Symmetrical Two-Mass System
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:
The coefficient matrix for a symmetrical coupled two-mass system is bf A pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix Derive the eigenvalues and angular frequencies and verify that hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix are the corresponding eigenvectors.
Solution:
The trace and determinant of the matrix bf A are tau -fracK+km Delta leftfracK+kmright^-leftfrackmright^ fracK^+Kk+k^-k^mfracKK+km It follows for the eigenvalues lambda fractaupmsqrttau^-Delta^ frac-K+kpmsqrtK+k^-KK+km frac-K+kpmsqrtK^+Kk+k^-K^-Kkm frac-K+kpm km Longrightarrow lambda_ result-fracKm lambda_ result-fracK+km The angular frequencies are therefore omega_ sqrt-lambda_resultsqrtfracKm omega_ sqrt-lambda_resultsqrtfracK+km To verify whether hat x_ is an eigenvector we calculate bf A hat x_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix pmatrix pmatrix-fracK+km+frackm frackm-fracK+km pmatrix pmatrix-fracKm -fracKm pmatrix -fracKmpmatrix pmatrix i.e. hat x_ is the eigenvector for the eigenvalue lambda_. In the same way we can show that hat x_ is the eigenvector for the eigenvalue lambda_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix - pmatrix pmatrix-fracK+km-frackm frackm+fracK+km pmatrix pmatrix-fracK+km fracK+km pmatrix -fracK+kmpmatrix - pmatrix
The coefficient matrix for a symmetrical coupled two-mass system is bf A pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix Derive the eigenvalues and angular frequencies and verify that hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix are the corresponding eigenvectors.
Solution:
The trace and determinant of the matrix bf A are tau -fracK+km Delta leftfracK+kmright^-leftfrackmright^ fracK^+Kk+k^-k^mfracKK+km It follows for the eigenvalues lambda fractaupmsqrttau^-Delta^ frac-K+kpmsqrtK+k^-KK+km frac-K+kpmsqrtK^+Kk+k^-K^-Kkm frac-K+kpm km Longrightarrow lambda_ result-fracKm lambda_ result-fracK+km The angular frequencies are therefore omega_ sqrt-lambda_resultsqrtfracKm omega_ sqrt-lambda_resultsqrtfracK+km To verify whether hat x_ is an eigenvector we calculate bf A hat x_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix pmatrix pmatrix-fracK+km+frackm frackm-fracK+km pmatrix pmatrix-fracKm -fracKm pmatrix -fracKmpmatrix pmatrix i.e. hat x_ is the eigenvector for the eigenvalue lambda_. In the same way we can show that hat x_ is the eigenvector for the eigenvalue lambda_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix - pmatrix pmatrix-fracK+km-frackm frackm+fracK+km pmatrix pmatrix-fracK+km fracK+km pmatrix -fracK+kmpmatrix - pmatrix
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Exercise:
The coefficient matrix for a symmetrical coupled two-mass system is bf A pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix Derive the eigenvalues and angular frequencies and verify that hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix are the corresponding eigenvectors.
Solution:
The trace and determinant of the matrix bf A are tau -fracK+km Delta leftfracK+kmright^-leftfrackmright^ fracK^+Kk+k^-k^mfracKK+km It follows for the eigenvalues lambda fractaupmsqrttau^-Delta^ frac-K+kpmsqrtK+k^-KK+km frac-K+kpmsqrtK^+Kk+k^-K^-Kkm frac-K+kpm km Longrightarrow lambda_ result-fracKm lambda_ result-fracK+km The angular frequencies are therefore omega_ sqrt-lambda_resultsqrtfracKm omega_ sqrt-lambda_resultsqrtfracK+km To verify whether hat x_ is an eigenvector we calculate bf A hat x_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix pmatrix pmatrix-fracK+km+frackm frackm-fracK+km pmatrix pmatrix-fracKm -fracKm pmatrix -fracKmpmatrix pmatrix i.e. hat x_ is the eigenvector for the eigenvalue lambda_. In the same way we can show that hat x_ is the eigenvector for the eigenvalue lambda_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix - pmatrix pmatrix-fracK+km-frackm frackm+fracK+km pmatrix pmatrix-fracK+km fracK+km pmatrix -fracK+kmpmatrix - pmatrix
The coefficient matrix for a symmetrical coupled two-mass system is bf A pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix Derive the eigenvalues and angular frequencies and verify that hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix are the corresponding eigenvectors.
Solution:
The trace and determinant of the matrix bf A are tau -fracK+km Delta leftfracK+kmright^-leftfrackmright^ fracK^+Kk+k^-k^mfracKK+km It follows for the eigenvalues lambda fractaupmsqrttau^-Delta^ frac-K+kpmsqrtK+k^-KK+km frac-K+kpmsqrtK^+Kk+k^-K^-Kkm frac-K+kpm km Longrightarrow lambda_ result-fracKm lambda_ result-fracK+km The angular frequencies are therefore omega_ sqrt-lambda_resultsqrtfracKm omega_ sqrt-lambda_resultsqrtfracK+km To verify whether hat x_ is an eigenvector we calculate bf A hat x_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix pmatrix pmatrix-fracK+km+frackm frackm-fracK+km pmatrix pmatrix-fracKm -fracKm pmatrix -fracKmpmatrix pmatrix i.e. hat x_ is the eigenvector for the eigenvalue lambda_. In the same way we can show that hat x_ is the eigenvector for the eigenvalue lambda_: pmatrix-fracK+km & frackm frackm & -fracK+km pmatrix pmatrix - pmatrix pmatrix-fracK+km-frackm frackm+fracK+km pmatrix pmatrix-fracK+km fracK+km pmatrix -fracK+kmpmatrix - pmatrix
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