Two Masses (One-Sided)
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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\(\LaTeX\)
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Exercise:
For the system of two coupled masses in the figure below: abcliste abc Derive the system of differential s and the corresponding coefficient matrix. abc Determine the eigenvalues for the case m_m_m and k_k_k and verify that the eigenvectors are hat x_ pmatrix -frac+sqrt pmatrix hat x_ pmatrix -frac-sqrt pmatrix abc Determine the eigenvalues and the corresponding angular frequencies for m_maO m_mbO k_kaO and k_kbO. abcliste center includegraphicswidthtextwidth#image_path:two-masses-onsided-# center
Solution:
abcliste abc The system of differential s is given by m_ ddot x_ -k_ x_+k_x_-x_ -k_+k_x_+k_ x_ m_ ddot x_ k_ x_ - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_m_ & frack_m_ frack_m_ & -frack_m_ pmatrix abc The coefficient matrix for m_m_m and k_k_k is bf A pmatrix-frackm & frackm frackm & -frackm pmatrix The trace and determinant of this matrix are tau -frackm Delta frack^m^-frack^m^frack^m^ and the eigenvalues are thus lambda fracmleft-kpmsqrtk^-k^right frac-pmsqrtfrackmfrac-pmsqrtomega_^ with omega_^k/m. vspacemm Applying the matrix bf A to the first eigenvector hat x_ yields bf Ahat x_ pmatrix-frackm & frackm frackm & -frackm pmatrix pmatrix -frac+sqrt pmatrix omega_^pmatrix+sqrt+ -frac+sqrt- pmatrix omega_^ pmatrix +sqrt -frac+sqrtpmatrix -frac+sqrt omega_^pmatrix-frac+sqrt pmatrix where we use the following identiy in the last step: frac+sqrtfrac+sqrt frac+sqrt+sqrt+ frac+sqrt+sqrt In the same way it can be verified that hat x_ is the eigenvector for the second eigenvalue. abc The coefficient matrix is given by bf A pmatrix aP & bP cP & dP pmatrix where the unit radiansquaredperssquared is suppressed. abcliste It follows for the eigenvalues lambda fraclefttaPpmsqrttaP^-timesdePright lambda_ resultlaaP lambda_ resultlabP and for the angular frequencies omega_ omaF resultomaP omega_ ombF resultombP
For the system of two coupled masses in the figure below: abcliste abc Derive the system of differential s and the corresponding coefficient matrix. abc Determine the eigenvalues for the case m_m_m and k_k_k and verify that the eigenvectors are hat x_ pmatrix -frac+sqrt pmatrix hat x_ pmatrix -frac-sqrt pmatrix abc Determine the eigenvalues and the corresponding angular frequencies for m_maO m_mbO k_kaO and k_kbO. abcliste center includegraphicswidthtextwidth#image_path:two-masses-onsided-# center
Solution:
abcliste abc The system of differential s is given by m_ ddot x_ -k_ x_+k_x_-x_ -k_+k_x_+k_ x_ m_ ddot x_ k_ x_ - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_m_ & frack_m_ frack_m_ & -frack_m_ pmatrix abc The coefficient matrix for m_m_m and k_k_k is bf A pmatrix-frackm & frackm frackm & -frackm pmatrix The trace and determinant of this matrix are tau -frackm Delta frack^m^-frack^m^frack^m^ and the eigenvalues are thus lambda fracmleft-kpmsqrtk^-k^right frac-pmsqrtfrackmfrac-pmsqrtomega_^ with omega_^k/m. vspacemm Applying the matrix bf A to the first eigenvector hat x_ yields bf Ahat x_ pmatrix-frackm & frackm frackm & -frackm pmatrix pmatrix -frac+sqrt pmatrix omega_^pmatrix+sqrt+ -frac+sqrt- pmatrix omega_^ pmatrix +sqrt -frac+sqrtpmatrix -frac+sqrt omega_^pmatrix-frac+sqrt pmatrix where we use the following identiy in the last step: frac+sqrtfrac+sqrt frac+sqrt+sqrt+ frac+sqrt+sqrt In the same way it can be verified that hat x_ is the eigenvector for the second eigenvalue. abc The coefficient matrix is given by bf A pmatrix aP & bP cP & dP pmatrix where the unit radiansquaredperssquared is suppressed. abcliste It follows for the eigenvalues lambda fraclefttaPpmsqrttaP^-timesdePright lambda_ resultlaaP lambda_ resultlabP and for the angular frequencies omega_ omaF resultomaP omega_ ombF resultombP
