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Object falling from rest, solution of differential equation

  1. Jul 20, 2012 #1
    1. The problem statement, all variables and given/known data

    An object is falling from rest with air resistance modelled by $$kv_x$$.
    Where v_x is the object's velocity in the x-direction (downwards).
    Find an expression for the speed of the object as a function of time.

    2. Relevant equations

    $$mg-kv_x=m\frac{dv_x}{dt}$$

    3. The attempt at a solution

    I can solve differential equations using the separation of variables technique, but can't seem to separate these algebraically. I know that it is possible to start off with:

    $$\frac{mg-kv_x}{m}=\frac{dv_x}{dt}$$

    But from here I can't separate the v_x term and put it by the derivative. My teacher hinted that one solution (which pure mathematicians may not like) is to rearrange to:

    $$dt=\frac{dv_x}{\left(g-kv_x / m\right)}$$

    Then integrate wrt time. The left hand side gives t, but I'm not sure what happens on the right hand side. I know that it must produce a log, but if someone could point out how, I'd be most grateful. I'd also appreciate an explanation of why pure mathematicians don't like this approach, and how they would do it...but only if someone has time.
    Thanks
     
  2. jcsd
  3. Jul 20, 2012 #2
    That's not a nice way of thinking about it, however, you're almost there.

    If you get your equation in this form: f(x) dx/dt = g(t)
    Then you can integrate both sides with respect to t.

    Now the right hand side, I'm sure you're comfortable with, just integrate g(t) with respect to t as usual.

    The left hand side is more subtle, so the best way to understand it, is that you're using the chain rule (or integration by substitution).
    So integrating f(x) dx/dt with respect to t, is the same as integrating f(x) with respect to x.

    Just a further elaboration on that last point:
    d/dt (F(x)) = F'(x) dx/dt
    So integrating both sides with respect to t, shows that:
    integral of F'(x) dx/dt is just F(x) i.e. the integral of F'(x) with respect to x.


    I hope that makes sense and wasn't confusing!
    Now you should be able to apply that thinking to your problem :)
     
  4. Jul 20, 2012 #3

    HallsofIvy

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    I can't imagine why "pure mathematicians" wouldn't like that. That's a standard method for solving first order differential equations- typically the first method taught in Differential Equations courses.

    Not exactly. You integrate the left side, which is a function of t only, with respect to t and integrate the right side, which is a function of x only, with respect to x.

    The integral on the right side is [itex]\int dvx/(g- kv_x/m)[/itex]. Make the substitution ]][itex]u= g- kv_x/m[/itex] so that [itex]du= -(k/m)dvx[/itex] or [itex]dv_x= -(m/k)du[/itex] and the integral becomes
    [tex]-\frac{m}{k}\int\frac{du}{u}= -\frac{k}{m}ln(u)+ C= -\frac{k}{m}ln(g- kv_x/m)+ C[/tex]

    Who told you that "pure mathematicians don't like this approach"? A physics teacher? Is this a secondary school teacher and not a college teacher? Perhaps he/she had been told at one time that "dy/dx is not a fraction" and so thinks it is not "mathematically" right to separate it into "dy" and "dx'.

    It is true that "dy/dx" is NOT defined as a fraction but any Calculus course should introduct "differentials" which are defined specifically so that "dy" and "dx" can be used as separate entities. Differentials are especially important in "differential geometry" which is "higher mathematics" but it is important for a physicist to know at least some differential geometry for Relativity.

    (By the way, mathematicians are seldom "pure"- they're randy devils!)
     
  5. Jul 20, 2012 #4
    Oh, and with regards to pure mathematicians, stay faaaaaaaaar away from that!

    I'll put it this way, in pure maths, if you want to integrate something as simple as x^2 with respect to x, it takes like a whole A4. You have to construct a lower sum and and upper sum (basically by using rectangles), then you have to obtain what's called the upper and lower integral by taking limits of those sums as the boxes have infinitesimal width, show the two are the same so you can say it's "Riemann Integrable" and then finally you have your answer.

    If you really want to look at it go ahead, but I'd say just stick to the applied approach where you just use the Fundamental Theorem of Calculus i.e. integral of F'(x) with respect to x is F(x) along with rules you've proven about differentiation such as the chain rule (for integration by substitution) and the product rule (for integration by parts).
     
  6. Jul 20, 2012 #5
    Thanks for both of these replies.

    Firstly, to Marioeden: Yes, that's the separation of variables technique that I'm familiar with. I usually look to separate the variables so that they are on either sides of the equation, then perform the integration using the chain rule as you mentioned. However, in this particular case I was unclear as to how to separate the variables fully. The term v_x is not a function of x at all. It maybe the way I expressed it, but v_x is just the x-component of velocity which varies as a function of time. This is why I couldn't separate out the time component properly. There is no g(x) in this problem.

    HallsofIvy: Thanks, that substitution looks to do the trick quite nicely. I hadn't spotted that possibility. You mentioned that the right-hand side was being integrated with respect to x, but (based on the comments in the previous paragraph) it should be integrated with respect to v_x, shouldn't it?
    My teacher is a university physicist. He may have just been joking for the reasons you mentioned...he just said that pure mathematicians might not approve of this approach.

