Particle in a Potential

1. Apr 28, 2013

CAF123

1. The problem statement, all variables and given/known data
Consider the particle moving in the potential $$V = \frac{1}{2} mw^2(x^2-\ell^2)\,\,\text{for}\,\,|x| < \ell\,\,\text{and}\,\,0\,\,\text{for}\,\,|x| \geq \ell$$

The particle starts moving to the right at $x = -\ell$ with positive velocity $v$. How long will it take the particle to reach x=l in the cases v=0 and w=0? For the general case, evaluate the period by using a substitution of the form x = Asinθ, for some suitably chosen A such that $$t = \frac{2}{w}\arcsin\left[\frac{\ell w}{(v^2 + w^2 \ell^2)^{1/2}}\right]$$ Recover the result for v=0.

3. The attempt at a solution

The potential is parabolic with a minimum of V and intersecting x at ±l. The period t derived was $$t = \left(\frac{m}{2}\right)^{1/2} \int_a^b \frac{dx}{(E-V)^{1/2}},$$ with a<b. If v=0, then the particle was released with no kinetic energy T and so its energy is all potential. Since F conservative, E=V always. This makes the denominator 0 and so t tends to infinity (i.e the particle never reaches l). If w=0, V(x) = 0 and so E=T which gives a period of t = $\sqrt{2m/T}\ell$

In general, I reduced the expression for t down to $$t =\sqrt{2} \sqrt{\frac{m}{2}} \int_a^b \frac{dx}{\sqrt{2E - mw^2(x^2-\ell^2)}}$$ and then let $x =(l/\sqrt{2}) \sin \theta$. This gives $$\sqrt{m} \int_{\theta_1}^{\theta_2} \frac{\ell \cos \theta d \theta}{\sqrt{2E + \frac{1}{2} m w^2\ell^2 \cos^2 \theta}}$$ but I can't see how to progress with this at the moment.

Last edited: Apr 28, 2013
2. Apr 28, 2013

TSny

Why would E always equal V? If I drop a ball from rest at some height above the floor, then the ball starts with all potential energy. But it nevertheless picks up kinetic energy as it falls.

That looks right. You might want to express the answer in terms of the initial velocity.

I think the integral is correct. Before deciding what to substitute for $x$, try writing the denominator as

$\sqrt{2E - mw^2(x^2-\ell^2)} = B\sqrt{A^2 - x^2}$ where $A$ and $B$ are certain constants.

Last edited: Apr 28, 2013
3. Apr 28, 2013

CAF123

Oops, rather I meant that the total energy is given by the initial conditions since F is conservative. So since v=0 (in this case), the total energy is given by E = (1/2)mw^2(x^2-l^2).

$t = \sqrt{\frac{4}{v^2}} \ell$

$$\sqrt{2E - mw^2(x^2-\ell^2)} = \sqrt{2E + mw^2\ell^2 - mw^2x^2} = \sqrt{mw^2(\frac{2E}{mw^2} + \ell^2 - x^2}$$So, $$\sqrt{m} w \sqrt{\frac{2E}{mw^2} + \ell^2 - x^2},$$ with $A^2 = \frac{2E}{mw^2} + \ell^2$, $B = \sqrt{m} w$

With this, I reduce the integral to $$t = \frac{2}{w} \arcsin\left(\frac{\ell \sqrt{m}w}{\sqrt{2E + \ell^2 m w^2}}\right)$$ which is close but I need to reexpress E. Doing so, this just brings x = Asin (theta) back into play, which is not what I want.

Last edited: Apr 28, 2013
4. Apr 28, 2013

TSny

This will simplify.

Try Expressing E in terms of the initial velocity.

5. Apr 28, 2013

CAF123

Do you mean reexpress l as well using the below eqn for energy?

What I have for E is $\frac{m}{2}v^2 + \frac{1}{2}mw^2(x^2-\ell^2)$, however subbing this in will give me x (=Asinθ) back into the eqn.

6. Apr 28, 2013

TSny

Simplify $\sqrt{\frac{4}{v^2}}$ in the expression $\sqrt{\frac{4}{v^2}}\;l$

Express E in terms of the initial velocity. What is the value of x when v is the initial velocity?

