I_laff
- 41
- 2
I am aware that the negative derivative of potential energy is equal to force. Why is the max force found when the negative derivative of potential energy is equal to zero?
It doesn't. If there is a turning value of Potential Energy with distance, the force is zero. As the hyperphysics link tells you, the highest force is where the Potential Energy is changing fastest with distance.I_laff said:So if we set the derivative to zero, we can calculate the minima and maxima of a potential energy graph, how does that help us find the max force?
I_laff said:So if we set the derivative to zero, we can calculate the minima and maxima of a potential energy graph, how does that help us find the max force?
This is incorrect. When the negative gradient of the PE is zero then the force is zero.I_laff said:Why is the max force found when the negative derivative of potential energy is equal to zero?
I_laff said:
I didn't say that Hyperphysics stated that the max force is when the derivative was set to 0. I said that I was told that it was, which confused me since it didn't make sense. My only mistake was assuming that the person who told me that the max force could be calculated by setting the derivative to zero was correct.ZapperZ said:Where exactly in the hyperphysics link does it claim that "... max force found when the negative derivative of potential energy is equal to zero ... "? I do not see it, and you're asking us to correct an non-existing error.
Force is the gradient of the potential energy. This means that the quicker the potential energy changes over distance, the higher the absolute value of the force. So it is not the absolute value of the potential energy, but rather the change in the potential energy that corresponds to the force.
Zz.
I_laff said:I didn't say that Hyperphysics stated that the max force is when the derivative was set to 0. I said that I was told that it was, which confused me since it didn't make sense. My only mistake was assuming that the person who told me that the max force could be calculated by setting the derivative to zero was correct.
So the maximum occurs where the slope of the potential energy curve is steepest. You can find those points by differentiating the potential energy twice with respect to position and setting that equal to 0. Only in those points can the force be maximal. (there could also be a minimum, or an inflection point)I_laff said:I didn't say that Hyperphysics stated that the max force is when the derivative was set to 0. I said that I was told that it was, which confused me since it didn't make sense. My only mistake was assuming that the person who told me that the max force could be calculated by setting the derivative to zero was correct.
willem2 said:So the maximum occurs where the slope of the potential energy curve is steepest. You can find those points by differentiating the potential energy twice with respect to position and setting that equal to 0. Only in those points can the force be maximal. (there could also be a minimum, or an inflection point)
I would've thought the quoted post below would've clarified what information was and was not from the Hyperphysics website.ZapperZ said:Look at Posts 1, 2, and 3 and read them again in sequence to see why it appears that you are using the Hyperphysics link to justify what you were told.
Zz.
I_laff said:I was told by someone that when you set the derivative to zero, you get the max force.
I_laff said:I would've thought the quoted post below would've clarified what information was and was not from the Hyperphysics website.
I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.ZapperZ said:This is not correct either. 2nd derivative tells you how rapidly the slope is changing, and thus, how rapidly the force changes. It doesn't tell you how large the force is.
You can have an almost vertical straight line on the V vs x graph (i.e. 2nd derivative is zero but not at a max or min or inflection point), and yet, this is where a force can be a maximum.
I think your confusion is that you didn't just stop after the relationship was described properly with Maths. There is no point in 'talking around' something as straightforward as the way Hyperphysics states it. Whatever "somebody" said about it, it just adds confusion when that statement is included in the formal definition. If you understand the Maths then just use it and get familiar with the way it applies. The content of the first post is correct so why are we still discussing it, twenty posts later? Ignore your unhelpful, unofficial source.willem2 said:I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.
You are correct, @ZapperZ was confused.willem2 said:I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.
willem2 said:I don't get this at all. You say you disagree with me, but then you produce a case where the 2nd derivative is 0 and the force is a maximum, but this is not a max or min or inflection point? This appears to be a contradiction.
The force is the first derivative of the potential energy. The only place where the force can be a maximum is where the derivative of the force, or the second derivative of the potential energy is 0.
willem2 said:ou can find those points by differentiating the potential energy twice with respect to position and setting that equal to 0. Only in those points can the force be maximal. (there could also be a minimum, or an inflection point)
ZapperZ said:You can have an almost vertical straight line on the V vs x graph (i.e. 2nd derivative is zero but not at a max or min or inflection point), and yet, this is where a force can be a maximum.
ZapperZ said:I'll give you another example since there straight-line curve didn't sink in with you: U = ½ kx2
What is the 2nd derivative of that? Is it zero?
Yet, there IS a maximum force for this potential when we do a simple Hooke's law experiment.
willem2 said:If the potential is reallly ½ kx2 for all x, there is no place where the second derivative is 0, but there is also no place where the force has a maximum. Of course real springs will break at some point, but U = ½ kx2 won't be valid anymore before this happens.
The use of the term "maximum force" can mean one of two things. 1. The Force / Displacement Law has a local maximum or 2. You chose to limit the range of forces or displacements in your experiment.ZapperZ said:One can find maximum force even if the system has no local max/min...
sophiecentaur said:The use of the term "maximum force" can mean one of two things. 1. The Force / Displacement Law has a local maximum or 2. You chose to limit the range of forces or displacements in your experiment.
Why is this verbal jiggery pokery being involved with a perfectly good mathematical description? If you want to be realllllly precise then include some more formal definitions and caveats such a that the function is continuous and differentiable within the limits being considered etc. etc. but would that really help?
