Small oscillations of a simple pendulum placed on a moving block

[MatinSAR]

Homework Statement

A simple pendulum with mass $m$ and length $L$ is placed on a block with mass $M$. The block can move freely on a frictionless horizontal surface. Find the equation of motion and the frequency of small oscillations.

Relevant Equations

Hamilton's principle and Lagrangian equation.

Hello. This is the figure of the problem:

First, we should determine the Lagrangian of the system. I have already completed this part without any issues. To respect everyone’s time, I won’t go into the details of how I accomplished it.

$$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$ To find equation of motion I should directly put the above lagrangian in lagrange's equation? Or is it possible to first simplify it for small oscillations?

Edit 1: I forgot to mention that for small oscillations we can ignore the higher powers of theta and its derivatives.

 

[erobz]

Homework Statement: A simple pendulum with mass $m$ and length $L$ is placed on a block with mass $M$. The block can move freely on a frictionless horizontal surface. Find the equation of motion and the frequency of small oscillations.
Relevant Equations: Hamilton's principle and Lagrangian equation.

Hello. This is the figure of the problem:
View attachment 343055
First, we should determine the Lagrangian of the system. I have already completed this part without any issues. To respect everyone’s time, I won’t go into the details of how I accomplished it. View attachment 343056
$$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$ To find equation of motion I should directly put the above lagrangian in lagrange's equation? Or is it possible to first simplify it for small oscillations?

I think you would be able to substitute now for $\cos \theta$

 

[MatinSAR]

I think you would be able to substitute now for $\cos \theta$

I wanted to simplify it for small oscillations then substitute it in lagrange's equation. But I was not sure. So you don't see a problem in doing this?
I will try it and share results as soon as I can ... Thank you.

 

[erobz]

I wanted to simplify it for small oscillations then substitute it in lagrange's equation. But I was not sure. So you don't see a problem in doing this?
I will try it and share results as soon as I can ... Thank you.

I could be wrong though ( you're already getting out of my wheelhouse of comfort), maybe try it both ways and see if you come to agreement (sometimes doing it twice is the best way to convince yourself)?

 

[Orodruin]

Yes, you can expand the Lagrangian for small $\theta$ and then apply the EL equation, but if you want linear EoMs, you need to expand to second order in $\theta$.

 

[MatinSAR]

I've forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives.

I could be wrong though ( you're already getting out of my wheelhouse of comfort)

No problem. Thanks for your help.

maybe try it both ways and see if you come to agreement?

I will.

 

[MatinSAR]

Yes, you can expand the Lagrangian for small $\theta$ and then apply the EL equation, but if you want linear EoMs, you need to expand to second order in $\theta$.

Good news ...
I forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives according to the question.
So ... Can I ignore $\dot \theta$ and $\theta^2$ at first place?

 

[PeroK]

Good news ...
I forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives according to the question.
So ... Can I ignore $\dot \theta ^2$ and $\theta^2$ at first place?

Definitely not!!

 

[MatinSAR]

Definitely not!!

Then I should do a lot of math to find its equations of motion, sadly. Thanks, I'll try and post the results soon.

 

[kuruman]

Personally, I prefer to first obtain the equations of motion and then consider approximations only because I don't trust myself not to throw out the baby with the bath water.

 

[MatinSAR]

Lagrangian of the system is: $$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$
For coordiante $x$ we have:
$$\frac {\partial L} {\partial x}-\frac {d}{dt}\frac {\partial L} {\partial \dot x}=0 $$ $$\frac {\partial L} {\partial \dot x}=A$$ A is a constant. $$(M+m)\dot x+mL\dot \theta cos \theta=A$$$$\dot x = \dfrac {A-ml\dot \theta \cos \theta}{M+m}$$$$x=\dfrac {1}{M+m}(At-ml\sin \theta)$$
I'm not sure with last part. Did I solve for $x$ correctly? Many thanks.

Personally, I prefer to first obtain the equations of motion and then consider approximations only because I don't trust myself not to throw out the baby with the bath water.

