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If a spring has a load $m$ added to it (and so is extended by $x$), the gravitational potential energy lost by the mass will be $mgx$. The elastic potential energy gained by the spring is $\frac{1}{2}\,k x^2$; however $mg = kx$ at the equilbrium rest point, so the gravitational potential energy lost could be written as $kx^2$ .

Surely the gravitational potential energy lost by the mass should equal the elastic potential energy gained by the spring? I realise I've done something wrong here - what is it please?

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  • $\begingroup$ First: you are making confusion in the notation: if $x_0$ is the rest point for the spring, the spring energy in $x$ is $\propto k(x-x_0)^2$. Second: do not forget to include the kinetic energy term: the only constraint to fulfil is that the total mechanical energy must be preserved, which is the sum of three contributions as kinetic energy, gravitational and spring potential energy. The three may then change at will, only their sum has to remain constant. $\endgroup$ – gented Oct 3 '15 at 22:52
  • $\begingroup$ Thanks for prompt reply (and tidying up the fonts!) - I'm still not getting it though - surely the mass could theoretically settle (no k.e.) at a drop in GPE of mgx = kx2, while the spring gains 1/2kx2 elastic PE. $\endgroup$ – justcurious Oct 3 '15 at 23:12
  • $\begingroup$ If a mass m is loaded gently onto a spring the spring stretches a length x where it is at equilibrium such that mg=kx.During this m lost potential energy equal to mgx(x is the difference from the ground)and this was converted to elastic potential energy of the spring given by 1/2kx^2 ie mgx =1/2kx^2. Replacing the earlier equation into this gives LHS not = RHS. I too would like to know the flaw. $\endgroup$ – Chappy Dec 14 '17 at 18:14
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In SHM there is no damping so the oscillation would go on forever.In order to bring the system to equilibrium an external force would be needed and the energy stored in the spring would be diissipated to overcome this force.The energy mgx where x is the point of equilibrium to which the system has been brought by an external force equals 1/2 kx^2 (stored in the spring)+ the energy dissipated to bring the oscillation to a halt.The distance h where the energy mgh is converted totally to spring energy 1/2kh^2 is the point where the mass is at the maximum distance from equilibrium and instantaneosly at rest just before reversing direction.But this is mot equilibrium and here mgis not = kh.

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What you've discovered is the reason for oscillations!

Let's call the position where the string is not stretched $x=0$. Suppose we just hold the ball at this position. The totaly energy of the ball at this point is $0$ in our conventions.

Now, we leave it and let it go down. The forces balance at $x_{eq} = - \tfrac{mg}{k}$. And, as you correctly pointed out, the potential energy of the ball at this point is $$V_{eq} = mgx_{eq} + \tfrac{1}{2} k x_{eq}^2 = - \tfrac{1}{2} k x_{eq}^2.$$ And, your question: why is this value not $0$?

The answer is: the rest of the energy is in the kinetic energy at this point. The speed at this point will have the correct value such that $\tfrac{1}{2} m v_{eq}^2 = \tfrac{1}{2} k x_{eq}^2$, and therefore the ball continues going down despite the forces being balanced.

Now, the potential for confusion is that the ball will eventually come to rest, so where's the enegy then. The answer here is dissipation: there has to be some force damping the oscillations, otherwise they'll never stop and the energy will always be $0$.

There's an alternative picture where you never let the ball gain any velocity by putting a stage below it and slowly moving the stage down. I don't have a copletely clear understanding of this, but I think the stage eats up the difference in energy in some way. A slightly simpler case is when you move the stage not continuously but in tiny discrete steps; then clearly at each step the stage is eating up enough energy to damp the oscillations. I don't understand the limit of this discrete process as the step size goes to $0$, but that's all that needs to be understood.

Aside: the fact that the total energy at the end when the ball has come to rest at the equilibrium point is negative is, in some guise, the virial theorem of classical mechanics. I haven't unpacked exactly how, but let me flesh out the analogy in my mind a bit; if anyone can say it more precisely, pleas edit this answer.

The classical example of the virial theorem is a planet in orbit around a star. In this case, the gravitational potential energy is $-2$ times the kinetic energy of rotation. If you concentrate only on the radial direction, it looks an awful lot like there's a centrifugal force pushing away from the star and gravitational energy is pushing towards it. The analogy is centrifugal force $\leftrightarrow$ $mgx$ and gravitational force $\leftrightarrow \tfrac{1}{2} k x^2$, counter-intuitive as that is.

Step 1 on the way to making it precise is to notice that the gravitational force $\tfrac{G M m}{r^2}$ can be expanded for small displacements --- so that $r=r_0 + \delta r$ --- as $\tfrac{G M m}{r_0^2} (1 - 2 \delta r)$, giving a linear force. I don't know how to think of the centrifugal force though.

Acknowledgements for discussion: user128785 and another friend who's not on stakc exchange.

EDIT: Turns out Timaeus said the same thing first. Apologies for repeat.

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Surely the gravitational potential energy lost by the mass should equal the elastic potential energy gained by the spring?

On the contrary, if something fell down doesn't that imply it gained some downwards velocity and hence the change in kinetic energy should be related to the net work done on the object?

Here is a warning. Let's say you measured k and g and m. Don't put something valuable a distance x below the original spring position where kx=mg. Because just because the force is zero at that point doesn't mean the object won't keep moving if it has some speed.

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Kinetic energy can be taken out of this problem by using the following setup. Let the mass be sitting on a height-adjustable stage and attached to the spring, but not stretching it yet. I start lowering the height of the stage slowly, so the mass starts extending the spring. The mass will extend the spring till the forces of the spring and weight cancel each other, i.e. $mg=kx$, at which point it will lose contact with the stage and be suspended from the spring alone; there will be no motion beyond this point and hence no kinetic energy.

People who disagree with the above argument, can think of the extension of a wire attached to mass in a similar manner, within the Hooke's law regime. The mass at the end of the wire does not perform SHM.

The potential energy should be treated separately from the force at a point. This is very important. The potential energy is arrived at by integrating the force over displacement, therefore substituting the equality of forces at any one point into the answer arrived at by integrating the same force over displacement will clearly give you such contradiction!

You can see this contradiction just mathematically. Let there be two functions $f_{1}(x)$ and $f_{2}(x)$; now even if $$\int_{0}^{x_{f}} f_{1}(x)dx=\int_{0}^{x_{f}} f_{2}(x)dx,$$ the LHS and RHS (after integration) could be two different functions of $x$ (like in our case), and even if $f_{1}(x_{1})=f_{2}(x_{1})$ for some $x_{1}$($=x_{f}$ in our case), you should not substitute this into the result of the integral!

Why, you may ask? Because in fact they are not equal for most of the integral (i.e. the domain of x). The reason the spring is even stretching under the weight of the mass is that $mg>kx \ \forall x\in [0,x_{f}) $; if $mg=kx$ were true at every $x$, the spring would not stretch at all!

Too long an answer for a simple error, but let this be a caution against blind substitution. :)

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  • $\begingroup$ Any justification for the downvote? If it's a disagreement with my answer, we can discuss it here in the open. :) $\endgroup$ – user128785 Feb 23 '18 at 8:44
  • $\begingroup$ I discussed with a friend and found my mistake. The equations refer only to the energy of the block whereas I confused it to be that of the mass + string system. And the comment about substitution in the integral was just plain wrong. The downvote is justified, but I would appreciate a short comment. $\endgroup$ – user128785 Feb 24 '18 at 10:06

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