Finding the time period using the potential

[green-beans]

Homework Statement

A particle of unit mass moves in one dimension with potential V(x) = ½μ²x² + εx⁴ (ε>0). Discuss the motion of the particle.
If the particle released from rest at x=a (a>0) express the time period T for the particle to return to a in the form of an integral and show that when ε is small, T is reduced by approximately 3επa²/μ³.

Homework Equations

v = dx/dt = ±√{(2/m) * (E - V(x))}
Time period taken for a particle to move from point x1 to x2:
T = ∫1/{v} dx

The Attempt at a Solution

Discuss the motion:[/B]
If we sketch V(x) it will look alike to parabola with V(0) = 0. Depending on the value of E that the particle has, it will have a simple harmonic motion and will oscillate in the region enclosed between E and the x-axis.

Time period to return to a:
I obtained this and it matched with the answers (sorry for the long expression I wanted to upload a picture but it did not work)
T=(4/μ) ∫(dx)/(√{(a²-x²) +(2/μ²)ε(a⁴-x⁴)})

Show that T is reduced by approximately 3επa²/μ³
In this part I am not sure how the fact that ε is small will affect the integral. I am not sure how small the numerator will become and whether the denominator will be greater than numerator. In this case should I just integrate?

Thank you for your help!

 

[TSny]

$\varepsilon$ is assumed to be small enough such that $\frac{2}{\mu^2} \varepsilon (a^4 - x^4)$ is small compared to $a^2 - x^2$. Expand $\left[a^2 - x^2 + \frac{2}{\mu^2} \varepsilon (a^4 - x^4) \right]^{-1/2}$ to first order in $\varepsilon$.

 

[TSny]

Discuss the motion:
If we sketch V(x) it will look alike to parabola with V(0) = 0. Depending on the value of E that the particle has, it will have a simple harmonic motion and will oscillate in the region enclosed between E and the x-axis.

V(x) will look approximately like a parabola, but it is not a true parabola. So, although the particle will oscillate, is the motion simple harmonic motion?

 

[green-beans]

V(x) will look approximately like a parabola, but it is not a true parabola. So, although the particle will oscillate, is the motion simple harmonic motion?

Hi, thank you for your comments! Regarding the part with ε, could you please specify what do you mean by "expand to first order in ε"? Do you mean to expand (2/μ²) * ε(a4−x4) using Taylor series and then take the first two terms?
As for the description of the motion, I thought that the particle will oscillate between two points depending where it starts on the graph and so will oscillate in a similar way to a pendulum. However, I am not sure if it's SHM or not since when oscillating the particle will be likely to lose some of its energy and so the amplitude of its oscillations will reduce. At the same time the particle will lose some of its energy due to external forces and the question does not mention any and so I am not sure if I can assume that or not.
Thank you once again!

 

[haruspex]

Do you mean to expand (2/μ2) * ε(a4−x4)

No, expand

1/√{(a2-x2) +(2/μ2)ε(a4-x4)})

 

[green-beans]

No, expand

I am sorry but I do not quite understand whether you mean to expand using Taylor series or not and if yes, about what x_0? I tried expanding it around x_0=0 and my first derivative gave me 0 leaving me with only f(0) as an approximation, and obviously this cannot be the case since we will not be able to get pi in the final expression. I also tried opening the brackets and completing the square. Then I used arcsin to solve the integral and I got a very long scary expression which I can't really simplify :(
I am sorry for being slow but I can't really see what else I can do.

 

[PeroK]

I am sorry but I do not quite understand whether you mean to expand using Taylor series or not and if yes, about what x_0? I tried expanding it around x_0=0 and my first derivative gave me 0 leaving me with only f(0) as an approximation, and obviously this cannot be the case since we will not be able to get pi in the final expression. I also tried opening the brackets and completing the square. Then I used arcsin to solve the integral and I got a very long scary expression which I can't really simplify :(
I am sorry for being slow but I can't really see what else I can do.