You’re right. I just intended to solve the question faster and get to the rest of the tasks.

 

[Orodruin]

Good news ...
I forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives according to the question.
So ... Can I ignore $\dot \theta$ and $\theta^2$ at first place?

Huh? Definitely not!

As I said you need to keep terms up to second order in the variable in the Lagrangian to obtain the linearised equations of motion. $\theta^2$ is second order so you cannot ignore it.

One thing to note about your Lagrangian is that it can be easily separated into a translational part and an oscillatory part. This makes the actual problem one-dimensional instead and quite easy to handle.

 

[kuruman]

I'm not sure with last part. Did I solve for $x$ correctly?

It doesn't look right. $\theta$ and $\dot{\theta}$ are both functions of time and you don't know what they look like, in fact you are looking for $\theta(t)$.

 

[MatinSAR]

One thing to note about your Lagrangian is that it can be easily separated into a translational part and an oscillatory part. This makes the actual problem one-dimensional instead and quite easy to handle.

I think this way is hard for me. I'm not familiar with oscillation that much. So I prefer to find equation of motion using Lagrange equations then simplify it the way that was mentioned in question.

It doesn't look right. $\theta$ and $\dot{\theta}$ are both functions of time and you don't know what they look like, in fact you are looking for $\theta(t)$.

Cant I wrote that as $d \theta / dt$? Then I can simplify integral to $cos \theta d\theta$.

I've seen the edit just now. Can I use the other equation that I can find from lagrangian for $\theta$ coordinate?

 

[kuruman]

That you can do to get $\int \cos(\theta(t))d\theta.$ Then what? You have an implicit function of time under the integral sign that you don't know what it looks like.

I thnk @Orodruin's suggestion is worth looking into. Note that the center of mass stays fixed in the horizontla direction. I would try defining new generalized coordinates $$X=\frac{Mx+mL\sin\theta}{M+m}~;~~\xi=\frac{Mx-mL\sin\theta}{M+m}$$and see what happens. Note that $X$ describes the translatinal motion of the CM, while $\xi$ the departure from the equilibrium position. It's more work, but it should decouple the equations of motion.

 

[Orodruin]

That you can do to get $\int \cos(\theta(t))d\theta.$ Then what? You have an implicit function of time under the integral sign that you don't know what it looks like.

I thnk @Orodruin's suggestion is worth looking into. Note that the center of mass stays fixed in the horizontla direction. I would try defining new generalized coordinates $$X=\frac{Mx+mL\sin\theta}{M+m}~;~~\xi=\frac{Mx-mL\sin\theta}{M+m}$$and see what happens. Note that $X$ describes the translatinal motion of the CM, while $\xi$ the departure from the equilibrium position. It's more work, but it should decouple the equations of motion.

Effectively, I would just do it … effectively …

Use the constant of motion from translational invariance to eliminate $\dot x$ from the constant of motion from time translation invariance (ie, energy). Then you are basically done.

 

[MatinSAR]

That you can do to get $\int \cos(\theta(t))d\theta.$ Then what? You have an implicit function of time under the integral sign that you don't know what it looks like.

I thnk @Orodruin's suggestion is worth looking into. Note that the center of mass stays fixed in the horizontla direction. I would try defining new generalized coordinates $$X=\frac{Mx+mL\sin\theta}{M+m}~;~~\xi=\frac{Mx-mL\sin\theta}{M+m}$$and see what happens. Note that $X$ describes the translatinal motion of the CM, while $\xi$ the departure from the equilibrium position. It's more work, but it should decouple the equations of motion.

So I cannot use $x$ and $\theta$ to solve, right? How did you find those generalized coordinates? By guessing?
Is there any other way?

 

[kuruman]

Not really guessing. Just having seen this sort of problem before and some common sense. In this case, you note that the CM of the system does not accelerate horizontally because there are no horizontal external forces acting on the parts making it up.

