Small oscillations of a simple pendulum placed on a moving block

[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.

 

[PeroK]

Thanks a lot for your help ...

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.

You can only use them safely once you have done all the differentiation. For example:
$$x = l\cos \theta$$$$\dot x = -l(\sin \theta)\dot \theta \approx - l\theta \dot \theta$$But, NOT:
$$x \approx l$$$$\dot x \approx 0$$

 

[Orodruin]

I mean, ultimately it is what I already said, you need to include up to quadratic terms in the Lagrangian if you want to have the linearised equations of motion. This is relatively easy to see. Suppose you have a Lagrangian that you expand to linear order in $\theta$ and its derivative $\dot\theta$, then
$$
\mathcal L = \mathcal L_0 + \theta \mathcal L_1 + \dot\theta \mathcal L_1' + \mathcal O(\theta^2)
$$
where $\mathcal L_0$, $\mathcal L_1$, and $\mathcal L_1'$ do not depend on $\theta$ or its derivatives and the $\mathcal O(\theta^2)$ indicates not only second order terms or higher in $\theta$, but also in its derivatives. The Euler-Lagrange equation for $\theta$ is then on the form
$$
\frac{\partial \mathcal L}{\partial \theta} - \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot\theta}
= \mathcal L_1 - \frac{d}{dt}\mathcal L_1' + \mathcal O(\theta) = 0.
$$
Ignoring the $\mathcal O(\theta^2)$ terms in the Lagrangian is therefore going to lead to ignoring $\mathcal O(\theta)$ terms in the equations of motion - i.e., we will not have a proper linearisation of the EoM. However, if we do the same thing while keeping the $\theta^2$ terms but ignoring the $\mathcal O (\theta^3)$ terms, the terms missing from the EoM will all be $\mathcal O(\theta^2)$ and therefore irrelevant to the linearisation of the EoM.

 

[MatinSAR]

@Steve4Physics @erobz @PeroK
Hello. First of all, thank you for your time. You have helped a lot. I am writing this after reading your posts. If I am wrong again, I don’t want to waste your time with this question. So, if I am wrong, please just let me know. I prefer to wait for the TA to solve it next week.
I rewrite equations in post #40:$$ (M+m)\ddot x +ml(\ddot \theta \cos \theta - \dot \theta^2 \sin \theta) = 0$$$$mgl \sin \theta+ml\ddot x \cos \theta + ml^2 \ddot \theta=0$$ We have these approximations: $$\cos \theta \approx 1$$$$\sin \theta \approx \theta $$ $$ \theta ^2 = \dot \theta ^2 \approx 0$$
After substituting we get: $$(M+m)\ddot x +ml\ddot \theta = 0$$$$mgl\theta + ml\ddot x+ml^2\ddot \theta =0$$ After solving for $\ddot x$ in first equation and substituting in second, we get: $$ \ddot x =\dfrac {-ml\ddot \theta}{M+m}$$ $$ mgl\theta + ml\dfrac {-ml\ddot \theta}{M+m}+ml^2\ddot \theta =0 $$ Simplifying : $$ (ml^2 - \dfrac {m^2l^2}{M+m})\ddot \theta +mgl \theta = 0 $$
Now, I have equations of motion: $$\ddot \theta + \dfrac {mgl}{(ml^2 - \dfrac {m^2l^2}{M+m})}\theta = 0 $$ $$ \ddot x =\dfrac {-ml\ddot \theta}{M+m}$$
To find the frequencies of small oscillations, I use the formula $f = \omega / 2\pi$. I think $\omega$ is the coefficient of $\theta$ in the above equation.

 

[MatinSAR]

@Orodruin Thanks a lot for your help. I’m sorry if I’m not educated enough to understand your posts well.

I mean, ultimately it is what I already said, you need to include up to quadratic terms in the Lagrangian if you want to have the linearised equations of motion. This is relatively easy to see. Suppose you have a Lagrangian that you expand to linear order in $\theta$ and its derivative $\dot\theta$, then
$$
\mathcal L = \mathcal L_0 + \theta \mathcal L_1 + \dot\theta \mathcal L_1' + \mathcal O(\theta^2)
$$

Are you writing the Lagrangian near equilibrium points using Taylor series?

