Damped Mass on Spring
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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 -
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Exercise:
The mass oscillating on a spring is immersed in a liquid. In this situation the drag force is given by Stokes' law: vecF_d - pi eta R vecv_yt where R is the radius of the spherical mass and eta is the viscosity of the liquid. abcliste abc Derive the differential for the damped oscillation and show that the damping coefficient is abc A mass-spring system consists of spring with spring constant kO and an aluminium ball with radius rO oscillating in glycerole etaetO. Calculate the damping coefficient and the angular frequency. Discuss the result. abc For the same spring as in b what would be the radius of the steel ball that leads to critical damping? abcliste
Solution:
abcliste abc The net force acting on the mass is sscFnet -k yt - pieta R v_yt -k yt - pieta R dot yt Using Newton's second law this yields m ddot yt -k yt - pieta R dot yt Longrightarrow ddot yt -frackm yt - fracpieta Rm dot yt This is the differential for a damped oscillation with omega_ sqrtfrackm and delta fracpieta Rm The damping coefficient is thus delta fracpieta Rm Remark: For a fully immersed mass buoyancy does not affect the oscillation but only the equilibrium position analogously to gravitation. abc The mass can be expressed as m V rho fracpi R^ rho It follows for the angular frequency of the undamped oscillator omega_ sqrtfrackm omF sqrtfrac times kpi times r^ times rh om approx resultomP The damping coefficient is given by delta fracpieta Rm pieta R fracpi R^ rho deF frac times et times r^ times rh de approx resultdeP Since delta omega_ the oscillation is weakly damped. The angular frequency of the damped oscillation is omega omdF sqrtom^-de^ omd approx omdP which is as expected in good agreement with the angular frequency of the undamped oscillation. center includegraphicswidthtextwidth#image_path:damped-mass-on-spring-# center abc The damping ratio is zeta fracdeltaomega_ deF sqrtfracpi R^ rho k zeF Solving for the radius yields R frac pi eta^ k rho zeta^ For critical damping zeta this corresponds to a radius R_c RcF frac pi times et^ times k times rh Rc approx resultRcP- For a ball with a radius RR_c the oscillation would be overdamped. abcliste
The mass oscillating on a spring is immersed in a liquid. In this situation the drag force is given by Stokes' law: vecF_d - pi eta R vecv_yt where R is the radius of the spherical mass and eta is the viscosity of the liquid. abcliste abc Derive the differential for the damped oscillation and show that the damping coefficient is abc A mass-spring system consists of spring with spring constant kO and an aluminium ball with radius rO oscillating in glycerole etaetO. Calculate the damping coefficient and the angular frequency. Discuss the result. abc For the same spring as in b what would be the radius of the steel ball that leads to critical damping? abcliste
Solution:
abcliste abc The net force acting on the mass is sscFnet -k yt - pieta R v_yt -k yt - pieta R dot yt Using Newton's second law this yields m ddot yt -k yt - pieta R dot yt Longrightarrow ddot yt -frackm yt - fracpieta Rm dot yt This is the differential for a damped oscillation with omega_ sqrtfrackm and delta fracpieta Rm The damping coefficient is thus delta fracpieta Rm Remark: For a fully immersed mass buoyancy does not affect the oscillation but only the equilibrium position analogously to gravitation. abc The mass can be expressed as m V rho fracpi R^ rho It follows for the angular frequency of the undamped oscillator omega_ sqrtfrackm omF sqrtfrac times kpi times r^ times rh om approx resultomP The damping coefficient is given by delta fracpieta Rm pieta R fracpi R^ rho deF frac times et times r^ times rh de approx resultdeP Since delta omega_ the oscillation is weakly damped. The angular frequency of the damped oscillation is omega omdF sqrtom^-de^ omd approx omdP which is as expected in good agreement with the angular frequency of the undamped oscillation. center includegraphicswidthtextwidth#image_path:damped-mass-on-spring-# center abc The damping ratio is zeta fracdeltaomega_ deF sqrtfracpi R^ rho k zeF Solving for the radius yields R frac pi eta^ k rho zeta^ For critical damping zeta this corresponds to a radius R_c RcF frac pi times et^ times k times rh Rc approx resultRcP- For a ball with a radius RR_c the oscillation would be overdamped. abcliste