7. Apr 28, 2013

CAF123

Duh! $\frac{2}{v}\ell$

I got it and it makes sense, thank you. It says I should recover my expression for the case of v=o. However, I said that it would take an infinite amount of time for the particle to reach l but it seems from the general expression derived for t, that this is not the case.

8. Apr 28, 2013

TSny

Your conclusion of infinite time for initial v = 0 was based on the assumption that the velocity would always remain zero. The potential energy function has a sharp corner at $x = -l$. So, the force acting on the particle is not defined there. But, I think the problem wants you to assume that you are letting the particle start at $x = -l + \epsilon$ where $\epsilon$ is a positive infinitesimal quantity. Then, there will be a force acting on the particle at the point of release and the particle will accelerate toward x = 0 and reach $x = +l$ in a finite time.

9. Apr 28, 2013

CAF123

How would I go about finding the finite time? I know that the energy of the particle is $E = \frac{1}{2}mw^2(x^2-l^2)$ since F is conservative. This was of the same form as the potential energy so E-V = 0 and the denominator of the expression for t is zero and since the numerator is finite, the whole expression tends to infinity. (so I then concluded that t was infinite). Where is my error here?

10. Apr 28, 2013

TSny

No, that expression for E is incorrect.

The general expression for the total energy when the particle is between $x = -l$ and $x = l$ is $$E = T + V =\frac{1}{2} mv^2 + \frac{1}{2} mw^2(x^2-\ell^2)$$
where, here, $v$ is the velocity when the particle is at position $x$.

The energy is a constant, so it can be evaluated at any point of the motion. For the case where you let the particle start at rest at $x = -l$, what is the value of the total energy E?

11. Apr 28, 2013

CAF123

The energy would be zero there and so the total energy of the particle is zero always with the kinetic = -potential. So then E-V = -V = T. In my expression for t, I get $$\frac{1}{w} \int_{-l}^{l} \frac{dx}{\sqrt{l^2-x^2}}$$ which gives me $\frac{\pi}{w}$ not the 2/w I think I want.

12. Apr 28, 2013

TSny

$t = \pi/\omega$ is correct. Look at your potential energy function:

$V(x) = \frac{1}{2}m\;\omega^2(x^2-l^2) = \frac{1}{2}m\;\omega^2 x^2 - \frac{1}{2}m\;\omega^2 l^2$

The last term is just a constant that determines the zero point of potential energy. Ignoring that constant, do you see that $V(x)$ is the potential energy for a simple harmonic oscillator of mass $m$ and angular frequency $\omega$?

How should the time to go from $x = -l$ to $x = l$ (starting from rest) be related to the period of SHM?

13. Apr 29, 2013

CAF123

Yes, with 'spring constant' k = mw^2.

The time from -l to l will be exactly half the period of SHM. This gives me what I want, that is pi/w. (ignore what I said about 2/w). I was a bit sloppy with my notation earlier - the eqn I derived was not the period, it was half a period. Another quick question: when I got pi/w, I did the integral from 0 to l and multiplied by 2 because of symmetry. However, if you do it from -l to l, you get a negative pi/w. What is the physical reasoning behind this?
Thanks

Last edited: Apr 29, 2013
14. Apr 29, 2013

TSny

The integral from -l to l should be positive. If you want to post the details of how you got a negative value, we can look at it.

15. Apr 29, 2013

CAF123

So what I am integrating is $$\frac{1}{w} \int_{-l}^l \frac{dx}{\sqrt{l^2-x^2}} = \frac{1}{w} \int_{\theta_1}^{\theta_2} d\theta\,\text{with the substitution x = l \sin \theta}$$ When $x=l, \sin \theta = 1$ and so principal value for theta is $\pi/2$. Similarly, for x=-l, $\sin \theta = -1$ which gives $\theta = 3\pi/2$ as the principal value. Then $$\frac{1}{w}\left[\frac{\pi}{2} - \frac{3\pi}{2}\right] = -\frac{\pi}{w}$$

16. Apr 29, 2013

TSny

The principle value of arcsin(-1) is $-\pi/2$.

17. Apr 29, 2013

Thanks TSny,