Seeking a simple logical argument to an interesting statement (spring-mass motion)

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  • #51
haruspex
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however, in the context of the problem, "the mass's acceleration" refers to the conditions of the already postulated forces; not to some other unspecified source of tangential acceleration − I agree that it arguably was remiss of the testers to not say anything about how an orbit would be incipiated to begin with.
I have no idea what your point is.
The given solution claimed that when the velocity is vertical there is no vertical acceleration, it relied on this for completing the solution, yet it offered no reasoning to arrive at the claim. Yes, it can be shown to be true in the particular context of the problem, but that is a nontrivial step and is conspicuously absent from the given solution.
This has nothing to do with hypothesising other forces nor how the motion was initiated.
You had stated that only with the uniform circular motion would the argument in the first two sentences of the solution be correct ##-## given that on the way down, ##mg## is added, while on the way up, it is subtracted, the question whether the interaction of the spring force with the gravitational force keeps the velocity constant seems germane to me.
You are again confusing how to solve the original physics problem with whether the official solution is valid. If the official solution had first established that the motion is uniform, or if it had been given as a fact in the problem statement, I would have no complaint with those two sentences.
 
  • #52
haruspex
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Why does the periodic varying of the direction of ##\text P## between toward the ground and away from the ground not affect the speed of ##\text P##, as it would if the spring were a rigid tether, or if ##\text P## were like the bob of a pendulum in that regard?
Because the string has zero relaxed length (which is not so in the other scenarios you mention) it happens that the vertical component of the tension is determined by the vertical height of the bob and the horizontal component by the horizontal displacement. The motions in the two directions become independent.
Therefore each motion is SHM and the two have the same frequency. By arranging that they also have the same amplitude and are 90 degrees out of phase, we have circular motion. It is clear that the peak velocities will also be the same magnitude, and that this will be the speed at all times.
 
  • #53
TSny
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Why does the periodic varying of the direction of ##\text P## between toward the ground and away from the ground not affect the speed of ##\text P##, as it would if the spring were a rigid tether, or if ##\text P## were like the bob of a pendulum in that regard?
The spring in this problem is very different from a string (rigid tether). The important thing about the spring is that it has zero natural length.

In this problem, if you try to make the mass move as a pendulum, it just doesn't work. Suppose you let the mass hang at rest from point A. The mass will be located at point C. Now, by hand, move the mass along a circular arc (about A) to a new position as shown. Release the mass at rest from this point. Will it swing back down in a circular arc like a pendulum?

1595799305393.png



No, it won't move like that. P will move in SHM along the blue straight line segment shown, centered on C.
1595799571024.png
 
  • #54
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I think that in your postulated experiment the mass would move in a circular arc.
 
  • #55
haruspex
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I think that in your postulated experiment the mass would move in a circular arc.
I showed in post #52 that the general motion for a zero length spring is an ellipse centred at the resting equilibrium position. So in general there is no point where it is instantaneously stationary.
The exception is for a degenerate ellipse, i.e. a straight line. So if it is released from rest it will oscillate in a straight line as @TSny depicts.
 
  • #56
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I showed in post #52 that the general motion for a zero length spring is an ellipse centred at the resting equilibrium position. So in general there is no point where it is instantaneously stationary.
The exception is for a degenerate ellipse, i.e. a straight line. So if it is released from rest it will oscillate in a straight line as @TSny depicts.
I think that it would not move in a straight line.
 
  • #57
haruspex
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I think that it would not move in a straight line.
It is not enough to have intuition. Both @TSny and I have provided arguments why it is a straight line. If you believe otherwise either find the flaw in our arguments or present your own analysis. Or both.
 
  • #58
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It is not enough to have intuition. Both @TSny and I have provided arguments why it is a straight line. If you believe otherwise either find the flaw in our arguments or present your own analysis. Or both.
Well, quoting TSny:
The spring in this problem is very different from a string (rigid tether). The important thing about the spring is that it has zero natural length.

In this problem, if you try to make the mass move as a pendulum, it just doesn't work. Suppose you let the mass hang at rest from point A. The mass will be located at point C. Now, by hand, move the mass along a circular arc (about A) to a new position as shown. Release the mass at rest from this point. Will it swing back down in a circular arc like a pendulum?