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Exercise:
The mass oscillating on a spring is immersed in a liquid. In this situation the drag force is given by Stokes' law: vecF_d - pi eta R vecv_yt where R is the radius of the spherical mass and eta is the viscosity of the liquid. abcliste abc Derive the differential for the damped oscillation and show that the damping coefficient is abc A mass-spring system consists of spring with spring constant kO and an aluminium ball with radius rO oscillating in glycerole etaetO. Calculate the damping coefficient and the angular frequency. Discuss the result. abc For the same spring as in b what would be the radius of the steel ball that leads to critical damping? abcliste
Solution:
abcliste abc The net force acting on the mass is sscFnet -k yt - pieta R v_yt -k yt - pieta R dot yt Using Newton's second law this yields m ddot yt -k yt - pieta R dot yt Longrightarrow ddot yt -frackm yt - fracpieta Rm dot yt This is the differential for a damped oscillation with omega_ sqrtfrackm and delta fracpieta Rm The damping coefficient is thus delta fracpieta Rm Remark: For a fully immersed mass buoyancy does not affect the oscillation but only the equilibrium position analogously to gravitation. abc The mass can be expressed as m V rho fracpi R^ rho It follows for the angular frequency of the undamped oscillator omega_ sqrtfrackm omF sqrtfrac times kpi times r^ times rh om approx resultomP The damping coefficient is given by delta fracpieta Rm pieta R fracpi R^ rho deF frac times et times r^ times rh de approx resultdeP Since delta omega_ the oscillation is weakly damped. The angular frequency of the damped oscillation is omega omdF sqrtom^-de^ omd approx omdP which is as expected in good agreement with the angular frequency of the undamped oscillation. center includegraphicswidthtextwidth#image_path:damped-mass-on-spring-# center abc The damping ratio is zeta fracdeltaomega_ deF sqrtfracpi R^ rho k zeF Solving for the radius yields R frac pi eta^ k rho zeta^ For critical damping zeta this corresponds to a radius R_c RcF frac pi times et^ times k times rh Rc approx resultRcP- For a ball with a radius RR_c the oscillation would be overdamped. abcliste
The mass oscillating on a spring is immersed in a liquid. In this situation the drag force is given by Stokes' law: vecF_d - pi eta R vecv_yt where R is the radius of the spherical mass and eta is the viscosity of the liquid. abcliste abc Derive the differential for the damped oscillation and show that the damping coefficient is abc A mass-spring system consists of spring with spring constant kO and an aluminium ball with radius rO oscillating in glycerole etaetO. Calculate the damping coefficient and the angular frequency. Discuss the result. abc For the same spring as in b what would be the radius of the steel ball that leads to critical damping? abcliste
Solution:
abcliste abc The net force acting on the mass is sscFnet -k yt - pieta R v_yt -k yt - pieta R dot yt Using Newton's second law this yields m ddot yt -k yt - pieta R dot yt Longrightarrow ddot yt -frackm yt - fracpieta Rm dot yt This is the differential for a damped oscillation with omega_ sqrtfrackm and delta fracpieta Rm The damping coefficient is thus delta fracpieta Rm Remark: For a fully immersed mass buoyancy does not affect the oscillation but only the equilibrium position analogously to gravitation. abc The mass can be expressed as m V rho fracpi R^ rho It follows for the angular frequency of the undamped oscillator omega_ sqrtfrackm omF sqrtfrac times kpi times r^ times rh om approx resultomP The damping coefficient is given by delta fracpieta Rm pieta R fracpi R^ rho deF frac times et times r^ times rh de approx resultdeP Since delta omega_ the oscillation is weakly damped. The angular frequency of the damped oscillation is omega omdF sqrtom^-de^ omd approx omdP which is as expected in good agreement with the angular frequency of the undamped oscillation. center includegraphicswidthtextwidth#image_path:damped-mass-on-spring-# center abc The damping ratio is zeta fracdeltaomega_ deF sqrtfracpi R^ rho k zeF Solving for the radius yields R frac pi eta^ k rho zeta^ For critical damping zeta this corresponds to a radius R_c RcF frac pi times et^ times k times rh Rc approx resultRcP- For a ball with a radius RR_c the oscillation would be overdamped. abcliste
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