I took a look at this, but I must confess I don't understand the question. If $\epsilon$ is small, then $T$ is reduced by something compared with what?

Edit: is it compared with $\epsilon = 0$?

 

[ehild]

I took a look at this, but I must confess I don't understand the question. If $\epsilon$ is small, then $T$ is reduced by something compared with what?

Edit: is it compared with $\epsilon = 0$?

Yes, it is compared with the time period when $\epsilon = 0$.

 

[PeroK]

Thanks @ehild

@green-beans I would take out a factor of $(a^2 - x^2)$, if you haven't already. Also, it's easier to use the Binomial expansion than try to differentiate your function for the Taylor series.

 

[ehild]

Hi, thank you for your comments! Regarding the part with ε, could you please specify what do you mean by "expand to first order in ε"? !

You have to expand with respect to ε and stop at the first-order term.
As @PeroK suggested, first take out the factor a²-x² from $\left[a^2 - x^2 + \frac{2}{\mu^2} \varepsilon (a^4 - x^4) \right]^{-1/2}$
, and expand the other factor with respect to ε. Note that a⁴-x⁴=(a²-x²)(a²+x²).

 

[green-beans]

You have to expand with respect to ε and stop at the first-order term.
As @PeroK suggested, first take out the factor a²-x² from $\left[a^2 - x^2 + \frac{2}{\mu^2} \varepsilon (a^4 - x^4) \right]^{-1/2}$
, and expand the other factor with respect to ε. Note that a⁴-x⁴=(a²-x²)(a²+x²).

So, by saying to expand with respect to ε, do you mean to take x_0 = ε?

 

[ehild]

So, by saying to expand with respect to ε, do you mean to take x_0 = ε?

No. Take ε_0=0.

 

[green-beans]

No. Take ε_0=0.

Oh, so you differentiate as well in respect to epsilon and regard it as a function of ε and expand it using Taylor series around epsilon =0?
Sorry for asking so many questions. It's just I tried expanding it, and I did not get the right answer

 

[PeroK]

Oh, so you differentiate as well in respect to epsilon and regard it as a function of ε and expand it using Taylor series around epsilon =0?

That's another way to do it. But, the most obvious way is just to think of $\epsilon$ as some small number and find the difference between the two integrals: one with $\epsilon = 0$ and the other with some small number. You should recognise the set-up for using the Binomial expansion.

 

[green-beans]

That's another way to do it. But, the most obvious way is just to think of $\epsilon$ as some small number and find the difference between the two integrals: one with $\epsilon = 0$ and the other with some small number. You should recognise the set-up for using the Binomial expansion.

That's another way to do it. But, the most obvious way is just to think of $\epsilon$ as some small number and find the difference between the two integrals: one with $\epsilon = 0$ and the other with some small number. You should recognise the set-up for using the Binomial expansion.

Thank you for your help! I still keep on getting something close to the answer but I can't figure out what I am doing wrong. Here I attached the link to my solution: https://drive.google.com/open?id=0BzewpRKo76bDWWNld1U3SHNBVEk

 

[PeroK]

Thank you for your help! I still keep on getting something close to the answer but I can't figure out what I am doing wrong. Here I attached the link to my solution: https://drive.google.com/open?id=0BzewpRKo76bDWWNld1U3SHNBVEk

It looks like you just have a couple of algebraic lapses in there somewhere. Easily done and hard to spot!

I can't immediately pin down where you've gone wrong.

I got:

$T = \frac{2}{\mu} \int_{-a}^{a} (a^2 - x^2)^{-\frac12} [1 + \frac{2\epsilon}{\mu^2}(a^2 + x^2)]^{-\frac12} dx$

Do you get that?

 

[green-beans]

It looks like you just have a couple of algebraic lapses in there somewhere. Easily done and hard to spot!

I can't immediately pin down where you've gone wrong.

I got:

$T = \frac{2}{\mu} \int_{-a}^{a} (a^2 - x^2)^{-\frac12} [1 + \frac{2\epsilon}{\mu^2}(a^2 + x^2)]^{-\frac12} dx$

Do you get that?

