Strange Spring
About points...
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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
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Exercise:
The differential for a mass susped from an unusual spring is given by the coefficient matrix * mathbfM pmatrix & frackm & - delta pmatrix with the spring constant k the mass m and the damping coefficient delta all being positive constants. vspacemm This corresponds to a spring where the force is proportional to the displacement from the equilibrium position but poing away ! from the equlibrium. abcliste abc Derive the correponding second-order differential for the motion of the mass on the spring. hfill abc Derive the formal expressions for the eigenvalues of the coefficient matrix and discuss the stability of the fixed po at the origin. hfill abc Show that for delta the eigenvalues and eigenvectors are * lambda_ omega_ quad &textrmand quad lambda_ -omega_ vec v_ pmatrix omega_ omega_^ pmatrix quad &textrmand quad vec v_ pmatrix -omega_ omega_^ pmatrix with omega_sqrtk/m. hfill abc For the undamped case delta derive the solution with initial conditions hfill * y A dot y abcliste
Solution:
abcliste abc The system of first-order differential s is dot y v_y dot v_y frackm y - delta v_y As a second-order differential this corresponds to ddot y frackm y - delta dot y abc The trace and determinant are tau -delta Delta - frackm -frackm It follows for the eigenvalues lambda_ fractaupmsqrttau^-Delta frac-delta pm sqrtdelta^ + frackm -delta pm sqrtdelta^+frackm Since sqrtdelta^+frackmadelta the eigenvalue lambda_-delta+sqrtdelta^+frackm is positive while lambda_-delta-sqrtdelta^+frackm is negative. The fixed po is therefore a saddle po. abc We have to show that mathbfM vec v_i lambda_ivec v_i quad textrmfor iin For i we have mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix omega_ omega_^ pmatrix pmatrix omega_^ omega_^ pmatrix omega_ pmatrix omega_ omega_^ pmatrix lambda_ vec v_ and for i mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix -omega_ omega_^ pmatrix pmatrix omega_^ -omega_^ pmatrix -omega_ pmatrix -omega_ omega_^ pmatrix lambda_ vec v_ textbfRemark: Since eigenvectors are only defined up to a scalar we could also use the eigenvectors vec u_ pmatrix omega_ pmatrix quad textrmand quad vec u_ pmatrix - omega_ pmatrix to simplify the calculations. abc 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 omega_ omega_^ pmatrix e^omega_ t + a_ pmatrix -omega_ omega_^ pmatrix e^-omega_ t The coefficients a_ and a_ follow from the initial conditions: pmatrix A pmatrix a_ pmatrix omega_ omega_^ pmatrix + a_ pmatrix -omega_ omega_^ pmatrix The second line yields omega_^ a_ + a_ Longrightarrow a_ -a_ The first line then yields A omega_ a_ - a_ omega_ a_ Longrightarrow a_ -a_ fracAomega_ The solution can then be written as yt fracAomega_leftomega_ e^omega_ t +omega_ e^-omega_ t right fracA left e^omega_ t + e^-omega_ tright A coshomega_ t v_yt fracAomega_omega_^ lefte^omega_ t-e^-omega_ t right fracA omega_ lefte^omega_ t-e^-omega_ t right A omega_ sinhomega_ t abcliste
The differential for a mass susped from an unusual spring is given by the coefficient matrix * mathbfM pmatrix & frackm & - delta pmatrix with the spring constant k the mass m and the damping coefficient delta all being positive constants. vspacemm This corresponds to a spring where the force is proportional to the displacement from the equilibrium position but poing away ! from the equlibrium. abcliste abc Derive the correponding second-order differential for the motion of the mass on the spring. hfill abc Derive the formal expressions for the eigenvalues of the coefficient matrix and discuss the stability of the fixed po at the origin. hfill abc Show that for delta the eigenvalues and eigenvectors are * lambda_ omega_ quad &textrmand quad lambda_ -omega_ vec v_ pmatrix omega_ omega_^ pmatrix quad &textrmand quad vec v_ pmatrix -omega_ omega_^ pmatrix with omega_sqrtk/m. hfill abc For the undamped case delta derive the solution with initial conditions hfill * y A dot y abcliste
