Please help me with this mechanics problem -- pulling one end of a spring

  • #1

Homework Statement


A spring of unstretched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity v is?

2. Homework Equations

potential energy stored in a spring is 1/2kx^2


The Attempt at a Solution



the free end is pulled by a velocity v but that is not constant through out,and it is the equivalent velocity that is to be considered.since kinetic energy changes with length,we can use integration to find the total kinetic energy,we can consider an element dm and find it's kinetic energy but i don't know what is the relation between velocity of the element and the distance of that element from rigid end.if this is know i can calculate the kinetic energy.
 

Answers and Replies

  • #2
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but i don't know what is the relation between velocity of the element and the distance of that element from rigid end.if this is know i can calculate the kinetic energy.
i think real springs which has mass can oscillate with the velocity of different elements say at y the velocity will be proportional to the linear displacement as when y=0 u=0 and at y=L its the maximum so one can take a linear relationship and find out kinetic energy; for details see

https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system)
 
  • #3
i think real springs which has mass can oscillate with the velocity of different elements say at y the velocity will be proportional to the linear displacement as when y=0 u=0 and at y=L its the maximum so one can take a linear relationship and find out kinetic energy; for details see

https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system)
thank you very much sir for replying,but why do velocity and distance from one end have a linear relationship?
 
  • #4
haruspex
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thank you very much sir for replying,but why do velocity and distance from one end have a linear relationship?
It is not made clear, but you have to assume there are no oscillations going on. So no accelerations, and uniform tension. If the tension is uniform, what does that tell you about the relationship between original distance from the fixed and the displacement?
 
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  • #5
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thank you very much sir for replying,but why do velocity and distance from one end have a linear relationship?
In fact, for a non-uniform spring, the effective mass solely depends on its linear density along its length: so if the velocity is v at say y then in the term representing mass one gets m(eff.) = integral of {(y^2/L^2) . density.dy} between the limits zero to L
the kinetic energy gives a term m(eff.) rather than spring mass-m
where effective mass is less than m.- this effective mass being equal to mass of the spring will mean that the whole mass is at the farther end at L.
however the basic assumption is that the spring constant is taken to be uniform in the above situation (which may not be in real massive springs.)
 
  • #6
It is not made clear, but you have to assume there are no oscillations going on. So no accelerations, and uniform tension. If the tension is uniform, what does that you about the relationship between original distance from the fixed and and displacement?
thank you very much for replying sir, there exists a force but how to we know it is uniform?when the free end has moved a distance dx in time dt,then the adjoining particle exerts a force on the free end,so by newton's third law,the free end exerts a force on the next particle and this displaces and this phenomenon continues.but shouldn't the force be proportional to deformation,so this force depends on deformation of particle,so how can we say this force is uniform?
 
  • #7
In fact, for a non-uniform spring, the effective mass solely depends on its linear density along its length: so if the velocity is v at say y then in the term representing mass one gets m(eff.) = integral of {(y^2/L^2) . density.dy} between the limits zero to L
the kinetic energy gives a term m(eff.) rather than spring mass-m
where effective mass is less than m.- this effective mass being equal to mass of the spring will mean that the whole mass is at the farther end at L.
however the basic assumption is that the spring constant is taken to be uniform in the above situation (which may not be in real massive springs.)
sir,what is this effective mass and why is it equal to the expression given here?
 
  • #8
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sir,what is this effective mass and why is it equal to the expression given here?
The effective mass of spring (which is not massless) can be defined;

In a real spring it has a non-negligible mass m.
Since not all of the spring's elements move at the same velocity u as the suspended mass M,( its kinetic energy is not 1/2. mu^2.)

so m cannot be simply added to M to determine the frequency of oscillation,

and the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behavior of the system.

you can see details in the following reference:
https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system)
 
  • #9
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thank you very much for replying sir, there exists a force but how to we know it is uniform?when the free end has moved a distance dx in time dt,then the adjoining particle exerts a force on the free end,so by newton's third law,the free end exerts a force on the next particle and this displaces and this phenomenon continues.but shouldn't the force be proportional to deformation,so this force depends on deformation of particle,so how can we say this force is uniform?
If you were to apply a steady force to the free end of a massive spring, oscillations would arise. But here we are told that the force applied is keeping the free end moving at velocity v. That rules out a steady force, and it cannot have gone instantly from rest to speed v. So to have enough information we have to assume no oscillations, and that the spring was got into its initial state of motion by some magic.
For there to be no oscillations, the spring must be uniformly stretched at each instant. For that to be the case, the tension must be uniform.
 