Oh, I think I copied the initial integral incorrectly. I'll redo it again and I'll get back here :) thank you once again

 

[PeroK]

Oh, I think I copied the initial integral incorrectly. I'll redo it again and I'll get back here :) thank you once again

Just get that integral above and hit it with the Binomial! It's much simpler.

 

[green-beans]

Just get that integral above and hit it with the Binomial! It's much simpler.

So, I got a similar integral but in front of it I got 4/μ and the limit between 0 and a. So, that should be the same thing. When using Taylor series for ε I then obtain two integrals. The first one is 1/sqrt(a^2-x^2)^(-1/2). The second one (-ε/μ^2)*(a^2+x^2)/sqrt(a^2-x^2). When solving the integral, I still get the wrong answer which is (4/μ) (pi/2 - (ε/μ^2)*(3pia^2/4) which would have been the answer if I did not get pi/2 for the first term. So, I am not sure what I am doing wrong since I redid the integral several times :(

 

[PeroK]

So, I got a similar integral but in front of it I got 4/μ and the limit between 0 and a. So, that should be the same thing. When using Taylor series for ε I then obtain two integrals. The first one is 1/sqrt(a^2-x^2)^(-1/2). The second one (-ε/μ^2)*(a^2+x^2)/sqrt(a^2-x^2). When solving the integral, I still get the wrong answer which is (4/μ) (pi/2 - (ε/μ^2)*(3pia^2/4) which would have been the answer if I did not get pi/2 for the first term. So, I am not sure what I am doing wrong since I redid the integral several times :(

The first integral is the period when $\epsilon = 0$, so only the second integral is the difference when $\epsilon \ne 0$.

 

[green-beans]

The first integral is the period when $\epsilon = 0$, so only the second integral is the difference when $\epsilon \ne 0$.

But I got the first integral by multiplying (sqrt(a^2-x^2)^(-1/2)) by the first term of Taylor expansion which was 1 and the second integral - by multiplying by the second term of expansion (i.e. the one with first derivative) and in both cases I had ε_0 = 0 and so I do not quite see how only the second integral is the solution for small ε. Since I thought the whole expansion (first and second term) will demonstrate the behaviour of the function as ε gets closer to 0.

 

[ehild]

So, I got a similar integral but in front of it I got 4/μ and the limit between 0 and a. So, that should be the same thing. When using Taylor series for ε I then obtain two integrals. The first one is 1/sqrt(a^2-x^2)^(-1/2). The second one (-ε/μ^2)*(a^2+x^2)/sqrt(a^2-x^2). When solving the integral, I still get the wrong answer which is (4/μ) (pi/2 - (ε/μ^2)*(3pia^2/4) which would have been the answer if I did not get pi/2 for the first term. So, I am not sure what I am doing wrong since I redid the integral several times :(

(4/μ) (pi/2 - (ε/μ^2)*(3pia^2/4)) You dropped a parenthesis. The expression is the total time T, not the difference ΔT caused by the εx^4 term in the potential.
The problem asks:

show that when ε is small, T is reduced by approximately 3επa²/μ³

.

 

[green-beans]

(4/μ) (pi/2 - (ε/μ^2)*(3pia^2/4)) You dropped a parenthesis. The expression is the total time T, not the difference ΔT caused by the εx^4 term in the potential.
The problem asks: .

Ohhh, so, it asks for how much the time period is reduced by compared with the time period when epsilon is zero? So, I need to find delta T which is the difference between when epsilon is zero and when it's small.
Thank you so much and so sorry for asking that many questions. I think I now understand!

 

[PeroK]

Ohhh, so, it asks for how much the time period is reduced by compared with the time period when epsilon is zero? So, I need to find delta T which is the difference between when epsilon is zero and when it's small.
Thank you so much and so sorry for asking that many questions. I think I now understand!

See posts #7 and #8.