Solution:
abcliste abc The system of first-order differential s is dot y v_y dot v_y frackm y - delta v_y As a second-order differential this corresponds to ddot y frackm y - delta dot y abc The trace and determinant are tau -delta Delta - frackm -frackm It follows for the eigenvalues lambda_ fractaupmsqrttau^-Delta frac-delta pm sqrtdelta^ + frackm -delta pm sqrtdelta^+frackm Since sqrtdelta^+frackmadelta the eigenvalue lambda_-delta+sqrtdelta^+frackm is positive while lambda_-delta-sqrtdelta^+frackm is negative. The fixed po is therefore a saddle po. abc We have to show that mathbfM vec v_i lambda_ivec v_i quad textrmfor iin For i we have mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix omega_ omega_^ pmatrix pmatrix omega_^ omega_^ pmatrix omega_ pmatrix omega_ omega_^ pmatrix lambda_ vec v_ and for i mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix -omega_ omega_^ pmatrix pmatrix omega_^ -omega_^ pmatrix -omega_ pmatrix -omega_ omega_^ pmatrix lambda_ vec v_ textbfRemark: Since eigenvectors are only defined up to a scalar we could also use the eigenvectors vec u_ pmatrix omega_ pmatrix quad textrmand quad vec u_ pmatrix - omega_ pmatrix to simplify the calculations. abc 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 omega_ omega_^ pmatrix e^omega_ t + a_ pmatrix -omega_ omega_^ pmatrix e^-omega_ t The coefficients a_ and a_ follow from the initial conditions: pmatrix A pmatrix a_ pmatrix omega_ omega_^ pmatrix + a_ pmatrix -omega_ omega_^ pmatrix The second line yields omega_^ a_ + a_ Longrightarrow a_ -a_ The first line then yields A omega_ a_ - a_ omega_ a_ Longrightarrow a_ -a_ fracAomega_ The solution can then be written as yt fracAomega_leftomega_ e^omega_ t +omega_ e^-omega_ t right fracA left e^omega_ t + e^-omega_ tright A coshomega_ t v_yt fracAomega_omega_^ lefte^omega_ t-e^-omega_ t right fracA omega_ lefte^omega_ t-e^-omega_ t right A omega_ sinhomega_ t abcliste
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Exercise:
The differential for a mass susped from an unusual spring is given by the coefficient matrix * mathbfM pmatrix & frackm & - delta pmatrix with the spring constant k the mass m and the damping coefficient delta all being positive constants. vspacemm This corresponds to a spring where the force is proportional to the displacement from the equilibrium position but poing away ! from the equlibrium. abcliste abc Derive the correponding second-order differential for the motion of the mass on the spring. hfill abc Derive the formal expressions for the eigenvalues of the coefficient matrix and discuss the stability of the fixed po at the origin. hfill abc Show that for delta the eigenvalues and eigenvectors are * lambda_ omega_ quad &textrmand quad lambda_ -omega_ vec v_ pmatrix omega_ omega_^ pmatrix quad &textrmand quad vec v_ pmatrix -omega_ omega_^ pmatrix with omega_sqrtk/m. hfill abc For the undamped case delta derive the solution with initial conditions hfill * y A dot y abcliste
Solution:
abcliste abc The system of first-order differential s is dot y v_y dot v_y frackm y - delta v_y As a second-order differential this corresponds to ddot y frackm y - delta dot y abc The trace and determinant are tau -delta Delta - frackm -frackm It follows for the eigenvalues lambda_ fractaupmsqrttau^-Delta frac-delta pm sqrtdelta^ + frackm -delta pm sqrtdelta^+frackm Since sqrtdelta^+frackmadelta the eigenvalue lambda_-delta+sqrtdelta^+frackm is positive while lambda_-delta-sqrtdelta^+frackm is negative. The fixed po is therefore a saddle po. abc We have to show that mathbfM vec v_i lambda_ivec v_i quad textrmfor iin For i we have mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix omega_ omega_^ pmatrix pmatrix omega_^ omega_^ pmatrix omega_ pmatrix omega_ omega_^ pmatrix lambda_ vec v_ and for i mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix -omega_ omega_^ pmatrix