Period of Oscillation of Pendulum Varies with Amplitude

[armis]

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.

Homework Equations

T(E) = \sqrt(2m) \int^{x_2(E)}_{x_1(E)}\frac{dx}{\sqrt(E-U(x)}

where T is the period of oscillations

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 ) the energy of the pendulum

E = \frac{1}{2}ml^2{\phi}^2 - mgl\cos{\phi} = -mgl\cos{\phi_o}

where \phi is the angle between the string and the vertical and \phi_0 the maximum value of \phi

Any hints?

thanks

 

[tiny-tim]

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 ) the energy of the pendulum

E = \frac{1}{2}ml^2{\phi}^2 - mgl\cos{\phi} = -mgl\cos{\phi_o}

where \phi is the angle between the string and the vertical and \phi_0 the maximum value of \phi

Hi armis!

KE + PE = constant. So …

Hint: at what angle does KE = 0?

 

[armis]

Hi Tiny-tim

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

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

Let's see:

x= l\sdot\sin{\phi}
y= l\sdot\cos{\phi}

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

\frac{dx}{dt}= l\sdot\cos{\phi}
\frac{dy}{dt}=-l\sdot\sin{\phi}

Thus KE=\frac{ml^2}{2} Which is totally ridicilous... It must depend on time! Where did I go wrong?

 

[armis]

After staring at my post for a while I realized that \phi actually does depend on time. Thus I forgot the derrivative of \phi ( correct me if my termionology is wrong please)
Oh, my godness... I am so terrible at this

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

But I still can't get the second part right

 

[tiny-tim]

Hi armis!

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).

 

[armis]

I think you need some sleep … :zzz:

I certainly do 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
x=l\sin{\phi}
y=l\cos{\phi}
then
\frac{dx}{dt}=l\sin{\phi}\frac{d\phi}{dt}
\frac{dy}{dt}=-l\cos{\phi}\frac{d\phi}{dt}
thus
KE=\frac{m}{2}({\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2)=\frac{m}{2}l^2(\frac{d\phi}{dt})^2
which is exactly what is given as the answer or I am completely wrong. How can I do that without derivatives?

lcosφ - lcosφo is the difference in height (for working out the PE).

I understand what you mean but I am still confused

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

 

[alphysicist]

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 -mg\cos{\phi} is PE. Now since \phi is the angle between the string and the vertical then once \phi is 0 then the potential energy has the highest value? I just can't get it. Isn't it supposed to be 0 at \phi=0?

When \phi is 0, the magnitude of that term ( -mg\ell \cos{\phi}) [/tex] is a maximum, but since it is negative, the potential energy will have it&#039;s smallest (most negative) value of the motion.<br /> <br /> You don&#039;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.

 

[armis]

Hi Alphysicist

Well

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

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

That's as standart as I could figure it out

As for the difference in PE:

- mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} )

Is this correct?

thanks

 

[alphysicist]

Hi Alphysicist

Well

KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2
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
F = - \nabla PE(q) thus in this case PE=-Fr

That's as standart as I could figure it out

As for the difference in PE:

- mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} )

Is this correct?

thanks

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

- mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} )

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

 

[tiny-tim]

Hi armis!

I know you've got it now, but let's just clarify what you did here:

Hmm, why? If
x=l\sin{\phi}
y=l\cos{\phi}
then
\frac{dx}{dt}=l\sin{\phi}\frac{d\phi}{dt}
\frac{dy}{dt}=-l\cos{\phi}\frac{d\phi}{dt}
thus
KE=\frac{m}{2}({\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2)=\frac{m}{2}l^2(\frac{d\phi}{dt})^2
which is exactly what is given as the answer or I am completely wrong. How can I do that without derivatives?

As you say, that's the correct formula for the KE.

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

But the formula without derivatives gave you the PE!

 

[armis]

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 \phi 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 -m g l \cos\phi_0 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

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

!Exactly... 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

 

[alphysicist]

One question though. If I choose the generalised coordinate as \phi and l then I will not arrive at the KE formula I did previously. What am I missing?

The specific formula you entered
KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2

where m is the mass, is not true when the generalized coordinate does not have units of length. If you want to use \phi as your coordinate, then you will end up with the rotational kinetic energy formula (1/2) I \dot\phi^2. 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 v=r\dot\phi in the kinetic energy formula (1/2) m v^2

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

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).

 

[armis]

Oh, I see. Thanks a lot!