Mods please help and put this thread out of its misery.
That is out of context but not wrong. The restoring force is proportional to the displacement away from the rest position. Any given oscillator falls into category 2 in my post. The maximum displacement is arbitrary and is chosen of the particular experiment and is nothing to do with the Force Law. Confusion could be caused but only because of lack of description. The PE variation with time is sinusoidal and the maximum of force corresponds to the maximum of PE. But this is just rehearsing what we know already and the sheer number of these posts is contributing more to confusion than any statement in any particular post.ZapperZ said:Someone who reads Post #14, and then learn about harmonic oscillator, will get VERY confused.
sophiecentaur said:That is out of context but not wrong. The restoring force is proportional to the displacement away from the rest position. Any given oscillator falls into category 2 in my post. The maximum displacement is arbitrary and is chosen of the particular experiment and is nothing to do with the Force Law. Confusion could be caused but only because of lack of description. The PE variation with time is sinusoidal and the maximum of force corresponds to the maximum of PE. But this is just rehearsing what we know already and the sheer number of these posts is contributing more to confusion than any statement in any particular post.
As with many questions and solutions to many problems, the initial statement about the physical situation has to be clear and bombproof. Introducing SHM is a red herring because it introduces Time into the situation and the term "maximum" can either imply dy/dx = 0 or dy/dt = 0. There's your potential confusion.
Iirc, you introduced the possible problems when students come across the Harmonic Oscillator. Is there an earlier mention of time? If you use the term "derivative" then the independent variable needs to be stated. Up till your post the dx was all we were concerned with and it would be best to avoid mention (implied or otherwise) of dynamics.ZapperZ said:I did not introduce time. You did.
sophiecentaur said:Iirc, you introduced the possible problems when students come across the Harmonic Oscillator. Is there an earlier mention of time? If you use the term "derivative" then the independent variable needs to be stated. Up till your post the dx was all we were concerned with and it would be best to avoid mention (implied or otherwise) of dynamics.
You were right to challenge the use of the second derivative as it doesn't help but the OP was almost driven by the thread into over-thinking the issue.
Haha - it didn't help that the original statement in the first post was wrong.
I will get on my soap box and shout the praises of Worked Examples from The Book. Doing a couple of those can usually put people right, far better than any amount of chat.
If time is not an issue then how would a student be confused in the context of a harmonic oscillator? The maximum displacement in SHM is arbitrary and depends on where it's been let go.- That's my second category (2) of Maximum and it doesn't involve differentiation with respect to displacement. I am not confused about that. I really think that bringing in SHM was not helpful.ZapperZ said:Until you brought it up, the derivative with respect to any other variables, such as time, was NEVER an issue
sophiecentaur said:If time is not an issue then how would a student be confused in the context of a harmonic oscillator? The maximum displacement in SHM is arbitrary and depends on where it's been let go.- That's my second category (2) of Maximum and it doesn't involve differentiation with respect to displacement. I am not confused about that. I really think that bringing in SHM was not helpful.
Talking to the wrong person is a common problembut it doesn't make my Physics wrong.
Just one turning value. A minimum at x=0 and, of course, a finite maximum displacement but the derivative increases all the time. Would you call that "a maximum" at the arbitrary extremes of displacement?ZapperZ said:The harmonic oscillator has the potential form of U = 1/2 kx2.
In the displacement / force law there is no 'maximum' because there is no turning value up there. I think your confused student would just sit and think about it for a minute and realize the difference. Just because the oscillator only explores a part of the Hooke's law doesn't mean that it's found a maximum. If he/she is interested enough to find this a problem all that's necessary is to increase the displacement and prove that the force still increases. Hooke's Law doesn't predict a turning value anywhere until the spring breaks. I guess that would be a valid maximum.ZapperZ said:Does this mean that there are no values of "x" for the force to be a maximum?
Now now! That should be in a PM, I think.ZapperZ said:No, but it shows that you are equally confused in many of these conversations.
sophiecentaur said:Just one turning value. A minimum at x=0 and, of course, a finite maximum displacement but the derivative increases all the time. Would you call that "a maximum" at the arbitrary extremes of displacement?
In the displacement / force law there is no 'maximum' because there is no turning value up there. I think your confused student would just sit and think about it for a minute and realize the difference. Just because the oscillator only explores a part of the Hooke's law doesn't mean that it's found a maximum. If he/she is interested enough to find this a problem all that's necessary is to increase the displacement and prove that the force still increases. Hooke's Law doesn't predict a turning value anywhere until the spring breaks. I guess that would be a valid maximum.
Without using Maths, how would you know the Potential Energy for a given displacement?ZapperZ said:The mathematics does NOT tell you this.
That is entirely up to the choice of the experimenter and has nothing do do with the Physics of the system. And I think this is the nub of our disagreement. If one was interested in the Physics of the setup, one would very soon come to the conclusion (experimentally) that there is no 'Maximum' except either when the experimenter chickens out or the spring departs from the simple law and would you seriously call that a maximum? (Apart from it being the biggest number used).ZapperZ said:picking up the location of maximum force.
I totally agree and have made this same point several times higher up. And slope of PE against distance, as we all agree, tells you the Force. (Ye Gods - that's more than thirty posts.ZapperZ said:To me, this thread has been sufficiently answered PRIOR to Post #14.