Meta Information
Exercise:
For the system of two coupled masses in the figure below: abcliste abc Derive the system of differential s and the corresponding coefficient matrix. abc Determine the eigenvalues for the case m_m_m and k_k_k and verify that the eigenvectors are hat x_ pmatrix -frac+sqrt pmatrix hat x_ pmatrix -frac-sqrt pmatrix abc Determine the eigenvalues and the corresponding angular frequencies for m_maO m_mbO k_kaO and k_kbO. abcliste center includegraphicswidthtextwidth#image_path:two-masses-onsided-# center
Solution:
abcliste abc The system of differential s is given by m_ ddot x_ -k_ x_+k_x_-x_ -k_+k_x_+k_ x_ m_ ddot x_ k_ x_ - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_m_ & frack_m_ frack_m_ & -frack_m_ pmatrix abc The coefficient matrix for m_m_m and k_k_k is bf A pmatrix-frackm & frackm frackm & -frackm pmatrix The trace and determinant of this matrix are tau -frackm Delta frack^m^-frack^m^frack^m^ and the eigenvalues are thus lambda fracmleft-kpmsqrtk^-k^right frac-pmsqrtfrackmfrac-pmsqrtomega_^ with omega_^k/m. vspacemm Applying the matrix bf A to the first eigenvector hat x_ yields bf Ahat x_ pmatrix-frackm & frackm frackm & -frackm pmatrix pmatrix -frac+sqrt pmatrix omega_^pmatrix+sqrt+ -frac+sqrt- pmatrix omega_^ pmatrix +sqrt -frac+sqrtpmatrix -frac+sqrt omega_^pmatrix-frac+sqrt pmatrix where we use the following identiy in the last step: frac+sqrtfrac+sqrt frac+sqrt+sqrt+ frac+sqrt+sqrt In the same way it can be verified that hat x_ is the eigenvector for the second eigenvalue. abc The coefficient matrix is given by bf A pmatrix aP & bP cP & dP pmatrix where the unit radiansquaredperssquared is suppressed. abcliste It follows for the eigenvalues lambda fraclefttaPpmsqrttaP^-timesdePright lambda_ resultlaaP lambda_ resultlabP and for the angular frequencies omega_ omaF resultomaP omega_ ombF resultombP
For the system of two coupled masses in the figure below: abcliste abc Derive the system of differential s and the corresponding coefficient matrix. abc Determine the eigenvalues for the case m_m_m and k_k_k and verify that the eigenvectors are hat x_ pmatrix -frac+sqrt pmatrix hat x_ pmatrix -frac-sqrt pmatrix abc Determine the eigenvalues and the corresponding angular frequencies for m_maO m_mbO k_kaO and k_kbO. abcliste center includegraphicswidthtextwidth#image_path:two-masses-onsided-# center
Solution:
abcliste abc The system of differential s is given by m_ ddot x_ -k_ x_+k_x_-x_ -k_+k_x_+k_ x_ m_ ddot x_ k_ x_ - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_m_ & frack_m_ frack_m_ & -frack_m_ pmatrix abc The coefficient matrix for m_m_m and k_k_k is bf A pmatrix-frackm & frackm frackm & -frackm pmatrix The trace and determinant of this matrix are tau -frackm Delta frack^m^-frack^m^frack^m^ and the eigenvalues are thus lambda fracmleft-kpmsqrtk^-k^right frac-pmsqrtfrackmfrac-pmsqrtomega_^ with omega_^k/m. vspacemm Applying the matrix bf A to the first eigenvector hat x_ yields bf Ahat x_ pmatrix-frackm & frackm frackm & -frackm pmatrix pmatrix -frac+sqrt pmatrix omega_^pmatrix+sqrt+ -frac+sqrt- pmatrix omega_^ pmatrix +sqrt -frac+sqrtpmatrix -frac+sqrt omega_^pmatrix-frac+sqrt pmatrix where we use the following identiy in the last step: frac+sqrtfrac+sqrt frac+sqrt+sqrt+ frac+sqrt+sqrt In the same way it can be verified that hat x_ is the eigenvector for the second eigenvalue. abc The coefficient matrix is given by bf A pmatrix aP & bP cP & dP pmatrix where the unit radiansquaredperssquared is suppressed. abcliste It follows for the eigenvalues lambda fraclefttaPpmsqrttaP^-timesdePright lambda_ resultlaaP lambda_ resultlabP and for the angular frequencies omega_ omaF resultomaP omega_ ombF resultombP
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