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

[Davidllerenav]

In post #27 you correctly wrote mA= In one equation, A being the acceleration, but in the next equation you changed it to ma=.

Oh, sorry, I meant $mA=0+\frac{c}{2a^3}(x-a)$. Since I'm plugging in $x=-a$, would it be $mA=0-\frac{c}{2a^3}(-a-a)$?

 

[haruspex]

Oh, sorry, I meant $mA=0+\frac{c}{2a^3}(x-a)$. Since I'm plugging in $x=-a$, would it be $mA=0-\frac{c}{2a^3}(-a-a)$?

You don't want to plug in x=-a.
As I posted, it would be rather clearer if you were to rewrite the equation I am terms of displacement from -a, e.g. x=-a+y. Then you would be looking for a differential equation of the form $\ddot y+\beta y=0$ for some constant beta.
Try that approach and maybe it will fix your sign error.

 

[Davidllerenav]

You don't want to plug in x=-a.
As I posted, it would be rather clearer if you were to rewrite the equation I am terms of displacement from -a, e.g. x=-a+y. Then you would be looking for a differential equation of the form $\ddot y+\beta y=0$ for some constant beta.
Try that approach and maybe it will fix your sign error.

Sorry for taking so long answer. I don't undertand how should I should I do that, should I just replace $x$ by $-a+y$ on the taylor polynomial?

 

[haruspex]

Sorry for taking so long answer. I don't undertand how should I should I do that, should I just replace $x$ by $-a+y$ on the taylor polynomial?

I would use that substitution in your equation in post #17. Then simplify a bit and discard terms beyond the first nonzero term in each sum.

 

[Davidllerenav]

I would use that substitution in your equation in post #17. Then simplify a bit and discard terms beyond the first nonzero term in each sum.

Ok, so I should make that substitution in the expression of the acceleration, not in the taylor expansion, right?

 

[haruspex]

Ok, so I should make that substitution in the expression of the acceleration, not in the taylor expansion, right?

Yes.

 

[Davidllerenav]

Yes.

I got $A=\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}$. I need to expand this with Taylor, right?

 

[haruspex]

I got $A=\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}$. I need to expand this with Taylor, right?

Not really. As I posted, you are only interested in the leading term in each sum, one above the line and one below.

 

[Davidllerenav]

Not really. As I posted, you are only interested in the leading term in each sum, one above the line and one below.

I'm not sure if I undesrtand it correctly, but do you mean something like this: $\frac{2ayc}{2a^2}$?

 

[haruspex]

I'm not sure if I undesrtand it correctly, but do you mean something like this: $\frac{2ayc}{2a^2}$?

Close, but you forgot something in the denominator.

 

[Davidllerenav]

I'm not sure if I undesrtand it correctly, but do you mean something like this: $\frac{2ayc}{2a^2}$?

That it is squared, right? $\frac{2ayc}{4a^4}$?

 

[haruspex]

That it is squared, right? $\frac{2ayc}{4a^4}$?

Yes.
Now write out the differential equation entirely in terms of y. I.e. using $\ddot y$ instead of A. Careful with signs.

 

[Davidllerenav]

Yes.
Now write out the differential equation entirely in terms of y. I.e. using $\ddot y$ instead of A. Careful with signs.

Like this $\frac{2ayc}{4a^4}+\omega^2y=0$?

 

[haruspex]

Like this $\frac{2ayc}{4a^4}+\omega^2y=0$?

No. Don't try to bring omega into it yet.
A is the same as $\ddot x$. Using x=-a+y, write A in terms of $\ddot y$. Then equate it to the expression you found for A in post #41.

 

[Davidllerenav]

No. Don't try to bring omega into it yet.
A is the same as $\ddot x$. Using x=-a+y, write A in terms of $\ddot y$. Then equate it to the expression you found for A in post #41.

So $\ddot x= \frac {d^2}{dx^2}(-a+y)$, right?

 

[haruspex]

So $\ddot x= \frac {d^2}{dx^2}(-a+y)$, right?

No, the differentiation is wrt time. And simplify that. a is constant, so what is its derivative?

 

[Davidllerenav]

No, the differentiation is wrt time. And simplify that. a is constant, so what is its derivative?

So $\ddot x=\ddot y$.

 

[haruspex]

So $\ddot x=\ddot y$.

Right. Now write the differential equation that results from combining that with your expression for A in post #41.

 

[Davidllerenav]

Right. Now write the differential equation that results from combining that with your expression for A in post #41.

I can solve the differential equation for $\ddot x$ such that $\ddot x=-\omega^2 x$ and using the expression for A, I get $\frac{2ayc}{m4a^4}=-\omega^2 x$, right?

 

[haruspex]

I can solve the differential equation for $\ddot x$ such that
and using the expression for A, I get $\frac{2ayc}{m4a^4}=-\omega^2 x$, right?

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?

Oh, and it is not true that $\ddot x=-\omega^2 x$. There were other terms when the equation was all in x. This is why we switched to y.

 

[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.