• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Simple pendulum

  • Thread starter armis
  • Start date
103
0
1. Homework Statement
Determine the period of oscillations of a simple pendulum ( a particle of mass m suspended by a string of length l in a gravitational field) as a function of the amplitude of oscillations.

2. Homework Equations
[tex] T(E) = \sqrt(2m) \int^{x_2(E)}_{x_1(E)}\frac{dx}{\sqrt(E-U(x)} [/tex]

where T is the period of oscillations

3. The Attempt at a Solution
I only need the expression of E(x) and the problem is pretty much solved but I can't figure out why (which is rather embarassing :frown:) the energy of the pendulum

[tex] E = \frac{1}{2}ml^2{\phi}^2 - mgl\cos{\phi} = -mgl\cos{\phi_o} [/tex]

where [tex] \phi [/tex] is the angle between the string and the vertical and [tex] \phi_0 [/tex] the maximum value of [tex] \phi [/tex]

Any hints?

thanks
 
Last edited:

tiny-tim

Science Advisor
Homework Helper
25,790
249
I only need the expression of E(x) and the problem is pretty much solved but I can't figure out why (which is rather embarassing :frown:) the energy of the pendulum

[tex] E = \frac{1}{2}ml^2{\phi}^2 - mgl\cos{\phi} = -mgl\cos{\phi_o} [/tex]

where [tex] \phi [/tex] is the angle between the string and the vertical and [tex] \phi_0 [/tex] the maximum value of [tex] \phi [/tex]
Hi armis! :smile:

KE + PE = constant. So …

Hint: at what angle does KE = 0? :smile:
 
103
0
Hi Tiny-tim :smile:

Yeah, I knew that! At least that... :smile:

KE is equal to zero at maximum value of [tex] \phi [/tex], isn't it? And the velocity is highest at point where [tex] \phi [/tex] is zero thus the kinetic energy has the highest value as well.

Let's see:

[tex] x= l\sdot\sin{\phi} [/tex]
[tex] y= l\sdot\cos{\phi} [/tex]

let's assume l doesn't change ( metal or something). Then

[tex] \frac{dx}{dt}= l\sdot\cos{\phi} [/tex]
[tex] \frac{dy}{dt}=-l\sdot\sin{\phi} [/tex]

Thus [tex] KE=\frac{ml^2}{2} [/tex] :confused: Which is totally ridicilous... It must depend on time! Where did I go wrong?
 
103
0
After staring at my post for a while I realised that [tex]\phi[/tex] actually does depend on time. Thus I forgot the derrivative of [tex]\phi[/tex] ( correct me if my termionology is wrong please)
Oh, my godness... I am so terrible at this :smile:

Actually in the scanned copy of the book the answer part is scanned so badly that one can't see the dot sign above [tex]\phi[/tex]. I was more focused on how to get [tex]\phi[/tex] in there...:uhh:I guess I should look the answers less and focus on the problem instead, that's for sure :approve:

But I still can't get the second part right
 
Last edited:

tiny-tim

Science Advisor
Homework Helper
25,790
249
Hi armis! :smile:

I think you need some sleep … :zzz:

You don't need derivatives at all …

lcosφ - lcosφo is the difference in height (for working out the PE). :smile:
 
103
0
I think you need some sleep … :zzz:
I certainly do :smile: but I fell in love in physics just yesterday and just can't stop. This book by Landau by the way from which the exercise is taken is brilliant

You don't need derivatives at all …
Hmm, why? If
[tex] x=l\sin{\phi} [/tex]
[tex] y=l\cos{\phi} [/tex]
then
[tex] \frac{dx}{dt}=l\sin{\phi}\frac{d\phi}{dt} [/tex]
[tex] \frac{dy}{dt}=-l\cos{\phi}\frac{d\phi}{dt} [/tex]
thus
[tex] KE=\frac{m}{2}({\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2)=\frac{m}{2}l^2(\frac{d\phi}{dt})^2 [/tex]
which is exactly what is given as the answer or I am completely wrong. How can I do that without derivatives? :confused:

lcosφ - lcosφo is the difference in height (for working out the PE). :smile:
I understand what you mean but I am still confused :blushing:

By the way in their given formula if I understand correctly the part [tex] -mg\cos{\phi} [/tex] is PE. Now since [tex] \phi [/tex] is the angle between the string and the vertical then once [tex] \phi [/tex] is 0 then the potential energy has the highest value? I just can't get it. Isn't it supposed to be 0 at [tex] \phi=0 [/tex]?

:cry:
 
Last edited:

alphysicist

Homework Helper
2,238
1
Hi armis,

You don't need derivatives to figure out your energy formula because you can use the standard formulas for kinetic and potential energies. What are those formulas?



By the way in their given formula if I understand correctly the part [tex] -mg\cos{\phi} [/tex] is PE. Now since [tex] \phi [/tex] is the angle between the string and the vertical then once [tex] \phi [/tex] is 0 then the potential energy has the highest value? I just can't get it. Isn't it supposed to be 0 at [tex] \phi=0 [/tex]?

