Wilberforce Pendulum
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:
A Wilberforce pulum consists of a mass on a helical spring. The longitudinal motion and the torsional motion are coupled through the spring i.e. if the pulum is started with a purely torsional motion it will start to move up and down. center includegraphicswidthcm#image_path:wilberforcpulum-# center The differential s for the motion of a Wilberforce pulum are given by mddot z & -k z -epsilontheta Jddot theta -kappatheta - epsilon z where m and J are the mass and moment of inertia of the pulum bob k and kappa the linear and rotational spring constant and epsilon the coupling constant for the coupling between the longitudinal and torsional motion. abcliste abc Derive the general coefficient matrix. abc For appropriate parameters the longitudinal and torsional frequencies are equal i.e. k/mkappa/J. Calculate the eigenvalues and verify that the eigenvectors are hat x pmatrix pmfracrsqrt pmatrix abc For a spring with kkO and kappakaO calculate the radius for which the longitudinal and torsional frequency are the same. Calculate the frequencies of the two normal modes for a coupling constant epsilonepsO and a cylinder with mass mO. abcliste
Solution:
abcliste abc The coefficient matrix is bf A pmatrix -frackm & -fracepsilonm -fracepsilonJ & -frackappaJ pmatrix abc Since the two angular frequencies are equal we can write them as omega^ frackm frackappaJ The trace and the determinant of the matrix are then tau -omega^ Delta omega^-fracepsilon^mJ It follows for the eigenvalues lambda -omega^ pm sqrtomega^-omega^+fracepsilon^mJ -omega^pmfracepsilonsqrtmJ The moment of inertia for a cylinder with mass m and radius r is J fracmr^ The eigenvalues are thus lambda -omega^ pm fracsqrtepsilonsqrtm^ r^ -omega^ pm fracsqrtepsilonmr abc From the equality of the angular frequencies omega^ frackm frackappaJ frackappamr^ it follows for the radius r rF sqrtfractimeskak resultrP- The angular frequencies for the first normal modes is omega_ sqrtomega^-fracsqrtepsilonmr sqrtfrackm-fracsqrtepsilonmr sqrtfrackm-fracepsilonmsqrtfrackkappa omaF sqrtfrackmleft-fracepssqrtka times kright resultomaP For the second normal mode we find omega_ sqrtomega^+fracsqrtepsilonmr ombF sqrtfrackmleft+fracepssqrtka times kright resultombP abcliste
A Wilberforce pulum consists of a mass on a helical spring. The longitudinal motion and the torsional motion are coupled through the spring i.e. if the pulum is started with a purely torsional motion it will start to move up and down. center includegraphicswidthcm#image_path:wilberforcpulum-# center The differential s for the motion of a Wilberforce pulum are given by mddot z & -k z -epsilontheta Jddot theta -kappatheta - epsilon z where m and J are the mass and moment of inertia of the pulum bob k and kappa the linear and rotational spring constant and epsilon the coupling constant for the coupling between the longitudinal and torsional motion. abcliste abc Derive the general coefficient matrix. abc For appropriate parameters the longitudinal and torsional frequencies are equal i.e. k/mkappa/J. Calculate the eigenvalues and verify that the eigenvectors are hat x pmatrix pmfracrsqrt pmatrix abc For a spring with kkO and kappakaO calculate the radius for which the longitudinal and torsional frequency are the same. Calculate the frequencies of the two normal modes for a coupling constant epsilonepsO and a cylinder with mass mO. abcliste
Solution:
abcliste abc The coefficient matrix is bf A pmatrix -frackm & -fracepsilonm -fracepsilonJ & -frackappaJ pmatrix abc Since the two angular frequencies are equal we can write them as omega^ frackm frackappaJ The trace and the determinant of the matrix are then tau -omega^ Delta omega^-fracepsilon^mJ It follows for the eigenvalues lambda -omega^ pm sqrtomega^-omega^+fracepsilon^mJ -omega^pmfracepsilonsqrtmJ The moment of inertia for a cylinder with mass m and radius r is J fracmr^ The eigenvalues are thus lambda -omega^ pm fracsqrtepsilonsqrtm^ r^ -omega^ pm fracsqrtepsilonmr abc From the equality of the angular frequencies omega^ frackm frackappaJ frackappamr^ it follows for the radius r rF sqrtfractimeskak resultrP- The angular frequencies for the first normal modes is omega_ sqrtomega^-fracsqrtepsilonmr sqrtfrackm-fracsqrtepsilonmr sqrtfrackm-fracepsilonmsqrtfrackkappa omaF sqrtfrackmleft-fracepssqrtka times kright resultomaP For the second normal mode we find omega_ sqrtomega^+fracsqrtepsilonmr ombF sqrtfrackmleft+fracepssqrtka times kright resultombP abcliste