  • #10
If you were to apply a steady force to the free end of a massive spring, oscillations would arise. But here we are told that the force applied is keeping the free end moving at velocity v. That rules out a steady force, and it cannot have gone instantly from rest to speed v. So to have enough information we have to assume no oscillations, and that the spring was got into its initial state of motion by some magic.
For there to be no oscillations, the spring must be uniformly stretched at each instant. For that to be the case, the tension must be uniform.
sir,whatever you are saying is very complicated and much beyond my portions,is there any other way through which we can derive this?
 
  • #11
The effective mass of spring (which is not massless) can be defined;

In a real spring it has a non-negligible mass m.
Since not all of the spring's elements move at the same velocity u as the suspended mass M,( its kinetic energy is not 1/2. mu^2.)

so m cannot be simply added to M to determine the frequency of oscillation,

and the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behavior of the system.

you can see details in the following reference:
https://en.wikipedia.org/wiki/Effective_mass_(spring–mass_system)
sir,but how to derive equation for effective mass and in wikipedia also we are taking linear relationship between velocity of element and distance?
 
  • #12
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wikipedia also we are taking linear relationship between velocity of element and distance?
Since the spring is uniform,dm=(dy/L)m, where L is the length of the spring.
the displacement of any element is proportional to length at which the element is situated- at y=0 displacement is zero at y=L the displacement is maximum.
the acceleration of each element is proportional to length at which it is situated - as the force is proportional to length (F=-kx)
then why should not the velocity obey the same linear rule?
The velocity of each mass element of the spring is directly proportional to its length, i.e. u = v.y/L,
 
  • #13
Since the spring is uniform,dm=(dy/L)m, where L is the length of the spring.
the displacement of any element is proportional to length at which the element is situated- at y=0 displacement is zero at y=L the displacement is maximum.
the acceleration of each element is proportional to length at which it is situated - as the force is proportional to length (F=-kx)
then why should not the velocity obey the same linear rule?
The velocity of each mass element of the spring is directly proportional to its length, i.e. u = v.y/L,
thank you very much for replying sir but how is displacement of an element proportional to the length a which it is situated?please help.
 
  • #14
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thank you very much for replying sir but how is displacement of an element proportional to the length a which it is situated?please help.
Do you understand at least that this could be the case? Indeed, that it is the simplest possible arrangement here?
 
  • #15
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thank you very much for replying sir but how is displacement of an element proportional to the length a which it is situated?please help.
If you wish to find out the specific reason i remember you can get in the 'treatment of motion of helical spring " in the book on General properties of matter written by Newman and Searle / or another book by F.T. smith dealing with the spring motion -
However if the displacement will be power of x other than unity -one can not get F=-kx.; for example if its a relation x^n then the velcity will have dependence as n.x^(n-1) and Force will be having a factor n(n-1).x^(n-2) but that does not happen.
 
  • #16
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If you wish to find out the specific reason i remember you can get in the 'treatment of motion of helical spring " in the book on General properties of matter written by Newman and Searle / or another book by F.T. smith dealing with the spring motion -
However if the displacement will be power of x other than unity -one can not get F=-kx.; for example if its a relation x^n then the velcity will have dependence as n.x^(n-1) and Force will be having a factor n(n-1).x^(n-2) but that does not happen.
Not quite sure what you are saying there, but in general it would not be the case that displacement is proportional to distance from the fixed end. There could be oscillations going on within the spring. You have to assume there are no such oscillations. The problem then is to prove that this leads to the uniform stretching.
 
  • #17
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Not quite sure what you are saying there, but in general it would not be the case that displacement is proportional to distance from the fixed end. There could be oscillations going on within the spring. You have to assume there are no such oscillations. The problem then is to prove that this leads to the uniform stretching.
I think your contention is correct -the force displacement relationship in general is non-linear but an approximation is made as the the spring is uniform and stress -strain relations are taken to be same and continuous along the length under elastic limit..
 
