Overdamped Oscillation
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
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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
Meta Information
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
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