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Hey everyone! So I'm looking at the energy transfers in a mass-spring system oscillating with simple harmonic motion, and I'm a bit confused.
So, for the graph of total energy, potential energy and kinetic energy against displacement:
![Name: shm.gif
Views: 948
Size: 9.8 KB]()
I think it suggests that when the kinetic energy of the system is at a maximum value, the potential energy is 0.
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
Which bit of my thinking is wrong? Or does the graph only apply to simple pendulums ?
Cheers
So, for the graph of total energy, potential energy and kinetic energy against displacement:
I think it suggests that when the kinetic energy of the system is at a maximum value, the potential energy is 0.
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
Which bit of my thinking is wrong? Or does the graph only apply to simple pendulums ?
Cheers
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#2
(Original post by levi ackerman)
Hey everyone! So I'm looking at the energy transfers in a mass-spring system oscillating with simple harmonic motion, and I'm a bit confused.
So, for the graph of total energy, potential energy and kinetic energy against displacement:
![Name: shm.gif
Views: 948
Size: 9.8 KB]()
I think it suggests that when the kinetic energy of the system is at a maximum value, the potential energy is 0.
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
Which bit of my thinking is wrong? Or does the graph only apply to simple pendulums ?
Cheers
Hey everyone! So I'm looking at the energy transfers in a mass-spring system oscillating with simple harmonic motion, and I'm a bit confused.
So, for the graph of total energy, potential energy and kinetic energy against displacement:
I think it suggests that when the kinetic energy of the system is at a maximum value, the potential energy is 0.
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
Which bit of my thinking is wrong? Or does the graph only apply to simple pendulums ?
Cheers
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(Original post by Angels1234)
No it applies to vertical motion ie on a spring as well . With masses attached to form a vertical system . At max displacement for this arrangement gpe is max (obviously ) and at equilibrium position (initial point of release ) velocity is max and gpe is 0
No it applies to vertical motion ie on a spring as well . With masses attached to form a vertical system . At max displacement for this arrangement gpe is max (obviously ) and at equilibrium position (initial point of release ) velocity is max and gpe is 0
3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary? how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
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#4
(Original post by levi ackerman)
thanks for the reply 3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary?
how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
thanks for the reply
3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary? how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
I meant that gpe is 0 when the displacement is 0 . I mean displacement from the “starting point “ but this is equilibrium. You are always looking at displacement from “ equilibrium “ because the mass is continuously trying to get to equilibrium position . That’s why a mass goes up and down . It’s constantly trying to get to that position
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#5
(Original post by levi ackerman)
thanks for the reply 3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary?
how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
thanks for the reply
3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary? how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
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#7
(Original post by levi ackerman)
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
The rest of it is just pure conservation of energy, as there are no external forces acting on the system (air resistance negligible) which explains the shape of the graph and how the sum of the two energies stays constant.
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#8
(Original post by darkforest)
The elastic potential energy would actually be 0J at equilibrium position. If you think about it, at equilibrium position there is by definition 0N resultant force on the mass, which implies that no work is done on the spring and hence it has no potential energy. This is instantaneous btw so it only applies for the instant that the mass passes through it's resting position. You may be able to picture the system in your head and see that the velocity would be greatest at this point too, which accounts for the maximum kinetic energy.
The rest of it is just pure conservation of energy, as there are no external forces acting on the system (air resistance negligible) which explains the shape of the graph and how the sum of the two energies stays constant.
The elastic potential energy would actually be 0J at equilibrium position. If you think about it, at equilibrium position there is by definition 0N resultant force on the mass, which implies that no work is done on the spring and hence it has no potential energy. This is instantaneous btw so it only applies for the instant that the mass passes through it's resting position. You may be able to picture the system in your head and see that the velocity would be greatest at this point too, which accounts for the maximum kinetic energy.
The rest of it is just pure conservation of energy, as there are no external forces acting on the system (air resistance negligible) which explains the shape of the graph and how the sum of the two energies stays constant.
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(Original post by Angels1234)
Hope this helps
Hope this helps
so how is elastic potential energy 0J at equilibrium position? (sorry for being a pain haha
)
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#10
(Original post by levi ackerman)
but in this diagram, comparing the first and second illustrations, doesn't it show that the spring has an extension when the system is in the equilibrium position?
so how is elastic potential energy 0J at equilibrium position? (sorry for being a pain haha
)
but in this diagram, comparing the first and second illustrations, doesn't it show that the spring has an extension when the system is in the equilibrium position?
so how is elastic potential energy 0J at equilibrium position? (sorry for being a pain haha
)
If you are stuck still then please ask again . No point moving on till you are fully sure
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#11
(Original post by Angels1234)
I’m not really sure what you are asking for the first 2 . ??
