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Homework Help: Energy Question - Help Please

  1. Nov 28, 2004 #1
    Energy Question -- Help Please

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

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

    [tex] \frac{1}{2}\pi a \sqrt{\frac{ma}{2k}} [/tex]

    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.
  2. jcsd
  3. Nov 28, 2004 #2
    I tried divding by mass and integradting twice to get distance but it didn't work
  4. Nov 29, 2004 #3


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    The differential equation [itex]m\frac{dv}{dt}= -\frac{k}{x^2} [/itex] 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:
    [itex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}[/itex] by the chain rule.
    But [itex]\frac{dx}{dt}= v[/itex] so the equation becomes:
    [tex] mv\frac{dv}{dx}= -\frac{k}{x^2}[/tex]

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

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

    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
    [tex]2\sqrt{a-u^2}du= \sqrt{\frac{2k}{ma}}dt[/tex].

    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:
    [tex]2a cos^2\theta dx= \sqrt{\frac{2k}{ma}} dt[/tex]

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

    That can be integrated directly to get

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

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

    Wow! I hope you teacher had some simpler way of doing that!!
    Last edited by a moderator: Nov 30, 2004
  5. Nov 29, 2004 #4
    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.
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