-delta tleft-omega^+delta^ e^iomega t+omega^+delta^e^-iomega tright fraci v_omegaomega_^e^-delta tleftomega_^ e^-iomega t-omega_^ e^iomega tright fraci v_omega e^-delta t-isinomega t resultfracv_omegae^-delta tsinomega t The velocity is as follows: v_yt i v_fracdelta-iomegaomegaomega_^omega_^ e^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^omega_^ e^-delta te^-iomega t fraci v_omega e^-delta tleftdelta-iomegae^iomega t-delta+iomegae^-iomega tright fraci v_omega e^-delta tleftdelta lefte^iomega t-e^-iomega tright-iomegalefte^iomega t+e^-iomega trightright fraci v_omega e^-delta tleftdelta i sinomega t-iomega cosomega tright resultfracv_omega e^-delta tleftomegacosomega t-deltasinomega tright abcliste
A linearly damped oscillator can be described by the following system of differential s: fracddtleftmatrixy v_ymatrixright leftmatrix & -omega_^ & -deltamatrixright leftmatrixy v_ymatrixright abcliste abc Find the eigenvalues and eigenvectors. Show that the system corresponds to a stable spiral for deltaomega_ underdamped. abc Derive the solutions for the initial conditions yy_ v_y start from rest and y v_yv_ initial push. abcliste
Solution:
abcliste abc The trace tau and the determinant Delta for the matrix describing the system are tau -delta -delta Delta -delta--omega_^omega_^ Using the general expression for eigenvalues of a times matrix we find lambda_ fractaupmsqrttau^-Delta frac-deltapmsqrtdelta^-omega_^ result-deltapmsqrtdelta^-omega_^ For the underdamped case we have delta^-omega_^ and therefore the square root is an imaginary number: lambda_ -deltapm isqrtomega_^-delta^ -deltapm iomega where omegasqrtomega_^-delta^ is a real value. The eigenvectors are found to be bf v_ resultleftmatrix-delta-iomega omega_^ matrixright bf v_ resultleftmatrix-delta+iomega omega_^ matrixright Both eigenvalues are complex and have a negative real part so the system corresponds to a stable spiral. abc The fundamental solutions are bf xi_t bf v_ e^lambda_ t bf xi_t bf v_ e^lambda_ t The general solutions are superpositions of the fundamental solutions: leftmatrixyt v_ytmatrixright a_ bf v_ e^lambda_ t+a_ bf v_ e^lambda_ t The coefficients a_ and a_ are given by the initial conditions. In the first case we have leftmatrixy_ matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright The second tells us that a_-a_ so the first can be written as y_ a_left-delta-iomega+delta-iomegaright -a_ i omega with the solution a_ fracy_-iomegafraci y_omega-a_ The displacement for the case where the pulum is started from rest with an initial displacement y_ is then yt fraciy_omegaleft-delta-iomega e^-delta te^iomega t--delta+iomega e^-delta te^-iomega tright fraci y_omega e^-delta tleftdelta lefte^-iomega t-e^iomega tright-iomegalefte^-iomega t+e^iomega trightright The terms with the complex exponentials can be simplified as follows: e^-iomega t-e^iomega t cos-omega t+isin-omega t-cosomega t-isinomega t cosomega t-cosomega t-isinomega t-isinomega t -isinomega t e^-iomega t+e^iomega t cos-omega t+isin-omega t+cosomega t+isinomega t cosomega t+cosomega t-isinomega t+isinomega t cosomega t It follows for the displacement yt y_ e^-delta tleftfrac-delta i^omegasinomega t-fracomega i^omegacosomega tright resulty_ e^-delta tleftcosomega t+fracdeltaomegasinomega tright For the velocity we find v_yt fraci y_omega omega_^ e^-delta tlefte^iomega t-e^-iomega tright fraci y_omega omega_^ e^-delta t isinomega t result-y_ fracomega_^omega e^-delta tsinomega t For the pulum starting from the equilibrium position with an initial velocity v_ the coefficients are defined as follows: leftmatrix v_matrixright a_ leftmatrix-delta-iomega omega_^ matrixright + a_ leftmatrix-delta+iomega omega_^ matrixright From the first we get a_ -delta-iomega+a_-delta + iomega Longrightarrow a_ -a_fracdelta+iomegadelta-iomega The second is then v_ omega_^ a_+a_ omega_^ a_ left-fracdelta + iomegadelta-iomegaright omega_^ a_frac-iomegadelta-iomega Longrightarrow a_ -v_fracdelta-iomegaiomegaomega_^ i v_fracdelta-iomegaomegaomega_^ a_ -a_fracdelta+iomegadelta-iomega -i v_fracdelta-iomegaomegaomega_^fracdelta+iomegadelta-iomega -iv_fracdelta+iomegaomegaomega_^ The displacement is given by yt i v_fracdelta-iomegaomegaomega_^-delta-iomegae^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^-delta+iomegae^-delta te^-iomega t fraci v_omegaomega_^e^-delta tleft-delta-iomegadelta+iomegae^iomega t & quadquadquadquad-delta+iomega-delta+iomegae^-iomega tright fraci v_omegaomega_^e^-delta tleft-omega^+delta^ e^iomega t+omega^+delta^e^-iomega tright fraci v_omegaomega_^e^-delta tleftomega_^ e^-iomega t-omega_^ e^iomega tright fraci v_omega e^-delta t-isinomega t resultfracv_omegae^-delta tsinomega t The velocity is as follows: v_yt i v_fracdelta-iomegaomegaomega_^omega_^ e^-delta te^iomega t &quad - iv_fracdelta+iomegaomegaomega_^omega_^ e^-delta te^-iomega t fraci v_omega e^-delta tleftdelta-iomegae^iomega t-delta+iomegae^-iomega tright fraci v_omega e^-delta tleftdelta lefte^iomega t-e^-iomega tright-iomegalefte^iomega t+e^-iomega trightright fraci v_omega e^-delta tleftdelta i sinomega t-iomega cosomega tright resultfracv_omega e^-delta tleftomegacosomega t-deltasinomega tright abcliste
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