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Shortening the length of pendulum

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  • #26
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Sounds right. Final PE = mg(y+dy).
Thanks a lot for the help TSny! Doing that gives the correct answer. :smile:

But I am still confused on the signs. Sorry this is going to be dumb but why not replace every dL with (-dL)?

Also, you said that this problem is a well-known example of "adiabatic invariant" in mechanics. Do you know of any book which deals with this?
 
  • #27
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Hello Pranav...

From post #22 ,do you get (-3/4)(dL/L) = (dθmm)
 
  • #28
TSny
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But I am still confused on the signs. Sorry this is going to be dumb but why not replace every dL with (-dL)?
I don't see why you would want to do this. But I guess you could. I think it would confuse me.

Also, you said that this problem is a well-known example of "adiabatic invariant" in mechanics. Do you know of any book which deals with this?
This is sometimes covered in mechanics texts which treat the Hamiltonian formalism. I think the mechanics text by Landau and Lifshitz gives a formal treatment. Also, Goldstein's text (I believe). These are graduate level texts.

For the pendulum, the adiabatic invariant is the ratio energy/frequency. This ratio remains constant as the pendulum's length is slowly changed. Thus E/f = h for some constant h. [Hmm....E = hf.....]

Adiabatic invariants were important in the "Old Quantum Theory" of around 1920 where quantization conditions were imposed on certain adiabatic invariants. Max Born's The Mechanics of the Atom treats the old quantum theory and he derives the adiabatic invariant of the pendulum as an example. See here

I just found http://quantum-history.mpiwg-berlin.mpg.de/eLibrary/hq1_talks/old-qt/07_jordiPerez/perez_preprint.pdf [Broken] which I have only quickly scanned, but it looks interesting.
 
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  • #29
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Hello Pranav...

From post #22 ,do you get (-3/4)(dL/L) = (dθmm)
Hi Tanya! :)

No, I don't get (-3/4), I instead get (-1/4). Can you please show the steps?

This is sometimes covered in mechanics texts which treat the Hamiltonian formalism. I think the mechanics text by Landau and Lifshitz gives a formal treatment. Also, Goldstein's text (I believe). These are graduate level texts.
I doubt I will be ever using them. :biggrin:

For the pendulum, the adiabatic invariant is the ratio energy/frequency. This ratio remains constant as the pendulum's length is slowly changed. Thus E/f = h for some constant h. [Hmm....E = hf.....]

Adiabatic invariants were important in the "Old Quantum Theory" of around 1920 where quantization conditions were imposed on certain adiabatic invariants. Max Born's The Mechanics of the Atom treats the old quantum theory and he derives the adiabatic invariant of the pendulum as an example. See here

I just found this which I have only quickly scanned, but it looks interesting.
Ah, this quantum stuff is way above my current level. I did see the Max Born's derivation and its looks quite similar to what we just did. Thanks for those links TSny! :smile:
 
  • #30
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This is sometimes covered in mechanics texts which treat the Hamiltonian formalism. I think the mechanics text by Landau and Lifshitz gives a formal treatment. Also, Goldstein's text (I believe).
Goldstein has it, Landau does not (at least in the volume titled "Classical mechanics"). Another reference is Arnold's Mathematical methods in classical mechanics.

Speaking of the graduate level texts, I wonder whether there is a simpler solution, which would not involve averaging? I think the whole idea may be alien to an intro level course which I think Pranav is following.
 
  • #31
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I think the whole idea may be alien to an intro level course which I think Pranav is following.
Yes, I am unaware of this.

I had this problem bookmarked from quite a long time but I finally decided to post it here. The given solution also goes along the lines of averaging (I looked at the solution a year before, I should have the solution bookmarked too, should I post it here?) but since I had trouble understanding what's going on, I had to post it here. I face no problem in finding the average but this method of solving the problem is definitely new to me.
 
  • #32
TSny
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Goldstein has it, Landau does not (at least in the volume titled "Classical mechanics"). Another reference is Arnold's Mathematical methods in classical mechanics.
Landau's Mechanics (Vol. 1 of the Course in Theoretical Physics) has a brief treatment in section 49. See here This is not a section to just "jump into"!

Speaking of the graduate level texts, I wonder whether there is a simpler solution, which would not involve averaging? I think the whole idea may be alien to an intro level course which I think Pranav is following.
Yes, It would be nice to see a different way to solve it.
 
  • #33
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Hello Pranav...

From post #22 ,do you get (-3/4)(dL/L) = (dθmm)
That looks good to me.
 