So if one of the generalized coordinates is the position of the CM, its equation of motion should be $\ddot X =0$ and easy to solve. Furthermore, simple harmonic motion is expected to some approximation. It follows that the other generalized coordinate should be the departure from the equilibrium position in the horizontal direction. You should expect its equation of motion to be $\ddot{ \xi}+\omega^2~\xi=0$.

Similar considerations apply to the cousin of this problem, namely the block of mass $m$ sliding down on a frictionless wedge of mass $M$ placed on a frictionless horizontal surface.

 

[Orodruin]

Is there any other way?

Sure:

Use the constant of motion from translational invariance to eliminate x˙ from the constant of motion from time translation invariance (ie, energy). Then you are basically done.

 

[MatinSAR]

Use the constant of motion from translational invariance to eliminate $\dot x$ from the constant of motion from time translation invariance (ie, energy). Then you are basically done.

Do you mean that I can use conservation of momentum $\vec P$ in horizontal direction?(Or using conservation of $T+U$?)
I'm going to try this. Hope it works ...

Not really guessing. Just having seen this sort of problem before and some common sense. In this case, you note that the CM of the system does not accelerate horizontally because there are no horizontal external forces acting on the parts making it up.

So if one of the generalized coordinates is the position of the CM, its equation of motion should be $\ddot X =0$ and easy to solve. Furthermore, simple harmonic motion is expected to some approximation. It follows that the other generalized coordinate should be the departure from the equilibrium position in the horizontal direction. You should expect its equation of motion to be $\ddot{ \xi}+\omega^2~\xi=0$.

Similar considerations apply to the cousin of this problem, namely the block of mass $m$ sliding down on a frictionless wedge of mass $M$ placed on a frictionless horizontal surface.

Thanks again. However, this method is complicated for me. Given that there’s a chance my teacher might ask me to explain what I’ve written, I’d prefer to use a method that’s more understandable to me.

 

[MatinSAR]

@Orodruin @kuruman
This is my attempt at solving the problem. To save time, please skip the calculations. I just need help with the last part because I’m unsure if this method works.
Lagrangian of the system is given by: $$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$ For coordiante $x$ we have: $$\frac {\partial L} {\partial x}-\frac {d}{dt}\frac {\partial L} {\partial \dot x}=0 $$ $$\frac {\partial L} {\partial \dot x}=A$$ A is a constant. $$(M+m)\dot x+ml\dot \theta cos \theta=A$$
For coordiante $\theta$ we have :$$\dfrac {\partial L}{\partial \theta}=-ml \dot x \dot \theta \sin \theta - mgl \sin \theta$$ $$\dfrac {\partial L}{\partial \dot \theta}=ml\dot x \cos \theta+ml^2 \dot \theta$$ $$\frac {d}{dt} \dfrac {\partial L}{\partial \dot \theta}=ml\ddot x \cos \theta -ml\dot x \dot \theta \sin \theta +ml^2 \ddot \theta$$ Using lagrange equation I get: $$-mgl \sin \theta -ml\ddot x \cos \theta - ml^2 \ddot \theta=0$$

I have two equations now: $$(M+m)\dot x+ml\dot \theta cos \theta=A$$$$-g \sin \theta -\ddot x \cos \theta - l \ddot \theta=0$$ Can I find $\theta (t)$ from the second equation and substitute it into the first equation, then solve for $x(t)$?

Edit 1: I've just fixed a big problem in my solution. $\ddot x$ is not necessarily $0$.

 

[erobz]

I have two equations now: $$(M+m)\dot x+ml\dot \theta cos \theta=A$$$$\ddot \theta + \dfrac {g}{l} \sin \theta =0$$ Can I find $\theta (t)$ from the second equation and substitute it into the first equation, then solve for $x(t)$?

What makes you think $\ddot x = 0$?

When I apply conservation of momentum in the $x$ direction I'm not finding that $\ddot x = 0$. Maybe I'm messing that up. What do you get when you apply conservation of momentum?

 

[MatinSAR]

What makes you think $\ddot x = 0$?