 

[Orodruin]

Are you writing the Lagrangian near equilibrium points using Taylor series?

Yes.

 

[Orodruin]

To find the frequencies of small oscillations, I use the formula $f = \omega / 2\pi$. I think $\omega$ is the coefficient of $\theta$ in the above equation.

Not exactly. It has the wrong dimensions to be a frequency.

Are you familiar with the differential equation governing the harmonic oscillator?

 

[MatinSAR]

Yes.

So, if I write the Lagrangian this way… Can I substitute those approximations into this Lagrangian? Then, I apply the Euler-Lagrange equations to find $x(t)$ and $\theta (t)$.
Or should I apply EL equations before applying approximations?

Are you familiar with the differential equation governing the harmonic oscillator?

Yes. For small oscillations, we have: $$ \ddot \theta + \omega ^2 \theta = 0$$

 

[Orodruin]

So, if I write the Lagrangian this way… Can I substitute those approximations into this Lagrangian? Then, I apply the Euler-Lagrange equations to find $x(t)$ and $\theta (t)$.

The entire point of the post was to show that you need to include up to second order terms to het the correct linearised behaviour. If you include the quadratic terms, sure, as long as $\theta$ is small.

Yes. For small oscillations, we have: $$ \ddot \theta + \omega ^2 \theta = 0$$

 

[MatinSAR]

Or should I apply EL equations before applying approximations?

According to @PeroK, I should differentiate first. I am asking again because your way seems different to me…

 

[Orodruin]

According to @PeroK, I should differentiate first. I am asking again because your way seems different to me…

It doesn’t matter if you do it consistently, ie, keep one order higher in $\theta$ and derivatives if you expand before applying the EL equation.

 

[Steve4Physics]

I rewrite equations in post #40:$$ (M+m)\ddot x +ml(\ddot \theta \cos \theta - \dot \theta^2 \sin \theta) = 0$$$$mgl \sin \theta+ml\ddot x \cos \theta + ml^2 \ddot \theta=0$$

You should now simplify further by cancelling-out $ml$ in the second equation. (This then simplifies subsequent equations.)

We have these approximations: $$\cos \theta \approx 1$$$$\sin \theta \approx \theta $$ $$ \theta ^2 = \dot \theta ^2 \approx 0$$

I think you mentioned in an earlier post that the approximations were given as part of the question. The approximation that $\dot \theta ^2 \approx 0$ is not immediately obvious (to me, anyway) but makes sense.

After substituting we get: $$(M+m)\ddot x +ml\ddot \theta = 0$$$$mgl\theta + ml\ddot x+ml^2\ddot \theta =0$$ After solving for $\ddot x$ in first equation and substituting in second, we get: $$ \ddot x =\dfrac {-ml\ddot \theta}{M+m}$$ $$ mgl\theta + ml\dfrac {-ml\ddot \theta}{M+m}+ml^2\ddot \theta =0 $$ Simplifying : $$ (ml^2 - \dfrac {m^2l^2}{M+m})\ddot \theta +mgl \theta = 0 $$
Now, I have equations of motion: $$\ddot \theta + \dfrac {mgl}{(ml^2 - \dfrac {m^2l^2}{M+m})}\theta = 0 $$ $$ \ddot x =\dfrac {-ml\ddot \theta}{M+m}$$

You will get simpler equations if you cancel $ml$, as noted above.

I think $\omega$ is the coefficient of $\theta$ in the above equation.

No (as already noted by @Orodruin).

One useful check of your final answer is to consider the case where $M \gg m$. This corresponds to a basic simple pendulum with the top of the pendulum stationary; you should be familiar with the standard formula for the period in this case.

Edited - typo's

 

[MatinSAR]

You will get simpler equations if you cancel $ml$, as noted above.

Do you think my final equations are correct? After simplifying them the way you suggest…

No (as already noted by @Orodruin).

Yes.