    Before you pointed out the substitution, I had been attempting a solution like this:
    - converting instances of v_x into a derivatives of x with regard to time, of appropriate order.
    - Rearranging the expression into:
    $$\frac{d^2x}{dt^2}+\frac{k}{m}\frac{dx}{dt}=g$$
    - Solving the associated homogenous equation using the auxiliary equation to get:
    $$\lambda^2+\frac{k}{m}\lambda=0$$
    $$x(t)=Ce^{\lambda_1t} + De^{\lambda_2t}$$
    Where $$\lambda_1 = \frac{k}{m}$$ and $$\lambda_2=-\frac{k}{m}$$
    - Taking the derivative and second derivative of this, inserting back into the original homogenous equation, (which shows that C=0), finding a particular integral to account for the g in the original expression, and coming up with the expression
    $$\frac{dx}{dt}=-\frac{k}{m}De^{-kt/m}+g$$
    - Setting t=0 to find the value of D as:
    $$D=\frac{mg}{k}$$
    so
    $$v_x=g\left(1-e^{-kt/m}\right)$$

    So I guess I've gone wrong somewhere. Clearly, the substitution you offered avoids all this messing about though. If anyone has a handy hint to make my solution work, it would be greatly appreciated.
     
  7. Jul 20, 2012 #6
    A basic way to solve the equation is to realize the solution may be formed by adding a particular solution to the homogenous solution. In a case like this the particular solution can be 'eyeballed' or found by inspection because it is 'intuitively obvious' as is said in math texts. Things in math texts were not always intuitively obvious to me.

    You have

    dV/dt + (k/m)V = g

    The particular solution Vp is mg/k.

    The homogenous equation in Vh is

    dVh/dt = -k/mVh

    It's solution is

    Vh = Ce^((-k/m))t) where C is the constant of integration

    Adding Vp and Vh

    V = Ce^((-k/m))t) + mg/k

    Evaluate the constant of integration at t=0 yields the solution.
     
  8. Jul 20, 2012 #7

    ehild

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    Those lambda values are not right.


    ehild
     
  9. Jul 20, 2012 #8
    Ahhh, thanks. That makes sense now.

    So $$\lambda_1=\lambda_2=-\frac{k}{m}$$

    so

    $$x(t)=\left(C+Dt\right)e^{-\frac{k}{m}t}$$

    If I include the fact that I made a mistake with the particular integral (should have been mg/k, thanks Lawrence) I might be nearly there.

    EDIT:
    I put the wrong variable in
     
    Last edited: Jul 20, 2012
  10. Jul 20, 2012 #9
    Still not right:

    I'm getting

    $$x(t)=\left(C+Dt\right)e^{-\frac{k}{m}t}+\frac{mg}{k}t$$

    So taking out C, as I'll define the object to be at the origin at t=0, and differentiating:

    $$v(t)=e^{-\frac{k}{m}t}D\left(1-\frac{k}{m}t\right)+\frac{mg}{k}$$

    Using initial conditions (object at rest), D=-mg/k:

    $$v(t)=e^{-\frac{k}{m}t}\left(gt-\frac{mg}{k}\right)+\frac{mg}{k}$$


    Which seems far removed from the answer I get if I use the substitution at the beginning of this thread. Where did I go wrong?
     
    Last edited: Jul 20, 2012
  11. Jul 20, 2012 #10
    x is just a variable, call it whatever you want, v for velocity or u, or little boy, whatever you like. The separation of variables still works, only you have v_x as your variable.
     
  12. Jul 20, 2012 #11
    Ok, I take your point...I just thought maybe people had thought I was referring to v(x) which also gets used in kinematics quite a lot. The problem I had was really how to separate the variables.

    Anyway, having spent most of the day on this (!) I can now see how the separation of variables is done using substitution. What I can't do is make the answer it produces tie in with the answer I get if I make this a second order DE and use the homogenous equation + particular integral technique, nor can I get either of these answers to tie in with the method keeping it as a first order equation (as Lawrence indicated above). They should give the same result, really! Having said that, my answer using d^2x/dt^2 is getting close to Lawrence's result...I probably just made a little slip or something.

    I still don't see how the two techniques could give such different answers though...one must be wrong.
    EDIT:
    I've just realised that the substitution technique gives a log function equated to t, not to v. So in fact the techniques probably do give the same result. When I gather strength I'll just check it out for satisfaction.
    Thanks for all your help, guys.
     
  13. Jul 20, 2012 #12

    ehild

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    No. λ1=0, λ2=-k/m

    ehild
     
  14. Jul 20, 2012 #13
    Right, so let's see what happens with the eigenfunction solution:

    mg - kv = mv'

    Rearrange to get:

    v' + kv/m = g

    homogeneous problem is v' + kv/m = 0

    Try v = Ae^Lt, so we get
    LAe^Lt + (kAe^Lt)/m = 0

    So L + k/m = 0
    Hence L = -k/m

    so solution to homogenous problem is v = Ae^(-kt/m)

    An obvious particular solution is v = mg/k, so we have our general solution:

    v = mg/k + Ae^(-kt/m)

    If we seperate variables instead, we have:

    (mv')/(mg - kv) = 1

    Integrating both sides with respect to t gives us
    -(m/k)ln(mg - kv) = t + c
    (where c is some constant)

    Multiplying through by -k/m gives us:
    ln(mg - kv) = -kt/m + b
    (where b is some constant)

    Then taking exponentials:
    mg - kv = Be^(-kt/m)
    (Where B = e^b is some constant)

    Lastly, rearranging gives us:
    v = mg/k + Ae^(-kt/m)

    As expected :)
     
  15. Jul 20, 2012 #14
    Thanks a million.
    It's been a while since I've done this, and I seemed to be making a mistake at each stage...
     
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