I meant that gpe is 0 when the displacement is 0 . I mean displacement from the “starting point “ but this is equilibrium. You are always looking at displacement from “ equilibrium “ because the mass is continuously trying to get to equilibrium position . That’s why a mass goes up and down . It’s constantly trying to get to that position
I’m not really sure what you are asking for the first 2 . ??
I meant that gpe is 0 when the displacement is 0 . I mean displacement from the “starting point “ but this is equilibrium. You are always looking at displacement from “ equilibrium “ because the mass is continuously trying to get to equilibrium position . That’s why a mass goes up and down . It’s constantly trying to get to that position
It seems that you are confused with the “starting point” and “equilibrium point”. If you look at the diagram that you posted, at the equilibrium point, the spring force is equal to the weight of the mass attached to the spring. At the starting point, if the hand is not supporting the mass, the mass would fall due to the weight of the mass.
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#12
(Original post by Eimmanuel)
It seems that you are confused with the “starting point” and “equilibrium point”. If you look at the diagram that you posted, at the equilibrium point, the spring force is equal to the weight of the mass attached to the spring. At the starting point, if the hand is not supporting the mass, the mass would fall due to the weight of the mass.
It seems that you are confused with the “starting point” and “equilibrium point”. If you look at the diagram that you posted, at the equilibrium point, the spring force is equal to the weight of the mass attached to the spring. At the starting point, if the hand is not supporting the mass, the mass would fall due to the weight of the mass.
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#13
(Original post by levi ackerman)
Hey everyone! So I'm looking at the energy transfers in a mass-spring system oscillating with simple harmonic motion, and I'm a bit confused.
So, for the graph of total energy, potential energy and kinetic energy against displacement:
![Name: shm.gif
Views: 948
Size: 9.8 KB]()
I think it suggests that when the kinetic energy of the system is at a maximum value, the potential energy is 0.
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
Which bit of my thinking is wrong? Or does the graph only apply to simple pendulums ?
Cheers
Hey everyone! So I'm looking at the energy transfers in a mass-spring system oscillating with simple harmonic motion, and I'm a bit confused.
So, for the graph of total energy, potential energy and kinetic energy against displacement:
I think it suggests that when the kinetic energy of the system is at a maximum value, the potential energy is 0.
But, for vertical (and horizontal? not sure) mass-spring systems, when the kinetic energy is maximum as the mass passes through the equilibrium position, the potential energy isn't 0, correct? Since, at that instant there is a non-0 value for extension, no?
Which bit of my thinking is wrong? Or does the graph only apply to simple pendulums ?
Cheers
(Original post by levi ackerman)
thanks for the reply 3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary?
how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
thanks for the reply
3 qs - how is the equilibrium position the initial point of release? surely if you release the mass at this point, the system continues to be stationary?how can gpe be 0 at the equilibrium position if the mass has a non-0 height with respect to the lowest point (i.e. with respect to the maximum downwards displacement)?
and what about elastic potential energy stored in the spring?
The graph that you posted is only valid for horizontal mass-spring system.
The various types of energy versus displacement for the vertical mass-spring system is shown in the diagram below.
Note that the horizontal axis is the displacement of the mass from the starting point. With reference to the diagram posted by Angels1234, this is the energy vs displacement graph for d0 = A.
The energy vs displacement graphs for the case of A < d0 or A > d0 is slightly different from the graph shown above.
I would ask my students “practice” their understanding of the above graph to sketch the graphs for the two different cases.
If you can sketch the various types of energy with time, you may be surprised by what you get.
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#14
(Original post by Angels1234)
I mean yea the mass would fall which is why there is a picture showing that the person is holding the weight . If the person wasn’t holding the weight then obviously the weight will fall ? I thought this was obvious from the picture so I didn’t explain it . Yeah so the balancing act of the forces as I said means it in equilibrium?
I mean yea the mass would fall which is why there is a picture showing that the person is holding the weight . If the person wasn’t holding the weight then obviously the weight will fall ? I thought this was obvious from the picture so I didn’t explain it . Yeah so the balancing act of the forces as I said means it in equilibrium?
Your previous answer seems to imply that equilibrium point is the same as the starting point. This is incorrect.
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#15
(Original post by Eimmanuel)
Your previous answer seems to imply that equilibrium point is the same as the starting point. This is incorrect.
Your previous answer seems to imply that equilibrium point is the same as the starting point. This is incorrect.
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#16
(Original post by Angels1234)
So are you saying the starting point is where there are no masses attached ?
So are you saying the starting point is where there are no masses attached ?
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#17
(Original post by Eimmanuel)
I think this is what you are saying or implying. I am sorry that if I misinterpret it.
I think this is what you are saying or implying. I am sorry that if I misinterpret it.
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#18
(Original post by Angels1234)
Could you explain what you mean because I’m confused now ?
Could you explain what you mean because I’m confused now ?
The starting point has to be at the maximum displacement if it is being made to oscillate freely. I.e. you either drop it from the maximum positive amplitude or you release it from the maximum negative amplitude.
The equilibrium position is defines as the point of zero displacement in the system about which the body will oscillate. Therefore, the potential energy is zero. In reality the equilibrium position is not on the floor and so in reality GPE is not zero but for THE SYSTEM it is classed as the point of zero PE (since Gravitational potential energy = Elastic potential energy here and as they are acting in opposite directions at this point they cancel).
During oscillation, when the body passes through equilibrium it will have maximum velocity, minimum acceleration, zero potential energy and so maximum kinetic energy.
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#19
(Original post by splitter2017)
Equilibrium position is when the forces are balanced. Spring force = Weight
The starting point has to be at the maximum displacement if it is being made to oscillate freely. I.e. you either drop it from the maximum positive amplitude or you release it from the maximum negative amplitude.
The equilibrium position is defines as the point of zero displacement in the system about which the body will oscillate. Therefore, the potential energy is zero. In reality the equilibrium position is not on the floor and so in reality GPE is not zero but for THE SYSTEM it is classed as the point of zero PE (since Gravitational potential energy = Elastic potential energy here and as they are acting in opposite directions at this point they cancel).
During oscillation, when the body passes through equilibrium it will have maximum velocity, minimum acceleration, zero potential energy and so maximum kinetic energy.
Equilibrium position is when the forces are balanced. Spring force = Weight
The starting point has to be at the maximum displacement if it is being made to oscillate freely. I.e. you either drop it from the maximum positive amplitude or you release it from the maximum negative amplitude.
The equilibrium position is defines as the point of zero displacement in the system about which the body will oscillate. Therefore, the potential energy is zero. In reality the equilibrium position is not on the floor and so in reality GPE is not zero but for THE SYSTEM it is classed as the point of zero PE (since Gravitational potential energy = Elastic potential energy here and as they are acting in opposite directions at this point they cancel).
During oscillation, when the body passes through equilibrium it will have maximum velocity, minimum acceleration, zero potential energy and so maximum kinetic energy.
so basically starting point is actually from max displacement
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(Original post by Eimmanuel)
The graph that you posted is only valid for horizontal mass-spring system.
The various types of energy versus displacement for the vertical mass-spring system is shown in the diagram below.
![Name: mass-spring_energy_displacement.JPG
Views: 475
Size: 42.6 KB]()
Note that the horizontal axis is the displacement of the mass from the starting point. With reference to the diagram posted by Angels1234, this is the energy vs displacement graph for d0 = A.
The energy vs displacement graphs for the case of A < d0 or A > d0 is slightly different from the graph shown above.
I would ask my students “practice” their understanding of the above graph to sketch the graphs for the two different cases.
If you can sketch the various types of energy with time, you may be surprised by what you get.
The graph that you posted is only valid for horizontal mass-spring system.
The various types of energy versus displacement for the vertical mass-spring system is shown in the diagram below.
Note that the horizontal axis is the displacement of the mass from the starting point. With reference to the diagram posted by Angels1234, this is the energy vs displacement graph for d0 = A.
The energy vs displacement graphs for the case of A < d0 or A > d0 is slightly different from the graph shown above.
I would ask my students “practice” their understanding of the above graph to sketch the graphs for the two different cases.
If you can sketch the various types of energy with time, you may be surprised by what you get.
just for clarity - at the equilibrium position, do we sort of "say" that the potential energy (apparently defined as the system's 0 point of potential energy) is 0, even though it isn't? I think this is similar to what splitter2017 was saying (thank you btw!). Or is it better to say that potential energy is a minimum?
I was also interested in the equations describing the curves, and had a go at finding them.
http://i66.tinypic.com/2d7zw3o.jpg
Are these correct? (this is more for interest than anything, so no worries if you don't have time to have a look)
thanks!
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