Shortening the length of pendulum

[Saitama]

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.

 

[TSny]

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 https://archive.org/stream/Mechanics_541/LandauLifshitz-Mechanics#page/n161/mode/2up 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.

 

[TSny]

Hello Pranav...

From post #22 ,do you get (-3/4)(dL/L) = (dθ_m/θ_m)

That looks good to me.

 

[voko]

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 :)

 

[Saitama]

That looks good to me.

But how does Tanya finds this from post #22 which has got a sign error?

 

[TSny]

But how does Tanya finds this from post #22 which has got a sign error?

Good question. Maybe she accounted for the sign error.

 

[Tanya Sharma]

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 ?

 

[TSny]

Yes, that's right.

 

[Tanya Sharma]

Okay...another thing I wanted to ask is shouldn't the E_SHM = y + (1/2)mgLθ_m² since we have chosen reference point to be the floor?

It is another thing that it won't make a difference .

 

[Tanya Sharma]

The energy in SHM is (1/2)mω²A² where ω = √(g/L) .This gives SHM energy (1/2L)mgθ_m² .

But it should be (1/2)mgLθ_m² .

Why this apparent contradiction ?

 

[TSny]

Okay...another thing I wanted to ask is shouldn't the E_SHM = y + (1/2)mgLθ_m² 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?

 

[TSny]

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.

 

[Tanya Sharma]

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 E_SHM ?

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

Is it correct ?

 

[TSny]

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)Iω²Θ_m² where I is the appropriate moment of inertia.

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

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

Is it correct ?

Yes, I believe that's right.

 

[Tanya Sharma]

I guess a fairly general expression for the energy of angular SHM would be E = (1/2)Iω²Θ_m² 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 = E_SHM + 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 E_SHM 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?

 

[TSny]

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 + E_SHM.

Here, E_SHM 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).

 

[Tanya Sharma]

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.