Small oscillations of a simple pendulum placed on a moving block

[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.