  • #18
I guess your main problem is why velocity of an element is proportional to the length of spring behind it. So I have come up with a intuitive way to understand it.
Imagine the spring to be made of many little elements. Since the elongation is uniform, each elements of the spring is elongated the same with respect to their neighbour. With respect to their neighbour is important. Take an element at y distance from the wall. Now our element's neighbour's neighbour is also elongated by dx wrt it's neighbour. So w.r.t second neighbour our element is elongated by dx+dx=2dx. Now, to get elongation with respect to ground, you have to keep adding the elongation of all the neighbours.
Since number of elements is proportional to the length of the spring(more length means more elements), so we can write number of elements as ky(where y is the length of the spring, k is any arbitrary constant). So now you keep adding dx +dx +dx ..... Which will be kydx elements ( as this is the number of elements before our element).

Let's summarise. We have just found out that elongation (displacement) of a particle is proportional to y, which is the length of spring behind it. Since v is nothing but dx/dt, v is ALSO proportional to y.

DISCLAIMER this is just an intuitive way to understand the given problem, please do not take it as a formal proof of the given questions.
 
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  • #19
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Since the elongation is uniform
This thread is over a year old,so I doubt vijayramakrishnan is still interested.

However, there is no guarantee that the elongation is uniform. Yes, there is a solution in which that is the case and the spring is being a extended at a steady rate, but more generally there could be internal oscillations superimposed on this.
Indeed, it is hard to see how such a vibrationless motion could arise in practice. If it were attempted to go immediately from rest to a steady speed by pulling on one end then the leading end would be subjected to a very high acceleration. Points further from the leading end would have delayed acceleration. Some way would have to be found of accelerating all parts of the spring in unison and in proportion.
 
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  • #20
Yeah you are right about vijayramakrishnan not being interested... Still I have posted the answer in case he checks back

As for the second part, this is a question from JEE Mains, a highschool engineering entrance examination conducted in India. Since I myself am a engineering aspirant, I know what level of physics has been taught to us. Since we haven't been taught anything about vibrations in a non-massless strings, we just assume that the velocity was v at the starting itself. Heck, we are just told to take the equivalent mass of a spring as m/3. So I am sure such high level of physics has not been applied here, and all that ramakrishnan wanted to know(judging from his replies) is the relation between velocity and of an element and velocity of the end point.
 
  • #21
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Yeah you are right about vijayramakrishnan not being interested... Still I have posted the answer in case he checks back

As for the second part, this is a question from JEE Mains, a highschool engineering entrance examination conducted in India. Since I myself am a engineering aspirant, I know what level of physics has been taught to us. Since we haven't been taught anything about vibrations in a non-massless strings, we just assume that the velocity was v at the starting itself. Heck, we are just told to take the equivalent mass of a spring as m/3. So I am sure such high level of physics has not been applied here, and all that ramakrishnan wanted to know(judging from his replies) is the relation between velocity and of an element and velocity of the end point.
I quite agree that the question expects you to assume velocity varies uniformly along the spring. I wrote that in post #4. But it seemed to me that your post #18 claimed to prove, sort of, that it would be the case. In reality, it would not.
 
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  • #22
I quite agree that the question expects you to assume velocity varies uniformly along the spring. I wrote that in post #4. But it seemed to me that your post #18 claimed to prove, sort of, that it would be the case. In reality, it would not.
Sure I have used some questionable assumptions, like the fact that tension, and hence elongation, is constant everywhere. But that's what the question wants us to assume. So for JEE Mains aspirants it's enough. And I have also mentioned a Disclaimer so that all those who aren't actually engineering aspirants (like yourself) are not confused with this answer.
 
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  • #23
This thread is over a year old,so I doubt vijayramakrishnan is still interested.

However, there is no guarantee that the elongation is uniform. Yes, there is a solution in which that is the case and the spring is being a extended at a steady rate, but more generally there could be internal oscillations superimposed on this.
Indeed, it is hard to see how such a vibrationless motion could arise in practice. If it were attempted to go immediately from rest to a steady speed by pulling on one end then the leading end would be subjected to a very high acceleration. Points further from the leading end would have delayed acceleration. Some way would have to be found of accelerating all parts of the spring in unison and in proportion.
Thank you very much sir for clearing my doubt
 

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