# Deceptively Difficult Physics Integration Problem (Restorative Forces)?

1. Jul 18, 2004

### cj

The following seemed simple enough to me . . . I'm somewhat sure about the requisite physics, but shakey on the integrals:

A particle of mass m is released from rest a distance b from a fixed origin of force that attracts a particle according to the inverse square law:

F(x) = -kx-2

Show that the time required for the particle to reach the origin is:

π(mb3/8k)1/2

And then I reviewed the hints provided by the author -- the very first of which completely stumped me. My rusty calc skills not withstanding, how is the following hint true:

Show that dx/dt = -(2k/m)1/2 · (1/x - 1/b)1/2
the negative sign results from the physical situation

the subsequent hint is also a mystery to me:

Show that t = sqrt(mb3/2k) · ∫sqrt[y/(1-y)]dy
where y = x/b (evaluated from 1 to 0)

the 3rd hint is likewise elusive to me:

Show that setting y=sin2θ results in t = sqrt(mb3/2k) · ∫2sin2θdθ (evaluated from π/2 to 0)

let alone the final result of π(mb3/8k)

Last edited: Jul 18, 2004
2. Jul 18, 2004

### AKG

$$F(x) = -kx^{-2}$$

$$ma = -kx^{-2}$$

$$dv/dt = -(k/m)x^{-2}$$

$$(dv/dx)(dx/dt) = -(k/m)x^{-2}$$

$$vdv = -(k/m)x^{-2}dx$$

$$\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )$$

I believe you can figure out the first hint from here. Note that the integration was done from time=0 to time=t. Given what you have, see if you can work with it to get the second hint.

3. Jul 18, 2004

### AKG

Actually, it's quite simple:

$$dx/dt = -(2k/m)^{1/2}(1/x - 1/b)^{1/2}$$

$$(1/x - 1/b)^{-1/2}dx = -(2k/m)^{1/2}dt$$

$$\left ( \frac{bx}{b-x} \right ) ^{1/2}dx = -(2k/m)^{1/2}dt$$

$$\sqrt {\frac{y}{1-y}}dy = -(2k/mb^3)^{1/2}dt$$

Now, integrate from $t=0$ to $t=t_f$. You know that $y(t_f) = x(t_f)/b = 0/b = 0$ and $y(0) = x(0)/b = b/b = 1$. So, you'll have:

$$\int _1 ^0 \sqrt{\frac{y}{1-y}}dy = -\sqrt{(2k/mb^3)}t_f$$

The rest should be pretty simple, unless you're just rusty on the calculus. They're suggesting a substitution: y = sin²$\theta$.

$$\int _1 ^0 \sqrt{\frac{y}{1-y}}dy$$

$$= \int _{\pi /2} ^0 \sqrt{\frac{\sin ^2 \theta}{1 - \sin ^2 \theta}}(2\sin \theta \cos \theta d\theta )$$

$$= 2\int _{\pi /2} ^0 \sqrt{\frac{\sin ^2 \theta}{\cos ^2 \theta}} \sin \theta \cos \theta d\theta$$

$$= 2\int _{\pi /2} ^0 \frac{\sin \theta}{\cos \theta} \sin \theta \cos \theta d\theta$$

$$= 2\int _{\pi /2} ^0 \sin ^2 \theta d\theta$$

$$= \int _{\pi /2} ^0 1 - \cos (2\theta ) d\theta$$

$$= \theta - \frac{1}{2}\sin (2\theta )$$

$$= (-\pi /2)$$

EDIT: no, this looks fine. This is the integral on the left side. Now isolate $t_f$ and you're done.

Last edited: Jul 18, 2004
4. Jul 21, 2004

### cj

Deceptively Difficult Physics Integration Problem (Restorative Forces)? Reply to T

Thanks very much -- this is quite helpful.

However (with regard to the last line in your
post below), if you integrate from t = 0 to t = t
is the order not (1/b - 1/x) as opposed to (1x - 1/b)??

Thanks!

5. Jul 21, 2004

### AKG

cj

The second last line of what you quoted was:

$$vdv = -(k/m)x^{-2}dx$$

Now, showing all my steps:

$$\int _{v(0)} ^{v(t)} vdv = \int _{x(0)} ^{x(t)} -(k/m)x^{-2}dx$$

$$\int _0 ^{v(t)} vdv = (k/m) \int _b ^{x(t)} -x^{-2}dx$$

$$\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}$$

$$\frac{v(t)^2}{2} = (k/m)\left (\frac{1}{x(t)} - \frac{1}{b} \right )$$

And the last line was:

$$\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)} - \frac{1}{b} \right )$$

There's no difference between the last two lines here, just changed a $(k/m)$ to a $\frac{k}{m}$.

6. Jul 21, 2004

### cj

I certainly understand your reasoning -- thanks again.

What is still perplexing me is that, since the initial conditions have the
object located at x=b, shouldn't -- technically -- the interval be taken as:

$$\frac{v^2}{2}|_{v(t)}^{(0)} = (k/m)(1/x)|_{x(t)} ^{(b)}$$

rather than:

$$\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}$$

??

7. Jul 21, 2004

### AKG

There's no difference:

$$\frac{v^2}{2}|_{v(t)}^{(0)} = (k/m)(1/x)|_{x(t)} ^{(b)}$$

$$\frac{0^2}{2} - \frac{v(t)^2}{2} = (k/m)[1/b - 1/x(t)]$$

$$-\frac{v(t)^2}{2} = (k/m)[1/b - 1/x(t)]$$

$$\frac{v(t)^2}{2} = (k/m)[1/x(t) - 1/b]$$

Work out this one, and you'll see it's the same:

$$\frac{v^2}{2}|_{0}^{v(t)} = (k/m)(1/x)|_{b} ^{x(t)}$$

8. Jul 22, 2004

### cj

Got it -- thanks for answering this, as well as all my
other questions.

I was thinking there was a convention that said
something like the integration interval should
be taken as from the final or "end" state to the
initial state.

Again, thanks for all your answers!

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook