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Deceptively Difficult Physics Integration Problem (Restorative Forces)? 
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#1
Jul1804, 05:51 PM

P: 66

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(mb^{3}/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(mb^{3}/2k) · ∫sqrt[y/(1y)]dy where y = x/b (evaluated from 1 to 0) the 3rd hint is likewise elusive to me: Show that setting y=sin^{2}θ results in t = sqrt(mb^{3}/2k) · ∫2sin^{2}θdθ (evaluated from π/2 to 0) let alone the final result of π(mb^{3}/8k) 


#2
Jul1804, 08:43 PM

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[tex]F(x) = kx^{2}[/tex]
[tex]ma = kx^{2}[/tex] [tex]dv/dt = (k/m)x^{2}[/tex] [tex](dv/dx)(dx/dt) = (k/m)x^{2}[/tex] [tex]vdv = (k/m)x^{2}dx[/tex] [tex]\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)}  \frac{1}{b} \right )[/tex] 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
Jul1804, 09:01 PM

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Actually, it's quite simple:
[tex]dx/dt = (2k/m)^{1/2}(1/x  1/b)^{1/2}[/tex] [tex](1/x  1/b)^{1/2}dx = (2k/m)^{1/2}dt[/tex] [tex]\left ( \frac{bx}{bx} \right ) ^{1/2}dx = (2k/m)^{1/2}dt[/tex] [tex]\sqrt {\frac{y}{1y}}dy = (2k/mb^3)^{1/2}dt[/tex] Now, integrate from [itex]t=0[/itex] to [itex]t=t_f[/itex]. You know that [itex]y(t_f) = x(t_f)/b = 0/b = 0[/itex] and [itex]y(0) = x(0)/b = b/b = 1[/itex]. So, you'll have: [tex]\int _1 ^0 \sqrt{\frac{y}{1y}}dy = \sqrt{(2k/mb^3)}t_f[/tex] The rest should be pretty simple, unless you're just rusty on the calculus. They're suggesting a substitution: y = sinē[itex]\theta[/itex]. [tex]\int _1 ^0 \sqrt{\frac{y}{1y}}dy[/tex] [tex]= \int _{\pi /2} ^0 \sqrt{\frac{\sin ^2 \theta}{1  \sin ^2 \theta}}(2\sin \theta \cos \theta d\theta )[/tex] [tex]= 2\int _{\pi /2} ^0 \sqrt{\frac{\sin ^2 \theta}{\cos ^2 \theta}} \sin \theta \cos \theta d\theta[/tex] [tex]= 2\int _{\pi /2} ^0 \frac{\sin \theta}{\cos \theta} \sin \theta \cos \theta d\theta[/tex] [tex]= 2\int _{\pi /2} ^0 \sin ^2 \theta d\theta[/tex] [tex]= \int _{\pi /2} ^0 1  \cos (2\theta ) d\theta[/tex] [tex]= \theta  \frac{1}{2}\sin (2\theta )[/tex] [tex]= (\pi /2)[/tex] EDIT: no, this looks fine. This is the integral on the left side. Now isolate [itex]t_f[/itex] and you're done. 


#4
Jul2104, 06:06 AM

P: 66

Deceptively Difficult Physics Integration Problem (Restorative Forces)?
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
Jul2104, 10:37 AM

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cj
The second last line of what you quoted was: [tex]vdv = (k/m)x^{2}dx[/tex] Now, showing all my steps: [tex]\int _{v(0)} ^{v(t)} vdv = \int _{x(0)} ^{x(t)} (k/m)x^{2}dx[/tex] [tex]\int _0 ^{v(t)} vdv = (k/m) \int _b ^{x(t)} x^{2}dx[/tex] [tex]\frac{v^2}{2}_{0}^{v(t)} = (k/m)(1/x)_{b} ^{x(t)}[/tex] [tex]\frac{v(t)^2}{2} = (k/m)\left (\frac{1}{x(t)}  \frac{1}{b} \right )[/tex] And the last line was: [tex]\frac{v(t)^2}{2} = \frac{k}{m} \left (\frac{1}{x(t)}  \frac{1}{b} \right )[/tex] There's no difference between the last two lines here, just changed a [itex](k/m)[/itex] to a [itex]\frac{k}{m}[/itex]. 


#6
Jul2104, 12:43 PM

P: 66

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: [tex]\frac{v^2}{2}_{v(t)}^{(0)} = (k/m)(1/x)_{x(t)} ^{(b)}[/tex] rather than: [tex]\frac{v^2}{2}_{0}^{v(t)} = (k/m)(1/x)_{b} ^{x(t)}[/tex] ?? 


#7
Jul2104, 04:10 PM

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There's no difference:
[tex]\frac{v^2}{2}_{v(t)}^{(0)} = (k/m)(1/x)_{x(t)} ^{(b)}[/tex] [tex]\frac{0^2}{2}  \frac{v(t)^2}{2} = (k/m)[1/b  1/x(t)][/tex] [tex]\frac{v(t)^2}{2} = (k/m)[1/b  1/x(t)][/tex] [tex]\frac{v(t)^2}{2} = (k/m)[1/x(t)  1/b][/tex] Work out this one, and you'll see it's the same: [tex]\frac{v^2}{2}_{0}^{v(t)} = (k/m)(1/x)_{b} ^{x(t)}[/tex] 


#8
Jul2204, 10:26 AM

P: 66

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! 


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