Small oscillations of a simple pendulum placed on a moving block

[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.