1595812930601.png


No, it won't move like that. P will move in SHM along the blue straight line segment shown, centered on C.

1595812982124.png
That is not a proof. It's not really even an argument. @TSny did not present it as such.
 
  • #59
TSny
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I think that in your postulated experiment the mass would move in a circular arc.
It is a simple consequence of what was proven in post #22 that if you release P from rest at any point other than point C, P will move in SHM along a straight line through C. Is there a specific part of the argument of post #22 that you do not accept?
 
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  • #60
haruspex
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That is not a proof. It's not really even an argument.
You are deflecting again, ignoring the two posts that do present the argument (#22 and #52) and electing to respond to one that doesn’t.
 
  • #61
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let be ##\vec r ## the position vector of the particle. A the spring has zero relaxed length, the spring force is -k ##\vec r##. the total force also includes gryvity. WE set the coordinate system with horizontal x axis and vertical down y axis. The equation of motion in coordinates is
##m\ddot x=-kx##
##m\ddot y=-ky +mg##
This is a linear , constant coeffient, inhomogeneous DE. the solution of the homogeneous part is a two-dimensional vibration with angular frequancy
##\omega= \sqrt{k/m}## A particular solution of the inhomogeneous equation is ##h= mg/k## that corresponds to the steady state. For the new variables X=x and Y= y-h, the differential equation becomes ##m\ddot X = - kX## and ##m\ddot Y=-kY##. The solution can be a circular motion with angular frequeny ##\omega= \sqrt{k/m}##. and arbitrary radius R. Then the speed is constant, ##\omega R##..
The problem maker might have took that straightforward, when assuming constant speed.
 
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  • #62
haruspex
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The problem maker might have took that straightforward, when assuming constant speed.
I don't buy that. It's far too big a jump to be made knowingly in a "textbook solution".
 
  • #63
ehild
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I don't buy that. It's far too big a jump to be made knowingly in a "textbook solution".
I agree. Assuming constant speed was not at all straightforward.
 
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Is there a specific part of the argument of post #22 that you do not accept?
That post included:
The net force on ##P## is ##\vec F_{net} = m \vec g -k \vec R =- k(- \vec h + \vec R) = -k \vec r##.
It seems to that it should be ##+m \vec g## on the way down, and ##-m \vec g## on the way up.
ehild said:
let be ##\vec r ## the position vector of the particle. A the spring has zero relaxed length, the spring force is -k ##\vec r##. the total force also includes gryvity. WE set the coordinate system with horizontal x axis and vertical down y axis. The equation of motion in coordinates is
##m\ddot x=-kx##
##m\ddot y=-ky +mg##
It seems to that it should be ##+mg## on the way down, and ##-mg## on the way up.
 
  • #65
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OK. Let me try this. Do you believe that the constant speed is true if there were no gravity (g=0)?
Yes, but then the rotation would be circular about ##\text A##, rather than about ##\text C##.
 
  • #66
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But gravity can be eliminated from the equation of motion by simply moving the origin : y→ y+##\frac {mg} k## This constant translation cannot change any velocities. It does change the origin to C

I am frankly mystified that there is an argument here.
 
  • #67
TSny
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It seems to that it should be ##+m \vec g## on the way down, and ##-m \vec g## on the way up.
The force of gravity acting on an object is the same whether the object is standing still, moving upwards, downwards, sideways, or in any other direction. I wrote the force as ##m \vec g## where ##\vec g## is a vector pointing downwards with magnitude equal to the acceleration due to gravity.
 
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  • #68
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The OP was answered completely with a three line explanation before post #10.
Perhaps it would serve us all well to start a new post with a precise statement of the outstanding issues.
I have no idea what they are, and so look forward to the new post.
 
  • #69
haruspex
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Yes, but then the rotation would be circular about ##\text A##, rather than about ##\text C##.
No, in the absence of gravity A and C would be the same point.
And as @TSny reminds you, gravity doesn't suddenly change sign according to which way an object moves.
 