:cry:
When [itex]\phi[/itex] is 0, the magnitude of that term [itex]( -mg\ell \cos{\phi}) [/tex] is a maximum, but since it is negative, the potential energy will have it's smallest (most negative) value of the motion.

You don't need the potential energy to have a specific value of 0 at any particular point, because all that matters is the difference in energy between two points.
 
103
0
Hi Alphysicist :smile:

Well

[tex] KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2 [/tex]
where q are the generalised coordinates and s are the degrees of freedom.

And if the force is conservative as it is here then
[tex] F = - \nabla PE(q) [/tex] thus in this case [tex] PE=-Fr [/tex]

That's as standart as I could figure it out :frown:

As for the difference in PE:

[tex] - mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} ) [/tex]

Is this correct?

thanks
 
Last edited:

alphysicist

Homework Helper
2,238
1
Hi Alphysicist :smile:

Well

[tex] KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2 [/tex]
where q are the generalised coordinates and s are the degrees of freedom.
That's right; and so by knowing that in this problem only the angular coordinate is changing with time you can just write down the kinetic energy term.

And if the force is conservative as it is here then
[tex] F = - \nabla PE(q) [/tex] thus in this case [tex] PE=-Fr [/tex]

That's as standart as I could figure it out :frown:

As for the difference in PE:

[tex] - mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} ) [/tex]

Is this correct?

thanks
Yes, so the potential energy at any particular point you want is given by that formula:

[tex] - mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} ) [/tex]

and remember that you can add any constant value you like to the potential energy; so to simplify add the constant [itex]-m g l \cos\phi_0[/itex] to that and you'll get the original energy formula.
 

tiny-tim

Science Advisor
Homework Helper
25,790
249
Hi armis! :smile:

I know you've got it now, but let's just clarify what you did here:
Hmm, why? If
[tex] x=l\sin{\phi} [/tex]
[tex] y=l\cos{\phi} [/tex]
then
[tex] \frac{dx}{dt}=l\sin{\phi}\frac{d\phi}{dt} [/tex]
[tex] \frac{dy}{dt}=-l\cos{\phi}\frac{d\phi}{dt} [/tex]
thus
[tex] KE=\frac{m}{2}({\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2)=\frac{m}{2}l^2(\frac{d\phi}{dt})^2 [/tex]
which is exactly what is given as the answer or I am completely wrong. How can I do that without derivatives? :confused:
As you say, that's the correct formula for the KE.

(Though you could have used the formula for converting angular velocity to speed: [itex]v = \omega r[/itex].)

But the formula without derivatives gave you the PE! :smile:
 
103
0
That's right; and so by knowing that in this problem only the angular coordinate is changing with time you can just write down the kinetic energy term.
One question though. If I choose the generalised coordinate as [tex]\phi[/tex] and l then I will not arrive at the KE formula I did previously. What am I missing?

and remember that you can add any constant value you like to the potential energy; so to simplify add the constant [itex]-m g l \cos\phi_0[/itex] to that and you'll get the original energy formula.
That's still a little bit confusing but I am getting there. At least by now I already understand that all what matters is the difference in potential energy but I still didn't figure the physical interpretation of this :redface:

(Though you could have used the formula for converting angular velocity to speed: [itex]v = \omega r[/itex].)
!!Exactly... :uhh: This is harder than I thought but I have learned a lot so far. Thanks for your help

By the way I am free to book suggestions on classical mechanics as by now it's rather clear I should read one. Landau is a pretty tough one
 
Last edited:

alphysicist

Homework Helper
2,238
1
One question though. If I choose the generalised coordinate as [tex]\phi[/tex] and l then I will not arrive at the KE formula I did previously. What am I missing?
The specific formula you entered
[tex] KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2 [/tex]

where m is the mass, is not true when the generalized coordinate does not have units of length. If you want to use [itex]\phi[/itex] as your coordinate, then you will end up with the rotational kinetic energy formula [itex](1/2) I \dot\phi^2[/itex]. Since l is not changing with time it does not have a kinetic energy term.

The alternative, which I would do, is to follow tiny-tim's good advice about using [itex]v=r\dot\phi[/itex] in the kinetic energy formula [itex](1/2) m v^2[/itex]


That's still a little bit confusing but I am getting there. At least by now I already understand that all what matters is the difference in potential energy but I still didn't figure the physical interpretation of this :redface:
The potential energy function here is mgh. By adding the constant mentioned earlier, h is the vertical position relative to the center of the circular path (so that h=0 when the string is horizontal).
 
103
0
Oh, I see. Thanks a lot!
 

Related Threads for: Simple pendulum

  • Posted
Replies
1
Views
943
  • Posted
Replies
8
Views
2K
  • Posted
Replies
2
Views
2K
  • Posted
Replies
8
Views
2K
  • Posted
Replies
3
Views
673
  • Posted
Replies
2
Views
3K
  • Posted
Replies
16
Views
598
  • Posted
Replies
3
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top