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Yes.Davidllerenav said:Ok, so I should make that substitution in the expression of the acceleration, not in the taylor expansion, right?
Yes.Davidllerenav said:Ok, so I should make that substitution in the expression of the acceleration, not in the taylor expansion, right?
I got ##A=\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}##. I need to expand this with Taylor, right?haruspex said:Yes.
Not really. As I posted, you are only interested in the leading term in each sum, one above the line and one below.Davidllerenav said:I got ##A=\frac{2ayc-y^2c}{(2a^2-2ay+y^2)^2}##. I need to expand this with Taylor, right?
I'm not sure if I undesrtand it correctly, but do you mean something like this: ##\frac{2ayc}{2a^2}##?haruspex said:Not really. As I posted, you are only interested in the leading term in each sum, one above the line and one below.
Close, but you forgot something in the denominator.Davidllerenav said: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}##?Davidllerenav said:I'm not sure if I undesrtand it correctly, but do you mean something like this: ##\frac{2ayc}{2a^2}##?
Yes.Davidllerenav said:That it is squared, right? ##\frac{2ayc}{4a^4}##?
Like this ##\frac{2ayc}{4a^4}+\omega^2y=0##?haruspex said:Yes.
Now write out the differential equation entirely in terms of y. I.e. using ##\ddot y## instead of A. Careful with signs.
No. Don't try to bring omega into it yet.Davidllerenav said:Like this ##\frac{2ayc}{4a^4}+\omega^2y=0##?
So ##\ddot x= \frac {d^2}{dx^2}(-a+y)##, right?haruspex said: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.
No, the differentiation is wrt time. And simplify that. a is constant, so what is its derivative?Davidllerenav said:So ##\ddot x= \frac {d^2}{dx^2}(-a+y)##, right?
So ##\ddot x=\ddot y##.haruspex said:No, the differentiation is wrt time. And simplify that. a is constant, so what is its derivative?
Right. Now write the differential equation that results from combining that with your expression for A in post #41.Davidllerenav said:So ##\ddot x=\ddot y##.
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 said:Right. Now write the differential equation that results from combining that with your expression for A in post #41.
You need to have only y or x, not both. Use the equation you had in post #47.Davidllerenav said: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?
It should be ##\frac{2ayc}{m4a^4}=-\omega^2 y##? Or do i need to itegrate two times the expression of A?haruspex said: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?
The equation should readDavidllerenav said:##\ddot{x}=\omega^2x=0##
I don't understad, can you please explain a bit more?haruspex said: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.)
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.Davidllerenav said:I don't understad, can you please explain a bit more?
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 said: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.
Looks good!Davidllerenav said: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?
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 said:Looks good!
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.Davidllerenav said: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?