CO2 Molecule
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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.
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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
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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 -
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Exercise:
The molecular vibrations of a linear triatomic molecule e.g. mathrmCO_ can be modelled as a system of three masses coupled through springs see figure. center includegraphicswidth.cm#image_path:co-molecul# center abcliste abc Derive the differential s for the longitudinal vibrations and the corresponding coefficient matrix. abc Determine the normale modes of oscillation. abc Calculate the ratio of the two nontrivial frequencies. abcliste
Solution:
abcliste abc The differential s for the three masses left to right are M ddot x_ kx_-x_ -k x_ + k x_ m ddot x_ -kx_-x_+kx_-x_ k x_ - k x_ + k x_ M ddot x_ -kx_-x_ k x_ -k x_ where M is the mass of an oxygen and m the mass of a carbon atom. The coefficient matrix is thus bf A pmatrix -frackM & frackM & frackm & -frackm & frackm & frackM & -frackM pmatrix abc The eigenvalues are the zeros of the characteristic polynomial: detbf A-lambda bf I pmatrix -frack+lambda MM & frackM & frackm & -frack+lambda mm & frackm & frackM & -frack+lambda MM pmatrix frac-k+lambda M^k+lambda m+k^k+lambda MM^ m Longrightarrow k^+k^ M lambda & quad - k^+k M lambda+M^lambda^k+lambda m k^+k^ M lambda & quad -k^-k^ M lambda-k M^ lambda^ & quad -k^ m lambda - k M m lambda^ - M^ m lambda^ -lambda k^ M+k^ m + k M^ lambda+k M m lambda + M^ m lambda^ -lambda leftk^M+m+k MM+mlambda+M^ m lambda^right The first zero is trivial: lambda_ The other two zeros follow from k^M+m+k MM+mlambda+M^ m lambda^ so lambda_ frac-k MM+mpmsqrtk^M^M+m^-k^M^mM+mM^m frac-kM+mpm ksqrtM^+Mm+m^-Mm-m^Mm frac-kM+mpm kMMm Longrightarrow lambda_ -frackM lambda_ -frackM+mMm The nontrivial angular frequencies are thus omega_ sqrtkM omega_ sqrtkM+mMm The eigenvectors are hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix hat x_ pmatrix -fracMm pmatrix The first normal mode is no vibration but a uniform motion of all three atoms. In the second normal mode the carbon atom remains at rest while the oxygen atoms oscillate in opposite directions. This explains why the mass of the carbon atom does not appear in the expression for the frequency. The third normal mode corresponds to the two oxygen atoms moving in phase while the carbon atom moves in the opposite direction. The amplitudes are such that the molecule's centre of mass remains at rest. abc The ratio of omega_ and omega_ is fracomega_omega_ sqrtfrackM+mMmfracMk ratioF sqrtfractimesM+mm resultratioS abcliste
The molecular vibrations of a linear triatomic molecule e.g. mathrmCO_ can be modelled as a system of three masses coupled through springs see figure. center includegraphicswidth.cm#image_path:co-molecul# center abcliste abc Derive the differential s for the longitudinal vibrations and the corresponding coefficient matrix. abc Determine the normale modes of oscillation. abc Calculate the ratio of the two nontrivial frequencies. abcliste
Solution:
abcliste abc The differential s for the three masses left to right are M ddot x_ kx_-x_ -k x_ + k x_ m ddot x_ -kx_-x_+kx_-x_ k x_ - k x_ + k x_ M ddot x_ -kx_-x_ k x_ -k x_ where M is the mass of an oxygen and m the mass of a carbon atom. The coefficient matrix is thus bf A pmatrix -frackM & frackM & frackm & -frackm & frackm & frackM & -frackM pmatrix abc The eigenvalues are the zeros of the characteristic polynomial: detbf A-lambda bf I pmatrix -frack+lambda MM & frackM & frackm & -frack+lambda mm & frackm & frackM & -frack+lambda MM pmatrix frac-k+lambda M^k+lambda m+k^k+lambda MM^ m Longrightarrow k^+k^ M lambda & quad - k^+k M lambda+M^lambda^k+lambda m k^+k^ M lambda & quad -k^-k^ M lambda-k M^ lambda^ & quad -k^ m lambda - k M m lambda^ - M^ m lambda^ -lambda k^ M+k^ m + k M^ lambda+k M m lambda + M^ m lambda^ -lambda leftk^M+m+k MM+mlambda+M^ m lambda^right The first zero is trivial: lambda_ The other two zeros follow from k^M+m+k MM+mlambda+M^ m lambda^ so lambda_ frac-k MM+mpmsqrtk^M^M+m^-k^M^mM+mM^m frac-kM+mpm ksqrtM^+Mm+m^-Mm-m^Mm frac-kM+mpm kMMm Longrightarrow lambda_ -frackM lambda_ -frackM+mMm The nontrivial angular frequencies are thus omega_ sqrtkM omega_ sqrtkM+mMm The eigenvectors are hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix hat x_ pmatrix -fracMm pmatrix The first normal mode is no vibration but a uniform motion of all three atoms. In the second normal mode the carbon atom remains at rest while the oxygen atoms oscillate in opposite directions. This explains why the mass of the carbon atom does not appear in the expression for the frequency. The third normal mode corresponds to the two oxygen atoms moving in phase while the carbon atom moves in the opposite direction. The amplitudes are such that the molecule's centre of mass remains at rest. abc The ratio of omega_ and omega_ is fracomega_omega_ sqrtfrackM+mMmfracMk ratioF sqrtfractimesM+mm resultratioS abcliste
