A position of stable equilibrium, and the period of small oscillations

[Davidllerenav]

You need to have only y or x, not both. Use the equation you had in post #47.
And you can simplify the expression on the left a little.
Can you now deduce an expression for ω?

It should be $\frac{2ayc}{m4a^4}=-\omega^2 y$? Or do i need to itegrate two times the expression of A?

 

[haruspex]

I don't seem to have pointed out this error:

$\ddot{x}=\omega^2x=0$

The equation should read
$\ddot{x}+\omega^2x=0$.
You need to write an equation of this form. It should not mention x or $\omega$. It should only contain y, the second derivative of y, and the constants m, a and c.
When you have that you can compare it with the equation above and discover what $\omega$ is.
(By trying to jump straight to the answer you have introduced a sign error.)

 

[Davidllerenav]

I don't seem to have pointed out this error:

The equation should read
$\ddot{x}+\omega^2x=0$.
You need to write an equation of this form. It should not mention x or $\omega$. It should only contain y, the second derivative of y, and the constants m, a and c.
When you have that you can compare it with the equation above and discover what $\omega$ is.
(By trying to jump straight to the answer you have introduced a sign error.)

I don't understad, can you please explain a bit more?

 

[haruspex]

I don't understad, can you please explain a bit more?

Rats! Just noticed you had a sign error in post #37 where you made the substitution with y. As a result your expression for A in post #41 has the wrong sign.

When you have corrected that, use the fact that A means $\ddot x$ and (post #47) that this equals $\ddot y$. Combining all that should produce an equation involving $\ddot y$, y, m, a, c and nothing else.
Post what you get.

 

[Davidllerenav]

Rats! Just noticed you had a sign error in post #37 where you made the substitution with y. As a result your expression for A in post #41 has the wrong sign.

When you have corrected that, use the fact that A means $\ddot x$ and (post #47) that this equals $\ddot y$. Combining all that should produce an equation involving $\ddot y$, y, m, a, c and nothing else.
Post what you get.

I see what the error is. The equation was $mA=-V'(x)$ right? So the correct equation in post #37 is $A=-\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}$ and since you said that I only need the leading term in both the numerator and the denominator, I have $mA=-\frac{2ayc}{4a^4}$. Using the fact that $A=\ddot x=\ddot y$ and $\ddot x=-\omega^2 x$ I can write $\ddot y=-\frac{2ayc}{4a^4}=-\frac {c}{2a^3m}y$ thus $\omega=\sqrt {\frac {c}{2a^3m}}=\frac {1}{a}\sqrt {\frac {c}{2am}}$. With that I can find the period using $\tau=\frac {2\pi}{\omega}\Rightarrow\tau=2\pi a\sqrt {\frac {2am}{c}}$. Am I right?

 

[haruspex]

I see what the error is. The equation was $mA=-V'(x)$ right? So the correct equation in post #37 is $A=-\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}$ and since you said that I only need the leading term in both the numerator and the denominator, I have $mA=-\frac{2ayc}{4a^4}$. Using the fact that $A=\ddot x=\ddot y$ and $\ddot x=-\omega^2 x$ I can write $\ddot y=-\frac{2ayc}{4a^4}=-\frac {c}{2a^3m}y$ thus $\omega=\sqrt {\frac {c}{2a^3m}}=\frac {1}{a}\sqrt {\frac {c}{2am}}$. With that I can find the period using $\tau=\frac {2\pi}{\omega}\Rightarrow\tau=2\pi a\sqrt {\frac {2am}{c}}$. Am I right?

Looks good!

 

[Davidllerenav]

Looks good!

Ok! I have some questions: Why do we need to use a Taylor approximation with the potential? Why did we used the substitution $x=-a+y$? Why did we need only the leading terms in the equation in post #37?

 

[haruspex]

Ok! I have some questions: Why do we need to use a Taylor approximation with the potential? Why did we used the substitution $x=-a+y$? Why did we need only the leading terms in the equation in post #37?

The motion is only approximately SHM for small perturbations about -a. It makes it much easier to see what is going on if you express x as -a+perturbation. This is also how Taylor series work.
From the expression you had for A in post #37 (but with corrected sign) you could have expanded the denominator using the Taylor expansion of (sum of terms)^-3 then multiplied that by the numerator, and only then discarding all but the first term. But it should be evident that this will produce the same result as discarding all but the first term in numerator and denominator separately, which is somewhat easier.