  • #34
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Landau's Mechanics (Vol. 1 of the Course in Theoretical Physics) has a brief treatment in section 49.
Oops, indeed. Not sure how I could miss that :)
 
  • #35
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That looks good to me.
But how does Tanya finds this from post #22 which has got a sign error? :confused:
 
  • #36
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But how does Tanya finds this from post #22 which has got a sign error? :confused:
Good question. Maybe she accounted for the sign error. :smile:
 
  • #37
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Thanks TSny...

Why is work done by tension -TdL ? Is it because the work done by tension is positive and since dL is negative ,we put a minus sign to make term positive ?
 
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  • #38
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Yes, that's right.
 
  • #39
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Okay...another thing I wanted to ask is shouldn't the ESHM = y + (1/2)mgLθm2 since we have chosen reference point to be the floor?

It is another thing that it won't make a difference .
 
  • #40
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The energy in SHM is (1/2)mω2A2 where ω = √(g/L) .This gives SHM energy (1/2L)mgθm2 .

But it should be (1/2)mgLθm2 .

Why this apparent contradiction ?
 
  • #41
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Okay...another thing I wanted to ask is shouldn't the ESHM = y + (1/2)mgLθm2 since we have chosen reference point to be the floor?

It is another thing that it won't make a difference .
I'm not sure I understand your question. Are you asking about why I wrote the energy as given in post #19?
 
  • #42
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The energy in SHM is (1/2)mω2A2 where ω = √(g/L) .
This is valid if the amplitude A is a spatial distance (having dimensions of length). But it's not valid if A represents an angular amplitude.
 
  • #43
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This is valid if the amplitude A is a spatial distance (having dimensions of length). But it's not valid if A represents an angular amplitude.
Okay...Thanks for clearing my misconception .Is there a standard result like (1/2)mω2A2 when we are dealing with angular displacements ? Or we have to derive the result according to the problem .

I'm not sure I understand your question. Are you asking about why I wrote the energy as given in post #19?
Sorry for being unclear . I will come back to this question later .

Please check my understanding...How do we derive ESHM ?

ESHM = Gravitational potential energy at the maximum displacement .So considering reference to be the lowest point of the pendulum ,GPE at θm = mgL(1-cosθm) .Now we approximate the function .

Is it correct ?
 
  • #44
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Is there a standard result like (1/2)mω2A2 when we are dealing with angular displacements ? Or we have to derive the result according to the problem .
I guess a fairly general expression for the energy of angular SHM would be E = (1/2)Iω2Θm2 where I is the appropriate moment of inertia.

Please check my understanding...How do we derive ESHM ?

ESHM = Gravitational potential energy at the maximum displacement .So considering reference to be the lowest point of the pendulum ,GPE at θm = mgL(1-cosθm) .Now we approximate the function .

Is it correct ?
Yes, I believe that's right.
 
  • #45
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I guess a fairly general expression for the energy of angular SHM would be E = (1/2)Iω2Θm2 where I is the appropriate moment of inertia.
Excellent .

Yes, I believe that's right.
In the small angle approximation, the energy of the system is E = ESHM + mgy where y is the height of the lowest point of the swing above the floor. Like any SHM, you should be able to express ESHM in terms of the amplitude of the motion. Then consider dE for a small change in length dL of the pendulum and relate it to the work done by the tension.
Now,back to my question in post#39 .In post#19 ,we chose reference of GPE to be at floor but while calculating Total energy of SHM ,we chose reference to be at the lowest point of the pendulum.

How are we considering two references simultaneously for calculating GPE?
 
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  • #46
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Now,back to my question in post#39 .In post#19 ,we chose reference of GPE to be at floor but while calculating Total energy of SHM ,we chose reference to be at the lowest point of the pendulum.

How are we considering two references simultaneously for calculating GPE?
I think we always used the frame of reference in which PE of gravity is measured from the floor. So, E = mgy + ESHM.

Here, ESHM includes only the PE due to height above the lowest point of the oscillation while mgy takes into account the height of the lowest point of the oscillation above the floor. So, together we account for the total PE relative to the floor.

The mgy term cancels out later on when setting up dE = -TdL. See post #14 for the expression of the average tension force. Thus, dE has a term mg(dy) while the work -TdL has a term -mg(dL) = -mg(-dy) = mg(dy).
 
  • #47
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Thank you very much TSny .You are simply brilliant.

Thanks Pranav for putting up an excellent question .Your follow up with the problem has been remarkable.
 

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