Small oscillations of a simple pendulum placed on a moving block

[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 : $$\omega = \sqrt {\dfrac {g}{l} \dfrac {m+M}{M}} $$ $$ f=\dfrac {1}{2\pi} \omega = \dfrac {1}{2\pi} \sqrt {\dfrac {g}{l} \dfrac {m+M}{M}}$$

 

[MatinSAR]

I've just found my last mistake(in finding $f$) using a friend's help.


@Orodruin @PeroK @Steve4Physics @erobz @kuruman
I appreciate your help and your patience.

 

[Orodruin]

Ok, so since you have reached a conclusion, let's discuss the expansion approach. You have the following Lagrangian:

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

Expanding to second order in $\theta$ around $\theta = 0$ results in
$$
L \simeq \dfrac{M + m}{2} \dot x^2 + ml \dot x \dot\theta + \dfrac 12 ml^2 \dot \theta^2 - \dfrac{mgl}{2}\theta^2
$$
(where I have dropped the constant term in the potential since it does not affect the equations of motion, $\simeq$ means equal up to constants and terms higher than quadratic in $\theta$\). Now complete the square for the quadratic term:
$$
\dfrac{M + m}{2} \dot x^2 + ml \dot x \dot\theta = \dfrac{M+m}{2}\left(\dot x^2 + 2\dfrac{ml}{M+m} \dot x \dot\theta \right)
= \dfrac{M+m}{2}\left[\left(\dot x + \dfrac{ml}{M+m} \dot \theta\right)^2 - \dfrac{m^2l^2}{(M+m)^2}\dot\theta^2\right] \equiv \dfrac{M+m}{2}\left[\dot X^2 - \dfrac{m^2l^2}{(M+m)^2}\dot\theta^2\right]
$$
where we have introduced the new coordinate $X = x + ml\theta/(M+m)$ to replace $x$. The expanded Lagrangian is now
$$
L \simeq \dfrac{M+m}2 \dot X^2 + \dfrac{\mu l^2\dot\theta^2}2 - \dfrac{mgl\theta^2}{2},
$$
where $\mu = mM/(M+m)$ is the reduced mass of the system. It is here clear that the problems for $X$ and $\theta$ decouple and so we can focus on the two last terms (the motion of $X$ just describes the CoM motion, which is linear). The Lagrangian for a harmonic oscillator with variable $\theta$ and mass $\bar m$ is
$$
L_{HO} = \dfrac{\bar m}{2}\left[ \dot\theta^2 - \omega^2 \theta^2\right].
$$
Identifying with the $\theta$ part of our Lagrangian immediately gives
$$
\bar m = \mu l^2, \qquad \bar m \omega^2 = mgl.
$$
Hence,
$$
\omega^2 = \dfrac{mgl}{\mu l^2} = \dfrac{M+m}{M} \dfrac{g}{l}.
$$
Take the square root to find $\omega$.

 

[MatinSAR]

Ok, so since you have reached a conclusion, let's discuss the expansion approach. You have the following Lagrangian:

Expanding to second order in $\theta$ around $\theta = 0$ results in
$$
L \simeq \dfrac{M + m}{2} \dot x^2 + ml \dot x \dot\theta + \dfrac 12 ml^2 \dot \theta^2 - \dfrac{mgl}{2}\theta^2
$$
(where I have dropped the constant term in the potential since it does not affect the equations of motion, $\simeq$ means equal up to constants and terms higher than quadratic in $\theta$\). Now complete the square for the quadratic term:
$$
\dfrac{M + m}{2} \dot x^2 + ml \dot x \dot\theta = \dfrac{M+m}{2}\left(\dot x^2 + 2\dfrac{ml}{M+m} \dot x \dot\theta \right)
= \dfrac{M+m}{2}\left[\left(\dot x + \dfrac{ml}{M+m} \dot \theta\right)^2 - \dfrac{m^2l^2}{(M+m)^2}\dot\theta^2\right] \equiv \dfrac{M+m}{2}\left[\dot X^2 - \dfrac{m^2l^2}{(M+m)^2}\dot\theta^2\right]
$$
where we have introduced the new coordinate $X = x + ml\theta/(M+m)$ to replace $x$. The expanded Lagrangian is now
$$
L \simeq \dfrac{M+m}2 \dot X^2 + \dfrac{\mu l^2\dot\theta^2}2 - \dfrac{mgl\theta^2}{2},
$$
where $\mu = mM/(M+m)$ is the reduced mass of the system. It is here clear that the problems for $X$ and $\theta$ decouple and so we can focus on the two last terms (the motion of $X$ just describes the CoM motion, which is linear). The Lagrangian for a harmonic oscillator with variable $\theta$ and mass $\bar m$ is
$$
L_{HO} = \dfrac{\bar m}{2}\left[ \dot\theta^2 - \omega^2 \theta^2\right].
$$
Identifying with the $\theta$ part of our Lagrangian immediately gives
$$
\bar m = \mu l^2, \qquad \bar m \omega^2 = mgl.
$$
Hence,
$$
\omega^2 = \dfrac{mgl}{\mu l^2} = \dfrac{M+m}{M} \dfrac{g}{l}.
$$
Take the square root to find $\omega$.

Fortunately, both methods yield the same results. I am reading and trying to understand this method.

Thanks once again for your invaluable assistance @Orodruin ...

 

[Orodruin]

Fortunately, both methods yield the same results.

Of course they do. As I showed, they are equivalent. The only question is which is faster for a particular problem and what you feel comfortable doing.