When I apply conservation of momentum in the $x$ direction I'm not finding that $\ddot x = 0$. Maybe I'm messing that up. What do you get when you apply C.o.M.?

Sorry. I've just edited that.

 

[erobz]

Sorry. I've just edited that.

I'm getting this from conservation of momentum:

$$M \dot x - m( \dot x - l \dot \theta \cos \theta ) = 0 $$

$$ \implies \dot x = -\frac{m}{M-m} l \dot \theta \cos \theta $$

That doesn't seem to be equivalent to your expression?

 

[MatinSAR]

I'm getting this from conservation of momentum:

$$M \dot x - m( \dot x - l \dot \theta \cos \theta ) = 0 $$

$$ \implies \dot x = -\frac{m}{M-m} l \dot \theta \cos \theta $$

That doesn't seem to be equivalent to your expression?

I've used lagrange equation. You've used conservation. Should our equations look the same? Or they should be different to help us find $x(t)$?

 

[erobz]

I've used lagrange equation. You've used conservation. Should our equations look the same? Or they should be different to help us find $x(t)$?

I think $\dot x$ is $\dot x$; if they don't agree, we have a problem.

 

[MatinSAR]

I think $\dot x$ is $\dot x$; if they don't agree, we have a problem.

So there should be a problem with the equation I derived from Lagrange equation. I will try to do it again.

 

[erobz]

In the inertial frame the little mass is swinging to the left faster than the large block is moving to the right. Does my approach with the signs make sense to you?

 

[MatinSAR]

In the inertial frame the little mass is swinging to the left faster than the large block is moving to the right. Does my approach with the signs make sense to you?

I think that $m \dot x$ should be positive. Because velocity of the mass $m$ is given by its velocity in block M frame + velocity of block M in inertial refrence frame.

 

[erobz]

Ok, so you are saying with $\dot \theta < 0$ we should have this equation:

$$ M \dot x + m( \dot x + m l \dot \theta \cos \theta ) = 0 $$

Which gives:

$$ \dot x = -\frac{ml}{M+m} \dot \theta \cos \theta $$

I can buy that.

 

[erobz]

Now differentiate your result and sub into your other equation ( which is correlates to applying conservation of energy), eliminating $\ddot x$.

 

[MatinSAR]

Now differentiate your result and sub into your other equation ( which is correlates to applying conservation of energy), eliminating $\ddot x$.

Good idea. Thanks. I'll try it tomorrow. It's too late here. (About 5 A.M.)

 

[Orodruin]

Do you mean that I can use conservation of momentum P→ in horizontal direction?(Or using conservation of T+U?)

Both.

 

[MatinSAR]

Now differentiate your result and sub into your other equation ( which is correlates to applying conservation of energy), eliminating $\ddot x$.

From first equation I get: $$\ddot x = \dfrac {ml\dot \theta^2\sin \theta-ml\ddot \theta \cos \theta}{M+m}$$Then I use it for second equation, I get: $$g\sin \theta + \dfrac {ml\dot \theta^2\sin \theta \cos \theta-ml\ddot \theta \cos ^2 \theta}{M+m}-l\ddot \theta=0 $$ Do you think can I solve for $\theta (t)$? Personally I don't see any way ...

Both.

Ok. Conservation of momentum gives us: $$M\dot x +m(\dot x + l\dot \theta \cos \theta)=A$$$$ \dot x = \dfrac {A-ml\dot \theta \cos \theta}{M+m} $$ And conservation of energy gives: $$ (M+m)\dot x^2+ml^2 \ddot \theta+ml\dot x \dot \theta \cos \theta - 2mgl\cos \theta=B $$ Should I substitute $\dot x$ from first equation to second equation, then solve for $\theta (t)$ ?

Is finding $\theta (t)$ and $x(t)$ really this hard? Or is the problem with my method? Because the TA said this is an easy question.

 

[Orodruin]

You can make things easier by just assuming the CoM frame. This gives a direct constraint on the relation between $x$ and $\theta$. You can also expand the Lagrangian to second order around the potential minimum. There are many tricks you can employ.

 

[Orodruin]

You also do not need to solve for $\theta$. Just looking at the total energy expression (expanded for small oscillations) will give you the frequency directly when comparing to the harmonic oscillator energy.

 

[MatinSAR]

You also do not need to solve for $\theta$. Just looking at the total energy expression (expanded for small oscillations) will give you the frequency directly when comparing to the harmonic oscillator energy.

I don't know about CoM frame. So I think I'm going to expand it for small oscillations. I think I should read about oscillations, then try to solve ...

 

[Orodruin]

I think I should read about oscillations, then try to solve ...

This should perhaps be your first step when trying to solve a problem regarding oscillations…

 

[MatinSAR]

This should perhaps be your first step when trying to solve a problem regarding oscillations…

I thought that I can find equation of motion without knowing g much about oscillations because this assignment is related to lagrangian mechanics. And I read lagrangian mechanics, not oscillations. Thanks for your help anyway. I will try again after reviewing about oscillations ...

 

[Steve4Physics]

@MatinSAR, I'm jumping into the thread a bit late in the day, but would like to add this…

In your post #21, I agree with your EL equation for $\theta$:
$-mgl \sin \theta -ml\ddot x \cos \theta - ml^2 \ddot \theta=0$
(but would drop the minus signs).

But you may have made a mistake with the EL equation for $x$.

Using your expression for $L$ (which I agree with), and at risk of being accused of doing too much:

$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta$

$\dfrac {\partial L} {\partial x}=0$

$\dfrac {\partial L} {\partial \dot x}= (M+m)\dot x +ml\dot \theta \cos \theta$

$\dfrac {d}{dt} \dfrac {\partial L}{\partial \dot x}=(M+m)\ddot x +ml(\ddot \theta \cos \theta - \dot \theta^2 \sin \theta)$
(Slightly fiddly; note you need to use both product and chain rules to differentiate the $\dot \theta \cos \theta$ term.)

$\dfrac {d}{dt} \dfrac {\partial L}{\partial \dot x} - \dfrac {\partial L} {\partial x} =0 \implies (M+m)\ddot x +ml(\ddot \theta \cos \theta - \dot \theta^2 \sin \theta) = 0$

Then, for $\theta \ll 1$, you can use $\sin \theta \approx \theta$ to get rid of $\sin$. It may be OK to use $\cos \theta \approx 1$ rather than $\cos \theta \approx 1 - \frac{\theta^2}2$ but that could be contentious.

 

[MatinSAR]

@MatinSAR, I'm jumping into the thread a bit late in the day, but would like to add this…

Thanks a lot for your help ...

Then, for $\theta \ll 1$, you can use $\sin \theta \approx \theta$ to get rid of $\sin$. It may be OK to use $\cos \theta \approx 1$ rather than $\cos \theta \approx 1 - \frac{\theta^2}2$ but that could be contentious.

Do you mean that I can substitute $\sin \theta \approx \theta$ and $\cos \theta \approx 1$ into the Euler-Lagrange equations to find $x(t)$ and $\theta (t)$? These approximations were stated in the question, but I wasn’t sure at which part I could use them.

 

[Steve4Physics]

Do you mean that I can substitute $\sin \theta \approx \theta$ and $\cos \theta \approx 1$ into the Euler-Lagrange equations to find $x(t)$ and $\theta (t)$?

Well, that's what I'd do!

Providing $\theta$ is small (i.e. you are dealing only with small amplitudes) they are pretty good approximations. For example, consider $\theta = 0.1rad$ (a little under $6^o$).

$\sin (0.1) \approx 0.0998$. So replacing $\sin \theta$ by $\theta$ introduces an error of around 0.2%.

$\cos(0.1) \approx 0.995$ So replacing $\cos \theta$ by 1 introduces an error of about 0.5%

Even more accurate for smaller angles of course.

A better approximation is $\cos \theta \approx 1 - \frac{\theta^2}2$ if you wanted to try. But I think it will hinder rather than help here.