Meta Information
Exercise:
A Wilberforce pulum consists of a mass on a helical spring. The longitudinal motion and the torsional motion are coupled through the spring i.e. if the pulum is started with a purely torsional motion it will start to move up and down. center includegraphicswidthcm#image_path:wilberforcpulum-# center The differential s for the motion of a Wilberforce pulum are given by mddot z & -k z -epsilontheta Jddot theta -kappatheta - epsilon z where m and J are the mass and moment of inertia of the pulum bob k and kappa the linear and rotational spring constant and epsilon the coupling constant for the coupling between the longitudinal and torsional motion. abcliste abc Derive the general coefficient matrix. abc For appropriate parameters the longitudinal and torsional frequencies are equal i.e. k/mkappa/J. Calculate the eigenvalues and verify that the eigenvectors are hat x pmatrix pmfracrsqrt pmatrix abc For a spring with kkO and kappakaO calculate the radius for which the longitudinal and torsional frequency are the same. Calculate the frequencies of the two normal modes for a coupling constant epsilonepsO and a cylinder with mass mO. abcliste
Solution:
abcliste abc The coefficient matrix is bf A pmatrix -frackm & -fracepsilonm -fracepsilonJ & -frackappaJ pmatrix abc Since the two angular frequencies are equal we can write them as omega^ frackm frackappaJ The trace and the determinant of the matrix are then tau -omega^ Delta omega^-fracepsilon^mJ It follows for the eigenvalues lambda -omega^ pm sqrtomega^-omega^+fracepsilon^mJ -omega^pmfracepsilonsqrtmJ The moment of inertia for a cylinder with mass m and radius r is J fracmr^ The eigenvalues are thus lambda -omega^ pm fracsqrtepsilonsqrtm^ r^ -omega^ pm fracsqrtepsilonmr abc From the equality of the angular frequencies omega^ frackm frackappaJ frackappamr^ it follows for the radius r rF sqrtfractimeskak resultrP- The angular frequencies for the first normal modes is omega_ sqrtomega^-fracsqrtepsilonmr sqrtfrackm-fracsqrtepsilonmr sqrtfrackm-fracepsilonmsqrtfrackkappa omaF sqrtfrackmleft-fracepssqrtka times kright resultomaP For the second normal mode we find omega_ sqrtomega^+fracsqrtepsilonmr ombF sqrtfrackmleft+fracepssqrtka times kright resultombP abcliste
A Wilberforce pulum consists of a mass on a helical spring. The longitudinal motion and the torsional motion are coupled through the spring i.e. if the pulum is started with a purely torsional motion it will start to move up and down. center includegraphicswidthcm#image_path:wilberforcpulum-# center The differential s for the motion of a Wilberforce pulum are given by mddot z & -k z -epsilontheta Jddot theta -kappatheta - epsilon z where m and J are the mass and moment of inertia of the pulum bob k and kappa the linear and rotational spring constant and epsilon the coupling constant for the coupling between the longitudinal and torsional motion. abcliste abc Derive the general coefficient matrix. abc For appropriate parameters the longitudinal and torsional frequencies are equal i.e. k/mkappa/J. Calculate the eigenvalues and verify that the eigenvectors are hat x pmatrix pmfracrsqrt pmatrix abc For a spring with kkO and kappakaO calculate the radius for which the longitudinal and torsional frequency are the same. Calculate the frequencies of the two normal modes for a coupling constant epsilonepsO and a cylinder with mass mO. abcliste
Solution:
abcliste abc The coefficient matrix is bf A pmatrix -frackm & -fracepsilonm -fracepsilonJ & -frackappaJ pmatrix abc Since the two angular frequencies are equal we can write them as omega^ frackm frackappaJ The trace and the determinant of the matrix are then tau -omega^ Delta omega^-fracepsilon^mJ It follows for the eigenvalues lambda -omega^ pm sqrtomega^-omega^+fracepsilon^mJ -omega^pmfracepsilonsqrtmJ The moment of inertia for a cylinder with mass m and radius r is J fracmr^ The eigenvalues are thus lambda -omega^ pm fracsqrtepsilonsqrtm^ r^ -omega^ pm fracsqrtepsilonmr abc From the equality of the angular frequencies omega^ frackm frackappaJ frackappamr^ it follows for the radius r rF sqrtfractimeskak resultrP- The angular frequencies for the first normal modes is omega_ sqrtomega^-fracsqrtepsilonmr sqrtfrackm-fracsqrtepsilonmr sqrtfrackm-fracepsilonmsqrtfrackkappa omaF sqrtfrackmleft-fracepssqrtka times kright resultomaP For the second normal mode we find omega_ sqrtomega^+fracsqrtepsilonmr ombF sqrtfrackmleft+fracepssqrtka times kright resultombP abcliste
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