Large and Small Blocks
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:
Three masses M m_ and m_ are coupled with springs with elastic constants k_ and k_ as shown in the figure. center includegraphicswidthcm#image_path:largand-small-blocks-# center abcliste abc Derive the differential s and the corresponding coefficient matrix. abc Derive the eigenvalues and eigenvectors for the special case with m_m_m and k_k_. Describe the normal modes. abcliste
Solution:
abcliste abc The differential s are M ddot x k_ x_-x + k_ x_-x -k_+k_ x + k_ x_ + k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_M & frack_M & frack_M frack_m_ & -frack_m_ & frack_m_ & & -frack_m_ pmatrix abc The coefficient matrix for the special case is bf A pmatrix -frackM & frackM & frackM frackm & -frackm & frackm & & -frackm pmatrix The eigenvalues are the zeros of the characteristic polynomial: det pmatrix -frack+MlambdaM & frackM & frackM frackm & -frack+mlambdam & frackm & & -frack+mlambdam pmatrix frac-k+Mlambdak+mlambda^+k^k+mlambdaMm^ frac-k+Mlambdak^+kmlambda+m^lambda^+k^+k^mlambdaMm^ fraccancel-k^-k^mlambda-km^lambda^-k^Mlambda-kMmlambda^-Mm^lambda^cancel+k^+k^mlambdaMm^ Longrightarrow -lambdaleftMm^lambda^+kmM+mlambda+k^ M+mright The trivial eigenvalue is lambda_ and the nontrivial eigenvalues are the solutions of the quadratic Mm^lambda^+kmM+mlambda+k^ M+m so lambda_ frac-kmM+m pm sqrtk^m^M+m^-k^Mm^M+mMm^ -frackMmleftM+m pm sqrtM+m^-MM+mright -frackMmleftM+m pm sqrtM^+Mm+m^-M^-Mmright -frackM+mpm mMm Longrightarrow lambda_ -frackm lambda_ -frackM+mMm The angular frequencies of the normal modes are omega_ omega_ sqrtfrackm omega_ sqrtfrackM+mMm and the corresponding eigenvectors vecv_ pmatrix pmatrix vecv_ pmatrix - pmatrix vecv_ pmatrix -fracmM pmatrix The first normal mode is not an oscillation but a uniform motion of all three masses. In the second normal mode the large mass is at rest while the small masses oscillate in opposite directions. It makes sense that the frequency does not dep on the large mass M. In the third normal mode the two small masses oscillate in sync and in the opposite direction of the large mass. The amplitudes are such that the centre of mass of the system remains at rest. abcliste
Three masses M m_ and m_ are coupled with springs with elastic constants k_ and k_ as shown in the figure. center includegraphicswidthcm#image_path:largand-small-blocks-# center abcliste abc Derive the differential s and the corresponding coefficient matrix. abc Derive the eigenvalues and eigenvectors for the special case with m_m_m and k_k_. Describe the normal modes. abcliste
Solution:
abcliste abc The differential s are M ddot x k_ x_-x + k_ x_-x -k_+k_ x + k_ x_ + k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_M & frack_M & frack_M frack_m_ & -frack_m_ & frack_m_ & & -frack_m_ pmatrix abc The coefficient matrix for the special case is bf A pmatrix -frackM & frackM & frackM frackm & -frackm & frackm & & -frackm pmatrix The eigenvalues are the zeros of the characteristic polynomial: det pmatrix -frack+MlambdaM & frackM & frackM frackm & -frack+mlambdam & frackm & & -frack+mlambdam pmatrix frac-k+Mlambdak+mlambda^+k^k+mlambdaMm^ frac-k+Mlambdak^+kmlambda+m^lambda^+k^+k^mlambdaMm^ fraccancel-k^-k^mlambda-km^lambda^-k^Mlambda-kMmlambda^-Mm^lambda^cancel+k^+k^mlambdaMm^ Longrightarrow -lambdaleftMm^lambda^+kmM+mlambda+k^ M+mright The trivial eigenvalue is lambda_ and the nontrivial eigenvalues are the solutions of the quadratic Mm^lambda^+kmM+mlambda+k^ M+m so lambda_ frac-kmM+m pm sqrtk^m^M+m^-k^Mm^M+mMm^ -frackMmleftM+m pm sqrtM+m^-MM+mright -frackMmleftM+m pm sqrtM^+Mm+m^-M^-Mmright -frackM+mpm mMm Longrightarrow lambda_ -frackm lambda_ -frackM+mMm The angular frequencies of the normal modes are omega_ omega_ sqrtfrackm omega_ sqrtfrackM+mMm and the corresponding eigenvectors vecv_ pmatrix pmatrix vecv_ pmatrix - pmatrix vecv_ pmatrix -fracmM pmatrix The first normal mode is not an oscillation but a uniform motion of all three masses. In the second normal mode the large mass is at rest while the small masses oscillate in opposite directions. It makes sense that the frequency does not dep on the large mass M. In the third normal mode the two small masses oscillate in sync and in the opposite direction of the large mass. The amplitudes are such that the centre of mass of the system remains at rest. abcliste
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Exercise:
Three masses M m_ and m_ are coupled with springs with elastic constants k_ and k_ as shown in the figure. center includegraphicswidthcm#image_path:largand-small-blocks-# center abcliste abc Derive the differential s and the corresponding coefficient matrix. abc Derive the eigenvalues and eigenvectors for the special case with m_m_m and k_k_. Describe the normal modes. abcliste
Solution:
abcliste abc The differential s are M ddot x k_ x_-x + k_ x_-x -k_+k_ x + k_ x_ + k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_M & frack_M & frack_M frack_m_ & -frack_m_ & frack_m_ & & -frack_m_ pmatrix abc The coefficient matrix for the special case is bf A pmatrix -frackM & frackM & frackM frackm & -frackm & frackm & & -frackm pmatrix The eigenvalues are the zeros of the characteristic polynomial: det pmatrix -frack+MlambdaM & frackM & frackM frackm & -frack+mlambdam & frackm & & -frack+mlambdam pmatrix frac-k+Mlambdak+mlambda^+k^k+mlambdaMm^ frac-k+Mlambdak^+kmlambda+m^lambda^+k^+k^mlambdaMm^ fraccancel-k^-k^mlambda-km^lambda^-k^Mlambda-kMmlambda^-Mm^lambda^cancel+k^+k^mlambdaMm^ Longrightarrow -lambdaleftMm^lambda^+kmM+mlambda+k^ M+mright The trivial eigenvalue is lambda_ and the nontrivial eigenvalues are the solutions of the quadratic Mm^lambda^+kmM+mlambda+k^ M+m so lambda_ frac-kmM+m pm sqrtk^m^M+m^-k^Mm^M+mMm^ -frackMmleftM+m pm sqrtM+m^-MM+mright -frackMmleftM+m pm sqrtM^+Mm+m^-M^-Mmright -frackM+mpm mMm Longrightarrow lambda_ -frackm lambda_ -frackM+mMm The angular frequencies of the normal modes are omega_ omega_ sqrtfrackm omega_ sqrtfrackM+mMm and the corresponding eigenvectors vecv_ pmatrix pmatrix vecv_ pmatrix - pmatrix vecv_ pmatrix -fracmM pmatrix The first normal mode is not an oscillation but a uniform motion of all three masses. In the second normal mode the large mass is at rest while the small masses oscillate in opposite directions. It makes sense that the frequency does not dep on the large mass M. In the third normal mode the two small masses oscillate in sync and in the opposite direction of the large mass. The amplitudes are such that the centre of mass of the system remains at rest. abcliste
Three masses M m_ and m_ are coupled with springs with elastic constants k_ and k_ as shown in the figure. center includegraphicswidthcm#image_path:largand-small-blocks-# center abcliste abc Derive the differential s and the corresponding coefficient matrix. abc Derive the eigenvalues and eigenvectors for the special case with m_m_m and k_k_. Describe the normal modes. abcliste
Solution:
abcliste abc The differential s are M ddot x k_ x_-x + k_ x_-x -k_+k_ x + k_ x_ + k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ m_ ddot x_ -k_ x_-x k_ x - k_ x_ It follows for the coefficient matrix bf A pmatrix -frack_+k_M & frack_M & frack_M frack_m_ & -frack_m_ & frack_m_ & & -frack_m_ pmatrix abc The coefficient matrix for the special case is bf A pmatrix -frackM & frackM & frackM frackm & -frackm & frackm & & -frackm pmatrix The eigenvalues are the zeros of the characteristic polynomial: det pmatrix -frack+MlambdaM & frackM & frackM frackm & -frack+mlambdam & frackm & & -frack+mlambdam pmatrix frac-k+Mlambdak+mlambda^+k^k+mlambdaMm^ frac-k+Mlambdak^+kmlambda+m^lambda^+k^+k^mlambdaMm^ fraccancel-k^-k^mlambda-km^lambda^-k^Mlambda-kMmlambda^-Mm^lambda^cancel+k^+k^mlambdaMm^ Longrightarrow -lambdaleftMm^lambda^+kmM+mlambda+k^ M+mright The trivial eigenvalue is lambda_ and the nontrivial eigenvalues are the solutions of the quadratic Mm^lambda^+kmM+mlambda+k^ M+m so lambda_ frac-kmM+m pm sqrtk^m^M+m^-k^Mm^M+mMm^ -frackMmleftM+m pm sqrtM+m^-MM+mright -frackMmleftM+m pm sqrtM^+Mm+m^-M^-Mmright -frackM+mpm mMm Longrightarrow lambda_ -frackm lambda_ -frackM+mMm The angular frequencies of the normal modes are omega_ omega_ sqrtfrackm omega_ sqrtfrackM+mMm and the corresponding eigenvectors vecv_ pmatrix pmatrix vecv_ pmatrix - pmatrix vecv_ pmatrix -fracmM pmatrix The first normal mode is not an oscillation but a uniform motion of all three masses. In the second normal mode the large mass is at rest while the small masses oscillate in opposite directions. It makes sense that the frequency does not dep on the large mass M. In the third normal mode the two small masses oscillate in sync and in the opposite direction of the large mass. The amplitudes are such that the centre of mass of the system remains at rest. abcliste
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