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Exercise:
The molecular vibrations of a linear triatomic molecule e.g. mathrmCO_ can be modelled as a system of three masses coupled through springs see figure. center includegraphicswidth.cm#image_path:co-molecul# center abcliste abc Derive the differential s for the longitudinal vibrations and the corresponding coefficient matrix. abc Determine the normale modes of oscillation. abc Calculate the ratio of the two nontrivial frequencies. abcliste
Solution:
abcliste abc The differential s for the three masses left to right are M ddot x_ kx_-x_ -k x_ + k x_ m ddot x_ -kx_-x_+kx_-x_ k x_ - k x_ + k x_ M ddot x_ -kx_-x_ k x_ -k x_ where M is the mass of an oxygen and m the mass of a carbon atom. The coefficient matrix is thus bf A pmatrix -frackM & frackM & frackm & -frackm & frackm & frackM & -frackM pmatrix abc The eigenvalues are the zeros of the characteristic polynomial: detbf A-lambda bf I pmatrix -frack+lambda MM & frackM & frackm & -frack+lambda mm & frackm & frackM & -frack+lambda MM pmatrix frac-k+lambda M^k+lambda m+k^k+lambda MM^ m Longrightarrow k^+k^ M lambda & quad - k^+k M lambda+M^lambda^k+lambda m k^+k^ M lambda & quad -k^-k^ M lambda-k M^ lambda^ & quad -k^ m lambda - k M m lambda^ - M^ m lambda^ -lambda k^ M+k^ m + k M^ lambda+k M m lambda + M^ m lambda^ -lambda leftk^M+m+k MM+mlambda+M^ m lambda^right The first zero is trivial: lambda_ The other two zeros follow from k^M+m+k MM+mlambda+M^ m lambda^ so lambda_ frac-k MM+mpmsqrtk^M^M+m^-k^M^mM+mM^m frac-kM+mpm ksqrtM^+Mm+m^-Mm-m^Mm frac-kM+mpm kMMm Longrightarrow lambda_ -frackM lambda_ -frackM+mMm The nontrivial angular frequencies are thus omega_ sqrtkM omega_ sqrtkM+mMm The eigenvectors are hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix hat x_ pmatrix -fracMm pmatrix The first normal mode is no vibration but a uniform motion of all three atoms. In the second normal mode the carbon atom remains at rest while the oxygen atoms oscillate in opposite directions. This explains why the mass of the carbon atom does not appear in the expression for the frequency. The third normal mode corresponds to the two oxygen atoms moving in phase while the carbon atom moves in the opposite direction. The amplitudes are such that the molecule's centre of mass remains at rest. abc The ratio of omega_ and omega_ is fracomega_omega_ sqrtfrackM+mMmfracMk ratioF sqrtfractimesM+mm resultratioS abcliste
The molecular vibrations of a linear triatomic molecule e.g. mathrmCO_ can be modelled as a system of three masses coupled through springs see figure. center includegraphicswidth.cm#image_path:co-molecul# center abcliste abc Derive the differential s for the longitudinal vibrations and the corresponding coefficient matrix. abc Determine the normale modes of oscillation. abc Calculate the ratio of the two nontrivial frequencies. abcliste
Solution:
abcliste abc The differential s for the three masses left to right are M ddot x_ kx_-x_ -k x_ + k x_ m ddot x_ -kx_-x_+kx_-x_ k x_ - k x_ + k x_ M ddot x_ -kx_-x_ k x_ -k x_ where M is the mass of an oxygen and m the mass of a carbon atom. The coefficient matrix is thus bf A pmatrix -frackM & frackM & frackm & -frackm & frackm & frackM & -frackM pmatrix abc The eigenvalues are the zeros of the characteristic polynomial: detbf A-lambda bf I pmatrix -frack+lambda MM & frackM & frackm & -frack+lambda mm & frackm & frackM & -frack+lambda MM pmatrix frac-k+lambda M^k+lambda m+k^k+lambda MM^ m Longrightarrow k^+k^ M lambda & quad - k^+k M lambda+M^lambda^k+lambda m k^+k^ M lambda & quad -k^-k^ M lambda-k M^ lambda^ & quad -k^ m lambda - k M m lambda^ - M^ m lambda^ -lambda k^ M+k^ m + k M^ lambda+k M m lambda + M^ m lambda^ -lambda leftk^M+m+k MM+mlambda+M^ m lambda^right The first zero is trivial: lambda_ The other two zeros follow from k^M+m+k MM+mlambda+M^ m lambda^ so lambda_ frac-k MM+mpmsqrtk^M^M+m^-k^M^mM+mM^m frac-kM+mpm ksqrtM^+Mm+m^-Mm-m^Mm frac-kM+mpm kMMm Longrightarrow lambda_ -frackM lambda_ -frackM+mMm The nontrivial angular frequencies are thus omega_ sqrtkM omega_ sqrtkM+mMm The eigenvectors are hat x_ pmatrix pmatrix hat x_ pmatrix - pmatrix hat x_ pmatrix -fracMm pmatrix The first normal mode is no vibration but a uniform motion of all three atoms. In the second normal mode the carbon atom remains at rest while the oxygen atoms oscillate in opposite directions. This explains why the mass of the carbon atom does not appear in the expression for the frequency. The third normal mode corresponds to the two oxygen atoms moving in phase while the carbon atom moves in the opposite direction. The amplitudes are such that the molecule's centre of mass remains at rest. abc The ratio of omega_ and omega_ is fracomega_omega_ sqrtfrackM+mMmfracMk ratioF sqrtfractimesM+mm resultratioS abcliste
Contained in these collections:
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Harmonische Schwingung: Doppelfeder by TeXercises
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Coupled Oscillators by by

