How Does the Particle's Velocity and Potential Shape Affect its Transit Time?

[CAF123]

Homework Statement

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.

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.

 

[TSny]

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.

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.

If w=0, V(x) = 0 and so E=T which gives a period of t = $\sqrt{2m/T}\ell$

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

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 $x =(l/\sqrt{2}) \sin \theta$.

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.

 

[CAF123]

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.

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).

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

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

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.

$$\sqrt{2E - mw^2(x^2-\ell^2)} = \sqrt{2E + mw^2\ell^2 - mw^2x^2} = \sqrt{mw^2(\frac{2E}{mw^2} + \ell^2 - x^2}$$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.

 

[TSny]

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

This will simplify.

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.

Try Expressing E in terms of the initial velocity.

 

[CAF123]

This will simplify.

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

Try Expressing E in terms of the initial velocity.

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.

 

[TSny]

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

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

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.

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

 

[CAF123]

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

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

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

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.

 

[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.

 

[CAF123]

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.

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?

 

[TSny]

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.

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?

 

[CAF123]

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?

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.

 

[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?

 

[CAF123]

$t = \pi/\omega$ 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 $\omega$?

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

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

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

 

[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.

 

[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 $$\frac{1}{w}\left[\frac{\pi}{2} - \frac{3\pi}{2}\right] = -\frac{\pi}{w}$$

 

[TSny]

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

 

[CAF123]

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

Thanks TSny,