These approximations were stated in the question,

That suggests you are expected to use them (where appropriate).

It would have been helpful (and could still be helpful) if you posted the complete question so we had all relevant information!

but I wasn’t sure at which part I could use them.

Best not to use them early on, or you can lose information, notably when differentiating. But now you have the final 'exact' equations (in Post #40) it looks like a good time to make any valid approximations.

(By the way, I'm assuming that you are required to solve the problem with a simple/direct Lagrangian approach using $x$ and $\theta$. So you are not expected to use any of the alternative approaches others have suggested. )

 

[erobz]

These approximations were stated in the question, but I wasn’t sure at which part I could use them.

If the problem tells you to explicitly use the small angle approximation $\cos \theta \approx 1$, then I think you must use it?

I thought your question was when you could use a small angle approximation, and I think everyone @kuruman, @Orodruin were expecting $\cos \theta \approx 1 - \frac{\theta^2}{2}$ as the approximation when it was said it shouldn't matter whether it was used before or after the application of the Euler-Lagrange equations. As @Steve4Physics points out, you left out some pertinent info.

 

[Orodruin]

I thought your question was when you could use a small angle approximation, and I think everyone @kuruman, @Orodruin were expecting $\cos \theta \approx 1 - \frac{\theta^2}{2}$ as the approximation when it was said it shouldn't matter whether it was used before or after the application of the Lagrangian.

As I said early on:

Yes, you can expand the Lagrangian for small $\theta$ and then apply the EL equation, but if you want linear EoMs, you need to expand to second order in $\theta$.

If you want to end up with a linearized equation of motion you must keep the quadratic terms in the Lagrangian as those are exactly the terms that give you linear terms in thd eom.

 

[erobz]

As I said early on:

If you want to end up with a linearized equation of motion you must keep the quadratic terms in the Lagrangian as those are exactly the terms that give you linear terms in thd eom.

Yeah, I meant everyone except Matin thought… turning cosine into 1 before the application of the derivative seems obviously problematic. When I said "I think everyone"-and listed you in there, it was intended to be a quiet nod to the OP they should be more clear about the required approximation. I certainly wasn't underestimating anyone's understanding of the problem (if anyone doesn't understand its me), sorry if it came off that way.

To be clear. @MatinSAR The answer to your original question is no, don't turn cosine into the constant 1 before the derivative!

 

[MatinSAR]

Thank you to everyone for his help. I am trying to understand.

 

[erobz]

I am trying to understand.

Well, are you still on the fence about it?

 

[MatinSAR]

Well, are you still on the fence about it?

I have two equations which were mentioned in post #40. As I understand I can't directly use approximations.

 

[erobz]

I have two equations which were mentioned in post #40. As I understand I can't directly use approximations.

The two equations you have after you apply the partial derivatives ( Euler-Lagrange Equations) are exact. You can apply the approximation $\cos \theta \approx 1$ to those. What you can't do is apply that approximation of $\cos \theta \approx 1$ to the original Lagrangian ( i.e. the equation you begin with $L = T - U$).

 

[MatinSAR]

It would have been helpful (and could still be helpful) if you posted the complete question so we had all relevant information!

Everything was mentioned in original post except the approximations of $\cos \theta =1$ and $\sin \theta =\theta$. I am sorry that I didn't mention them.

But now you have the final 'exact' equations (in Post #40) it looks like a good time to make any valid approximations.

(By the way, I'm assuming that you are required to solve the problem with a simple/direct Lagrangian approach using $x$ and $\theta$. So you are not expected to use any of the alternative approaches others have suggested. )

Thanks for your time @Steve4Physics ...

The two equations you have after you apply the partial derivatives are exact. You can apply the approximation $\cos \theta \approx 1$ to those. What you can't do is apply that approximation of $\cos \theta \approx 1$ to the original Lagrangian ( i.e. the equation you begin with $L = T - U$).

This is what I wanted to do. I am going to try ... Thanks again.