pmatrix omega_^ -omega_^ pmatrix -omega_ pmatrix -omega_ omega_^ pmatrix lambda_ vec v_ textbfRemark: Since eigenvectors are only defined up to a scalar we could also use the eigenvectors vec u_ pmatrix omega_ pmatrix quad textrmand quad vec u_ pmatrix - omega_ pmatrix to simplify the calculations. abc 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 omega_ omega_^ pmatrix e^omega_ t + a_ pmatrix -omega_ omega_^ pmatrix e^-omega_ t The coefficients a_ and a_ follow from the initial conditions: pmatrix A pmatrix a_ pmatrix omega_ omega_^ pmatrix + a_ pmatrix -omega_ omega_^ pmatrix The second line yields omega_^ a_ + a_ Longrightarrow a_ -a_ The first line then yields A omega_ a_ - a_ omega_ a_ Longrightarrow a_ -a_ fracAomega_ The solution can then be written as yt fracAomega_leftomega_ e^omega_ t +omega_ e^-omega_ t right fracA left e^omega_ t + e^-omega_ tright A coshomega_ t v_yt fracAomega_omega_^ lefte^omega_ t-e^-omega_ t right fracA omega_ lefte^omega_ t-e^-omega_ t right A omega_ sinhomega_ t abcliste
The differential for a mass susped from an unusual spring is given by the coefficient matrix * mathbfM pmatrix & frackm & - delta pmatrix with the spring constant k the mass m and the damping coefficient delta all being positive constants. vspacemm This corresponds to a spring where the force is proportional to the displacement from the equilibrium position but poing away ! from the equlibrium. abcliste abc Derive the correponding second-order differential for the motion of the mass on the spring. hfill abc Derive the formal expressions for the eigenvalues of the coefficient matrix and discuss the stability of the fixed po at the origin. hfill abc Show that for delta the eigenvalues and eigenvectors are * lambda_ omega_ quad &textrmand quad lambda_ -omega_ vec v_ pmatrix omega_ omega_^ pmatrix quad &textrmand quad vec v_ pmatrix -omega_ omega_^ pmatrix with omega_sqrtk/m. hfill abc For the undamped case delta derive the solution with initial conditions hfill * y A dot y abcliste
Solution:
abcliste abc The system of first-order differential s is dot y v_y dot v_y frackm y - delta v_y As a second-order differential this corresponds to ddot y frackm y - delta dot y abc The trace and determinant are tau -delta Delta - frackm -frackm It follows for the eigenvalues lambda_ fractaupmsqrttau^-Delta frac-delta pm sqrtdelta^ + frackm -delta pm sqrtdelta^+frackm Since sqrtdelta^+frackmadelta the eigenvalue lambda_-delta+sqrtdelta^+frackm is positive while lambda_-delta-sqrtdelta^+frackm is negative. The fixed po is therefore a saddle po. abc We have to show that mathbfM vec v_i lambda_ivec v_i quad textrmfor iin For i we have mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix omega_ omega_^ pmatrix pmatrix omega_^ omega_^ pmatrix omega_ pmatrix omega_ omega_^ pmatrix lambda_ vec v_ and for i mathbfM vec v_ pmatrix & omega_^ & pmatrix pmatrix -omega_ omega_^ pmatrix pmatrix omega_^ -omega_^ pmatrix -omega_ pmatrix -omega_ omega_^ pmatrix lambda_ vec v_ textbfRemark: Since eigenvectors are only defined up to a scalar we could also use the eigenvectors vec u_ pmatrix omega_ pmatrix quad textrmand quad vec u_ pmatrix - omega_ pmatrix to simplify the calculations. abc 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 omega_ omega_^ pmatrix e^omega_ t + a_ pmatrix -omega_ omega_^ pmatrix e^-omega_ t The coefficients a_ and a_ follow from the initial conditions: pmatrix A pmatrix a_ pmatrix omega_ omega_^ pmatrix + a_ pmatrix -omega_ omega_^ pmatrix The second line yields omega_^ a_ + a_ Longrightarrow a_ -a_ The first line then yields A omega_ a_ - a_ omega_ a_ Longrightarrow a_ -a_ fracAomega_ The solution can then be written as yt fracAomega_leftomega_ e^omega_ t +omega_ e^-omega_ t right fracA left e^omega_ t + e^-omega_ tright A coshomega_ t v_yt fracAomega_omega_^ lefte^omega_ t-e^-omega_ t right fracA omega_ lefte^omega_ t-e^-omega_ t right A omega_ sinhomega_ t abcliste
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