Energy Question -- Help Please

[Tom McCurdy]

A particle of mass m moves on the x-axis under the influence of a forace of attraction towards the origin give by
F= \frac{-k}{x^2} i .

If the particle starts from rest at x=a prove that it will arrive at the origin in a time given by

\frac{1}{2}\pi a \sqrt{\frac{ma}{2k}}

Our teacher explained it to me in class but unforutently he wasn't able to finish it. I was wondering if someone could step me through this proof please.

[Tom McCurdy]

I tried divding by mass and integradting twice to get distance but it didn't work

[HallsofIvy]

The differential equation m\frac{dv}{dt}= -\frac{k}{x^2} can't just be "integrated twice" because you don't know x as a function of t.

However, since t does not appear explicitely in the equation, there is a standard "trick" for reducing to a first order equation which can be integrated:
\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt} by the chain rule.
But \frac{dx}{dt}= v so the equation becomes:
mv\frac{dv}{dx}= -\frac{k}{x^2}

That's a "separable" differential equation. We separate it to get
m v dv= -\frac{kdx}{x^2}
and integrate to get
\frac{1}{2}m v^2= \frac{k}{x}+ C
When t=0, v= 0 and x= a so we have 0= \frac{k}{a}+ C or C= -\frac{k}{a} which gives

v^2= \frac{2k}{m}(\frac{1}{x}-\frac{1}{a})
= \frac{2k}{ma}\frac{a-x}{x}
Which reduces to
v= \frac{dx}{dt}= \sqrt\(\frac{2k}{ma}\frac{a-x}{x}\)}
or
\sqrt{\frac{x}{a-x}}dx= \sqrt{\frac{2k}{ma}}dt

To integrate that let u= (a-x)1/2 so that du= (1/2)(a-x)-1/2dx and u2= a-x so x= a- u2. The equation becomes
2\sqrt{a-u^2}du= \sqrt{\frac{2k}{ma}}dt.

The left hand side is now a fairly standard trigonometric substitution:
Let u= √(a) sin(θ) so that du= √(a) cos(θ)dθ and √(a- u^2) becomes √(a) cos(θ) so the equation, in terms of θ is:
2a cos^2\theta dx= \sqrt{\frac{2k}{ma}} dt

To integrate that, use the trig identity cos2θ= (1/2)(1+ cos(2θ)) so that the equation becomes:
a(1+ cos(2\theta))d\theta= \sqrt{\frac{2k}{ma}}dt

That can be integrated directly to get

a(\theta+ \frac{1}{2}sin(2\theta)= \sqrt{\frac{2k}{ma}}t+ C

When t= 0, x= a so that u= 0 and θ= 0. The equation becomes
a(0+ 0)= 0+ C so C= 0.

When x= 0, u= \sqrt{a} and \theta= \frac{\pi}{2}. Of course, in that case 2\theta= \pi so sin(2\theta)= 0 and our formula becomes
\frac{a\pi}{2}= \sqrt{\frac{2k}{ma}}T
and
T= \frac{\pi a}{2}\sqrt{\frac{ma}{2k}}
as advertised!

Wow! I hope you teacher had some simpler way of doing that!!

[Tom McCurdy]

Thank you so much for your response... this looks almsost identical to the way he showed in class today... Its alright that i didn't get it done in time, only a few kids in my class had it done. This is very hard for me as I am just taking Calc BC this year, your response is very useful to me.