 

[Orodruin]

I would simplify the coefficient further:
$$
\frac{g}{l} \frac{M+m}{M}
$$

 

[MatinSAR]

I would simplify the coefficient further:
$$
\frac{g}{l} \frac{M+m}{M}
$$

Can I continue this way(post #53) instead of writing the Lagrangian the way you suggested?
So far, I have not expanded the Lagrangian around an equilibrium point, so it’s a bit difficult.
I will try it, but after solving other questions of my assignment…

 

[Orodruin]

Can I continue this way(post #53) instead of writing the Lagrangian the way you suggested?
So far, I have not expanded the Lagrangian around an equilibrium point, so it’s a bit difficult.
I will try it, but after solving other questions of my assignment…

Sure

 

[MatinSAR]

Sure

I was getting discouraged from solving the question. Thank you! I am going to try finding the frequency of small oscillations again.

 

[PeroK]

I think you mentioned in an earlier post that the approximations were given as part of the question. The approximation that $\dot \theta ^2 \approx 0$ is not immediately obvious (to me, anyway) but makes sense.

The term that gets dropped is $\theta \dot \theta^2$. Which is effectively taken to be small compared to $\ddot \theta$. If we expect SHM, then the order of $\ddot \theta$ is $\omega^2 \theta$. Whereas, the order of $\theta \dot \theta^2$ is $\omega^2 \theta^3$.

 

[MatinSAR]

Here is my new attempt at solving the problem.
$$ \ddot \theta + \dfrac {g}{l} \dfrac {m+M}{M} \theta =0$$ If we consider $\dfrac {g}{l} \dfrac {m+M}{M} = \alpha$ then we have general solution: $$\theta (t) = A\cos(\sqrt \alpha t)+B\sin(\sqrt \alpha t)$$ So : $$ \ddot \theta =-A\alpha \cos(\sqrt \alpha t)-B\alpha \sin(\sqrt \alpha t)$$
We have $\ddot x=\dfrac {-ml}{M+m} \ddot \theta$. $$\ddot x = \dfrac {-ml}{M+m}(-A\alpha \cos(\sqrt \alpha t)-B\alpha \sin(\sqrt \alpha t)) $$ $$\ddot x = \dfrac {m}{M}gA\cos(\sqrt \alpha t)+\dfrac {m}{M}gB\sin(\sqrt \alpha t) $$ We had $\dfrac {g}{l} \dfrac {m+M}{M} = \alpha$. Now we define $\beta = \dfrac {g}{l} \dfrac {m+M}{m}$. So: $$\alpha/ \beta = m/M$$ $$ \ddot x = g(\alpha/ \beta) (A\cos(\sqrt \alpha t)+B\sin(\sqrt \alpha t) )$$ $$ x(t) = A_1 + B_1t-\dfrac{g}{\beta}(A\cos(\sqrt \alpha t)+B\sin(\sqrt \alpha t)) $$
Can I find frequency of small oscillations by comparing $\ddot \theta + \dfrac {g}{l} \dfrac {m+M}{M} \theta =0$ with $\ddot \theta + \omega ^2 \theta =0$?

 

[Orodruin]

You don’t need to go through the hassle of solving the ODE. Just identify $\omega$ from the ODE.

 

[MatinSAR]

You don’t need to go through the hassle of solving the ODE.

I did not. AI did. I just defined constants like $\alpha$ and $\beta$ to make calculations easier. So I didn't waste time in solving for $x(t)$ and $\theta (t)$.

Just identify $\omega$ from the ODE.

Can I find frequency of small oscillations by comparing $\ddot \theta + \dfrac {g}{l} \dfrac {m+M}{M} \theta =0$ with $\ddot \theta + \omega ^2 \theta =0$?

Like what I've said in my last post?


.............................. Edit ............................................................................

$$ \ddot \theta + \dfrac {g}{l} \dfrac {m+M}{M} \theta =0$$ We compare it with $\ddot \theta +\omega^2 \theta=0$ , We find out that : $$\omega = \sqrt {\dfrac {g}{l} \dfrac {m+M}{M}} $$ $$ f=\dfrac {1}{2\pi} \omega = \dfrac {1}{2\pi} \sqrt {\dfrac {g}{l} \dfrac {m+M}{M}}$$