Critically Damped Oscillator
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
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?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
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?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
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
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
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
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
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 has a degenerate eigenvalue for deltaomega_ critical damping. 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 critically damped case we have delta^omega_^ and therefore the square root vanishes: lambda -delta The corresponding eigenvector is found to be bf v resultleftmatrix -omega_ matrixright abc For a degenerate eigenvalue the general solution can be written as leftmatrixyt v_ytmatrixright leftmatrixy_ v_matrixrighte^lambda t+bf v t e^lambda t The vector bf v is given by bf v bf A-lambda bf I leftmatrix y_ v_matrixright where bf I is the identity matrix. For lambda-delta we find bf v leftmatrix delta & -delta^ & -delta matrixright leftmatrixy_ v_matrixright leftmatrix delta y_ + v_ -delta^ y_-delta v_ matrixright This leads to the general solution yt y_ e^-delta t+delta y_+v_ t e^-delta t v_yt v_ e^-delta t-delta^ y_+delta v_ t e^-delta t so for the first case v_ we have yt resulty_ e^-delta tleft+delta tright v_yt result-delta^ y_ t e^-delta t and for the second case y_ yt resultv_ t e^-delta t v_yt resultv_ e^-delta tleft-delta 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 has a degenerate eigenvalue for deltaomega_ critical damping. 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 critically damped case we have delta^omega_^ and therefore the square root vanishes: lambda -delta The corresponding eigenvector is found to be bf v resultleftmatrix -omega_ matrixright abc For a degenerate eigenvalue the general solution can be written as leftmatrixyt v_ytmatrixright leftmatrixy_ v_matrixrighte^lambda t+bf v t e^lambda t The vector bf v is given by bf v bf A-lambda bf I leftmatrix y_ v_matrixright where bf I is the identity matrix. For lambda-delta we find bf v leftmatrix delta & -delta^ & -delta matrixright leftmatrixy_ v_matrixright leftmatrix delta y_ + v_ -delta^ y_-delta v_ matrixright This leads to the general solution yt y_ e^-delta t+delta y_+v_ t e^-delta t v_yt v_ e^-delta t-delta^ y_+delta v_ t e^-delta t so for the first case v_ we have yt resulty_ e^-delta tleft+delta tright v_yt result-delta^ y_ t e^-delta t and for the second case y_ yt resultv_ t e^-delta t v_yt resultv_ e^-delta tleft-delta tright abcliste
Meta Information
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 has a degenerate eigenvalue for deltaomega_ critical damping. 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 critically damped case we have delta^omega_^ and therefore the square root vanishes: lambda -delta The corresponding eigenvector is found to be bf v resultleftmatrix -omega_ matrixright abc For a degenerate eigenvalue the general solution can be written as leftmatrixyt v_ytmatrixright leftmatrixy_ v_matrixrighte^lambda t+bf v t e^lambda t The vector bf v is given by bf v bf A-lambda bf I leftmatrix y_ v_matrixright where bf I is the identity matrix. For lambda-delta we find bf v leftmatrix delta & -delta^ & -delta matrixright leftmatrixy_ v_matrixright leftmatrix delta y_ + v_ -delta^ y_-delta v_ matrixright This leads to the general solution yt y_ e^-delta t+delta y_+v_ t e^-delta t v_yt v_ e^-delta t-delta^ y_+delta v_ t e^-delta t so for the first case v_ we have yt resulty_ e^-delta tleft+delta tright v_yt result-delta^ y_ t e^-delta t and for the second case y_ yt resultv_ t e^-delta t v_yt resultv_ e^-delta tleft-delta 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 has a degenerate eigenvalue for deltaomega_ critical damping. 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 critically damped case we have delta^omega_^ and therefore the square root vanishes: lambda -delta The corresponding eigenvector is found to be bf v resultleftmatrix -omega_ matrixright abc For a degenerate eigenvalue the general solution can be written as leftmatrixyt v_ytmatrixright leftmatrixy_ v_matrixrighte^lambda t+bf v t e^lambda t The vector bf v is given by bf v bf A-lambda bf I leftmatrix y_ v_matrixright where bf I is the identity matrix. For lambda-delta we find bf v leftmatrix delta & -delta^ & -delta matrixright leftmatrixy_ v_matrixright leftmatrix delta y_ + v_ -delta^ y_-delta v_ matrixright This leads to the general solution yt y_ e^-delta t+delta y_+v_ t e^-delta t v_yt v_ e^-delta t-delta^ y_+delta v_ t e^-delta t so for the first case v_ we have yt resulty_ e^-delta tleft+delta tright v_yt result-delta^ y_ t e^-delta t and for the second case y_ yt resultv_ t e^-delta t v_yt resultv_ e^-delta tleft-delta tright abcliste
Contained in these collections: