Masses & Springs
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
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Three Masses are connected to each other and the walls through springs with identical spring constants k. The masses are m left and right and Mm centre. center includegraphicswidthmm#image_path:masses-and-springs-# center The differential s for the three masses are given by * m ddot x_ -k x_ + k x_ - x_ M ddot x_ -k x_ - x_ + k x_ - x_ m ddot x_ -k x_ - x_ - k x_ The eigenvalues are * lambda_ -omega_^ lambda_ -fracpmsqrtomega_^ with omega_^k/m. abcliste abc Derive the expressions for the angular frequencies of the three normal modes and order them from lowest to highest frequency. Describe the behaviour of the system for a solution that corresponds to the superposition of the two fastest normal modes. hfill abc Describe the normal modes in words and assign them to the three eigenvalues. hfill abc Derive the characteristic polynomial for the eigenvalues of the coefficient matrix. Simplify it as much as possible and verify the eigenvalues given above. hfill abcliste
Solution:
abcliste abc The angular frequencies follow directly from the eigenvalues: omega sqrt-lambda Longrightarrow omega_ sqrtomega_ approx omaPtimesomega_ omega_ sqrtfrac+sqrtomega_ approx ombPtimesomega_ omega_ sqrtfrac-sqrtomega_ approx omcPtimesomega_ It follows that omega_omega_omega_. vspacemm textbfRemark: The normal frequencies omega_ and omega_ can also be simplified to get rid of the double root: omega_ fracsqrt+ omega_ fracsqrt- This makes it easier to see the relations between the three normal frequencies. vspacemm For a superposition of normal modes and angular frequencies omega_ and omega_ the energy is transferred between the two modes at a frequency corresponding to the difference of omega_ and omega_ beats. abc The slowest mode omega_ corresponds to all three masses moving in sync amplitude of M larger than of m. Normal mode omega_ corresponds to the two smaller masses moving in opposite directions with equal amplitudes and the middle mass staying at rest. In the fastest mode the outer masses move in sync while the middle mass moves in the opposite direction. abc The coefficient matrix is bf M pmatrix -omega_^ & omega_^ & omega_^/ & -omega_^ & omega_^/ & omega_^ & - omega_^ pmatrix The characteristic polynomial for the eigenvalues is given by detM-lambda I abspmatrix -omega_^-lambda & omega_^ & omega_^/ & -omega_^ - lambda & omega_^/ & omega_^ & - omega_^ - lambda pmatrix -omega_^+lambda^omega_^+lambda+fracomega_^omega_^+lambda -omega_^+omega_^ lambda + lambda^omega_^+lambda+omega_^+omega_^ lambda -omega_^-omega_^ lambda - omega_^ lambda^ - lambda^ To verify whether the given eigenvalues are correct we can calculate the resulting polynomial: lambda + omega_^&lambda + frac+sqrt omega_^lambda + frac-sqrt omega_^ lambda+omega_^lambda^+omega_^lambda+omega_^ lambda^ + omega_^lambda^ + omega_^ lambda + omega_^ This is the same es the expression above but for an irrelevant global sign. abcliste
Three Masses are connected to each other and the walls through springs with identical spring constants k. The masses are m left and right and Mm centre. center includegraphicswidthmm#image_path:masses-and-springs-# center The differential s for the three masses are given by * m ddot x_ -k x_ + k x_ - x_ M ddot x_ -k x_ - x_ + k x_ - x_ m ddot x_ -k x_ - x_ - k x_ The eigenvalues are * lambda_ -omega_^ lambda_ -fracpmsqrtomega_^ with omega_^k/m. abcliste abc Derive the expressions for the angular frequencies of the three normal modes and order them from lowest to highest frequency. Describe the behaviour of the system for a solution that corresponds to the superposition of the two fastest normal modes. hfill abc Describe the normal modes in words and assign them to the three eigenvalues. hfill abc Derive the characteristic polynomial for the eigenvalues of the coefficient matrix. Simplify it as much as possible and verify the eigenvalues given above. hfill abcliste
Solution:
abcliste abc The angular frequencies follow directly from the eigenvalues: omega sqrt-lambda Longrightarrow omega_ sqrtomega_ approx omaPtimesomega_ omega_ sqrtfrac+sqrtomega_ approx ombPtimesomega_ omega_ sqrtfrac-sqrtomega_ approx omcPtimesomega_ It follows that omega_omega_omega_. vspacemm textbfRemark: The normal frequencies omega_ and omega_ can also be simplified to get rid of the double root: omega_ fracsqrt+ omega_ fracsqrt- This makes it easier to see the relations between the three normal frequencies. vspacemm For a superposition of normal modes and angular frequencies omega_ and omega_ the energy is transferred between the two modes at a frequency corresponding to the difference of omega_ and omega_ beats. abc The slowest mode omega_ corresponds to all three masses moving in sync amplitude of M larger than of m. Normal mode omega_ corresponds to the two smaller masses moving in opposite directions with equal amplitudes and the middle mass staying at rest. In the fastest mode the outer masses move in sync while the middle mass moves in the opposite direction. abc The coefficient matrix is bf M pmatrix -omega_^ & omega_^ & omega_^/ & -omega_^ & omega_^/ & omega_^ & - omega_^ pmatrix The characteristic polynomial for the eigenvalues is given by detM-lambda I abspmatrix -omega_^-lambda & omega_^ & omega_^/ & -omega_^ - lambda & omega_^/ & omega_^ & - omega_^ - lambda pmatrix -omega_^+lambda^omega_^+lambda+fracomega_^omega_^+lambda -omega_^+omega_^ lambda + lambda^omega_^+lambda+omega_^+omega_^ lambda -omega_^-omega_^ lambda - omega_^ lambda^ - lambda^ To verify whether the given eigenvalues are correct we can calculate the resulting polynomial: lambda + omega_^&lambda + frac+sqrt omega_^lambda + frac-sqrt omega_^ lambda+omega_^lambda^+omega_^lambda+omega_^ lambda^ + omega_^lambda^ + omega_^ lambda + omega_^ This is the same es the expression above but for an irrelevant global sign. abcliste
Meta Information
Exercise:
Three Masses are connected to each other and the walls through springs with identical spring constants k. The masses are m left and right and Mm centre. center includegraphicswidthmm#image_path:masses-and-springs-# center The differential s for the three masses are given by * m ddot x_ -k x_ + k x_ - x_ M ddot x_ -k x_ - x_ + k x_ - x_ m ddot x_ -k x_ - x_ - k x_ The eigenvalues are * lambda_ -omega_^ lambda_ -fracpmsqrtomega_^ with omega_^k/m. abcliste abc Derive the expressions for the angular frequencies of the three normal modes and order them from lowest to highest frequency. Describe the behaviour of the system for a solution that corresponds to the superposition of the two fastest normal modes. hfill abc Describe the normal modes in words and assign them to the three eigenvalues. hfill abc Derive the characteristic polynomial for the eigenvalues of the coefficient matrix. Simplify it as much as possible and verify the eigenvalues given above. hfill abcliste
Solution:
abcliste abc The angular frequencies follow directly from the eigenvalues: omega sqrt-lambda Longrightarrow omega_ sqrtomega_ approx omaPtimesomega_ omega_ sqrtfrac+sqrtomega_ approx ombPtimesomega_ omega_ sqrtfrac-sqrtomega_ approx omcPtimesomega_ It follows that omega_omega_omega_. vspacemm textbfRemark: The normal frequencies omega_ and omega_ can also be simplified to get rid of the double root: omega_ fracsqrt+ omega_ fracsqrt- This makes it easier to see the relations between the three normal frequencies. vspacemm For a superposition of normal modes and angular frequencies omega_ and omega_ the energy is transferred between the two modes at a frequency corresponding to the difference of omega_ and omega_ beats. abc The slowest mode omega_ corresponds to all three masses moving in sync amplitude of M larger than of m. Normal mode omega_ corresponds to the two smaller masses moving in opposite directions with equal amplitudes and the middle mass staying at rest. In the fastest mode the outer masses move in sync while the middle mass moves in the opposite direction. abc The coefficient matrix is bf M pmatrix -omega_^ & omega_^ & omega_^/ & -omega_^ & omega_^/ & omega_^ & - omega_^ pmatrix The characteristic polynomial for the eigenvalues is given by detM-lambda I abspmatrix -omega_^-lambda & omega_^ & omega_^/ & -omega_^ - lambda & omega_^/ & omega_^ & - omega_^ - lambda pmatrix -omega_^+lambda^omega_^+lambda+fracomega_^omega_^+lambda -omega_^+omega_^ lambda + lambda^omega_^+lambda+omega_^+omega_^ lambda -omega_^-omega_^ lambda - omega_^ lambda^ - lambda^ To verify whether the given eigenvalues are correct we can calculate the resulting polynomial: lambda + omega_^&lambda + frac+sqrt omega_^lambda + frac-sqrt omega_^ lambda+omega_^lambda^+omega_^lambda+omega_^ lambda^ + omega_^lambda^ + omega_^ lambda + omega_^ This is the same es the expression above but for an irrelevant global sign. abcliste
Three Masses are connected to each other and the walls through springs with identical spring constants k. The masses are m left and right and Mm centre. center includegraphicswidthmm#image_path:masses-and-springs-# center The differential s for the three masses are given by * m ddot x_ -k x_ + k x_ - x_ M ddot x_ -k x_ - x_ + k x_ - x_ m ddot x_ -k x_ - x_ - k x_ The eigenvalues are * lambda_ -omega_^ lambda_ -fracpmsqrtomega_^ with omega_^k/m. abcliste abc Derive the expressions for the angular frequencies of the three normal modes and order them from lowest to highest frequency. Describe the behaviour of the system for a solution that corresponds to the superposition of the two fastest normal modes. hfill abc Describe the normal modes in words and assign them to the three eigenvalues. hfill abc Derive the characteristic polynomial for the eigenvalues of the coefficient matrix. Simplify it as much as possible and verify the eigenvalues given above. hfill abcliste
Solution:
abcliste abc The angular frequencies follow directly from the eigenvalues: omega sqrt-lambda Longrightarrow omega_ sqrtomega_ approx omaPtimesomega_ omega_ sqrtfrac+sqrtomega_ approx ombPtimesomega_ omega_ sqrtfrac-sqrtomega_ approx omcPtimesomega_ It follows that omega_omega_omega_. vspacemm textbfRemark: The normal frequencies omega_ and omega_ can also be simplified to get rid of the double root: omega_ fracsqrt+ omega_ fracsqrt- This makes it easier to see the relations between the three normal frequencies. vspacemm For a superposition of normal modes and angular frequencies omega_ and omega_ the energy is transferred between the two modes at a frequency corresponding to the difference of omega_ and omega_ beats. abc The slowest mode omega_ corresponds to all three masses moving in sync amplitude of M larger than of m. Normal mode omega_ corresponds to the two smaller masses moving in opposite directions with equal amplitudes and the middle mass staying at rest. In the fastest mode the outer masses move in sync while the middle mass moves in the opposite direction. abc The coefficient matrix is bf M pmatrix -omega_^ & omega_^ & omega_^/ & -omega_^ & omega_^/ & omega_^ & - omega_^ pmatrix The characteristic polynomial for the eigenvalues is given by detM-lambda I abspmatrix -omega_^-lambda & omega_^ & omega_^/ & -omega_^ - lambda & omega_^/ & omega_^ & - omega_^ - lambda pmatrix -omega_^+lambda^omega_^+lambda+fracomega_^omega_^+lambda -omega_^+omega_^ lambda + lambda^omega_^+lambda+omega_^+omega_^ lambda -omega_^-omega_^ lambda - omega_^ lambda^ - lambda^ To verify whether the given eigenvalues are correct we can calculate the resulting polynomial: lambda + omega_^&lambda + frac+sqrt omega_^lambda + frac-sqrt omega_^ lambda+omega_^lambda^+omega_^lambda+omega_^ lambda^ + omega_^lambda^ + omega_^ lambda + omega_^ This is the same es the expression above but for an irrelevant global sign. abcliste
Contained in these collections:

