How to Calculate Velocity and Position of a Proton in an Electric Field?

[Gogsey]

A proton is initially (at time zero) at rest at the origin when a uniform, sinusoidally-varying electric field is turned on, exerting a force F(t) = Fo sinwt on the proton. Calculate the velocity and position as functions of time. Show that the proton moves farther and farther from the origin as time goes on. Calculate the numerical value of the average velocity (averaged over a whole number of cycles) if Fo = eE , w = 2 pi f , with E0 = 5×104 volts/metre, f = 20 MHz.

Firstly there is no velocity term in the equation so I was a little confused, but tried integrating mdv = Fo sinwt dt, and got, V = -Fo coswt/wm, and then again to get x = -Fo sin wt/w^2m.

Can anyone confirm this is right nor not. And if not, what am I doing wrong?

As you can tell, my mechanics is extremely rusty. Thats cause I haven't done any in 2 years, and there was no review or older material at the beginning of this course.

Thanks

 

[gabbagabbahey]

Firstly there is no velocity term in the equation so I was a little confused, but tried integrating mdv = Fo sinwt dt, and got, V = -Fo coswt/wm, and then again to get x = -Fo sin wt/w^2m.

Can anyone confirm this is right nor not. And if not, what am I doing wrong?

Shouldn't there be a constant of integration or two? Given that v(0)=x(0)=0, what must those constants be?

 

[Gogsey]

Well, those constants should be Vo and Xo, but both are zero at time = 0, so they don't contribute to the question at all.

 

[gabbagabbahey]

Well, those constants should be Vo and Xo, but both are zero at time = 0, so they don't contribute to the question at all.

Nope, you definitely need to consider the constants of integration. Without them you have:

$$v(t)=\frac{-F_o}{m\omega}\cos\omega t \implies v(0)=\frac{-F_o}{m\omega}\cos(0)=\frac{-F_o}{m\omega}\neq 0$$

and x(t) simply oscillates instead of increasing as expected.

So clearly something is wrong

 

[Gogsey]

Is this anything to do with phase angles? Like cos(wt + phi), and that sort of stuff?

 

[Gogsey]

Can you find the constants of integration by solving the initial value conditions? This seems like the logical thing to do.

 

[gabbagabbahey]

Can you find the constants of integration by solving the initial value conditions? This seems like the logical thing to do.

Yes, that's exactly what you need to do. Give it a shot and see what you get.

 

[Gogsey]

Ok, so the first one should be Fo/mw. The next ones a little trickier. First attempt I got, the constant to be zero for the seond integral.

 

[gabbagabbahey]

Ok, so the first one should be Fo/mw.

Good

The next ones a little trickier. First attempt I got, the constant to be zero for the seond integral.

Better show me your attempt at that one

 

[Gogsey]

Ok,

So for v(t)=-Fo sinw(t)/mw + Fo/mw.

Actually, th prof skipped the integration for finding X, although it was an equation without any constant, ie, this equation had a cos function, so there was no constant in this part. It was simply dv/dt=Fo sinw(t), and he skipped the integration, to find x. So I don't really know how to do this.

All I can think of is that it could be x=Fo sinw(t)/mw^2 + Fot/mw + C. And since t is zero,
C=0.

 

[gabbagabbahey]

Ok,

So for v(t)=-Fo sinw(t)/mw + Fo/mw(t).

You mean x(t) right?

$$v(t)=\frac{-F_0}{m\omega}\cos\omega t +\frac{F_0}{m\omega} \implies x(t)=\int v(t)dt=\frac{-F_0}{m\omega}\sin\omega t +\frac{F_0 t}{m\omega}+C$$

When you substitute the initial conditions, you should find that $C=0$.

What happens to x(t) as $t \to \infty$?

To average some periodic function f(t) over a period T, you use the formula $$f_{ave}=\frac{1}{T}\int_0^T f(t)dt$$

Apply that to v(t) to find your average velocity.

 

[Gogsey]

Sorry, a bit lost now, I haven't seen this before, at least that I can remeber. Definatley not seen it in my 2nd year Mechanics course.

 

[gabbagabbahey]

You should have covered time averages in Calc I or II. Try applying the formula and see what you get.

 

[Gogsey]

Does this mean that I inetegrate v(t), and then after the integration I get what I had before times 1/t, then c = this integral evaluated at t(cause at zero its zero), times T?

 

[gabbagabbahey]

Does this mean that I inetegrate v(t), and then after the integration I get what I had before times 1/t, then c = this integral evaluated at t(cause at zero its zero), times T?

This integral has nothing to do with determining $C$. You already showed that $C=0$.

This integral is what you need to compute in order to find the average velocity and answer this part of your question:

Calculate the numerical value of the average velocity (averaged over a whole number of cycles) if Fo = eE , w = 2 pi f , with E0 = 5×104 volts/metre, f = 20 MHz.

In the formula I gave above, capital T represents the period of v(t).

$$T=\frac{1}{f}=\frac{2\pi}{\omega}$$

 

[Gogsey]

This may sound stupid, and it shows how little I know about this, but when you find Vav, which turns out to be just Xo (fromT to 0) * 1/T, which is then is just x(T) * 1/T.

Do you sub in T for the sin wt part, ie for the w(t)?

I'm pretty sure you don't but I wouldn't want to make this silly mistake.

 

[gabbagabbahey]

Do you sub in T for the sin wt part, ie for the w(t)?

Of course. To find the value of x(T), you substitute t=T everywhere into the expression for x(t).

 

[Gogsey]

Maybe I said that wrong.

Vav = 1/T (Fosinwt/mw^2 + FoT/mw)

So 1/T cancles with the second term T to get Fo/mw.

Now should the first term be Fo sin wt/Tmw^2, or Fo sin wT/TmW^2?

 

[gabbagabbahey]

Fo sin wT/TmW^2

for the exact reason stated in my previous post.

 

[Gogsey]

Awesome., thanks. And wT still is 2 pi f, which is given in the equation, right? By going from wt to wT, nothing has changed in this sense.

 

[gabbagabbahey]

w=2pi f, so wT =2pi fT= 2pi f *(1/f)=2 pi

 

[Gogsey]

w=2pi f, so wT =2pi fT= 2pi f *(1/f)=2 pi

Oh, right, therefore we get sin (2 pi).

then x(t) = Fo sin(2pi)/(2pi/w)mw^2 + Fo/mw?

 

[gabbagabbahey]

Sure, but sin(2pi)=0 right?

 

[Gogsey]

Right, so after all that we didn't even need that term, lol. Although it was importnat for me to get to the answer, since if it was cos, then that term wouldn't be zero.

So, this is why X changes with t or T.

Thank you so much.

I really appreciate the help. At some points you were probably thinking "geez he knows nothing" and you still kept going.

Thanks for that.

 

[gabbagabbahey]

Welcome