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  • #70
haruspex
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The OP was answered completely with a three line explanation before post #10.
Well, there does seem to have been some confusion even there.

What the OP asked was whether the official solution was valid, and I think we all agree it was 'inadequate', i.e. either the solver had made a huge unjustified jump or had used flawed reasoning.

Maybe there was also an implied question of what the simplest solution would be.
The crucial step, I would say, is to show that because of the relaxed zero length the horizontal and vertical components of the tension are in (the same) proportion to the horizontal and vertical components, respectively, of displacement from the tethered point. That makes the two motions SHM at the same frequency.
The circularity then fixes them as having the same amplitude and being 90 degrees out of phase.

But I know you argued that it suffices to use the info that the motion is circular. I can see that in principle that may do it, but I am afraid I still haven’t understood how your argument works. Maybe take that into a private discussion.
 
  • #71
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No, in the absence of gravity A and C would be the same point.
That's a definitional difference. The mass rotates about point ##\text A## in either case. Removing the gravity would make the rotation about point ##\text A## circular with the center at point ##\text A##, instead of the rotation being circular about the point formerly called point ##\text C##. In discussing the effect of removing gravity, I regarded point ##\text C## as a static point already defined as in the original problem. You're using a dynamic definition of point ##\text C## to mean the center of the circle of rotation, wherever that is. The effect of removing gravity is the same by either description.
 
  • #72
haruspex
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That's a definitional difference. The mass rotates about point ##\text A## in either case. Removing the gravity would make the rotation about point ##\text A## circular with the center at point ##\text A##, instead of the rotation being circular about the point formerly called point ##\text C##. In discussing the effect of removing gravity, I regarded point ##\text C## as a static point already defined as in the original problem. You're using a dynamic definition of point ##\text C## to mean the center of the circle of rotation, wherever that is. The effect of removing gravity is the same by either description.
In post #65 you wrote ".... rather than about C", implying it would not be about C. I merely pointed out that in the absence of gravity, by definition of those points, A and C wouid be the same point. Thus, the correct way to view it is that the rotation always centres on C.
 
  • #73
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But I know you argued that it suffices to use the info that the motion is circular. I can see that in principle that may do it, but I am afraid I still haven’t understood how your argument works. Maybe take that into a private discussion.
What I said was much stronger than that and much simpler. With a simple change in origin, the equations of motion have complete circular symmetry and no longer contain the gravity constant g.

What is there to discuss? (If I seem vexed it is because I am! This is just a simple truth. If you feel the need to describe the springs in gory detail then OK we can do that but it cannot change the result)
 
  • #74
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In post #65 you wrote ".... rather than about C", implying it would not be about C. I merely pointed out that in the absence of gravity, by definition of those points, A and C wouid be the same point. Thus, the correct way to view it is that the rotation always centres on C.
What I said meant that it would no longer be about point ##\text C##, by which I meant the point about which mass ##\text P## would rotate circularly in the presence of gravity. That point would still exist without gravity, and I didn't change its label in referencing it as no longer the center of the circular rotation once gravity is removed. I could have called it e.g. 'point ##\text C_{old}##', but I think that which point I meant was not unclear.
 
  • #75
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What I said was much stronger than that and much simpler. With a simple change in origin, the equations of motion have complete circular symmetry and no longer contain the gravity constant g.

What is there to discuss? (If I seem vexed it is because I am! This is just a simple truth. If you feel the need to describe the springs in gory detail then OK we can do that but it cannot change the result)
In SHM the velocity constantly varies, unlike in uniform circular motion. I alluded to the variation in velocity in post #21, without using the term 'SHM' (I said that the variation in velocity of the rotation would be sinusoidally periodic). I think that strictly uniform rotation is not present, and that the constancy of the period and frequency suffices, along with the problem statement, to justify the argument in sentences 1 and 2 of the given solution.

From: https://en.wikipedia.org/wiki/Simple_harmonic_motion:

1595904607073.png

(edited to freeze the animation)

No-one who has done (or even witnessed) a bungee cord jump would suppose that the velocity is constant, but some might be surprised that the period and frequency are (ideally) constant

1595895271181.png
 

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