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Classical mechanics with a mass on a light elastic string

  1. Jul 21, 2011 #1
    Classical Mechanics Homework question

    Question - A light elastic string AB of natural length L and spring constant K, lies slack on a horisontal plane. A particle of mass m also at rest, is attached to end A of the string. The other end B is pulled along the plane with constant velocity V away from A in a straight line passing through A.

    Find the time that elapses after the string first becomes taut until it becomes slack again.

    Show that the particle will have traveled a total distance [itex]2L + V\pi({\frac{m}{k}})^{\frac{1}{2}}[/itex] by the time it catches up with end B.


    Thoughts - I think I have to start with Newton's 2nd law, I'm just not sure what the force is a function of. I think it could be F(x) where x is the distance traveled but it can also depend on the speed that the string is being pulled so it would be F(V)?

    If the force depends on the velocity of the end of te string then N's 2nd law would be

    F(V) = ma = -Kx = m(dv/dt)

    where v is the velocity of the particle at time t and V is the velocity of the end of the string. If I equate the two terms on the right I get...

    -Kx = m(dv/dt) and if I rearrange this and integrate it I get...

    t*(K/m)^0.5 = ln(x) + c

    But I don't know where to go from here. I need to somehow eliminate x or t so that I only have 1 unknown. Or maybe I have to split the problem up into sections from the start and deal with separate parts?
     
    Last edited: Jul 21, 2011
  2. jcsd
  3. Jul 21, 2011 #2
    When the string is on tension, [itex]m \ddot{x} = k (V t - x)[/itex], where [itex]x[/itex] is the position of the block. Rewrite equation of motion as [itex]m \ddot{x} + k x = k V t[/itex]. This is an inhomogeneous differential equation. So you look for a homogeneous solution and a particular solution: [itex]x(t) = x_h (t) + x_p(t)[/itex]. [itex]x_h(t)[/itex] is just the usual harmonic motion with two unknown coefficient. [itex]x_p(t)[/itex] is also very easy to find.

    Once you write down [itex]x(t)[/itex], use the initial conditions to solve for the two unknown coefficient. The string loses tension, when [itex]Vt -x < 0[/itex]. This condition allows one to solve for t.
     
  4. Jul 22, 2011 #3
    So I did what you said and used m(dv/dt) = K (Vt-x) to generate the differential equation

    m(d^2x/dt^2) + Kx = KVt

    The homogeneous solution that I got is

    xh(t) = Asin(t(K/m)^0.5) + Bcos(t(K/m)^0.5)

    I found the particular solution by guessing the solution for x to be of the form

    x = bt^2 + ct + e

    resulting in a particular solution of

    xp(t) = Vt

    So the general solution is

    x(t) = Asin(t(K/m)^0.5) + Bcos(t(K/m)^0.5) + Vt

    I'm not sure about the initial conditions, the only one I can see is that when t=0 the velocity of the particle is also v=0 so B must be 0.

    so the general solution becomes

    x(t) = Asin(t(K/m)^0.5) + Vt

    That is as far as I can get without figuring out a second boundary condition. Does this look like it's on the right track?

    Thanks very much for all your help!
     
  5. Jul 22, 2011 #4
    The particle started at rest, so [itex]x^{\prime}(0) = 0[/itex].
     
  6. Jul 22, 2011 #5
    Oh of course thank you!

    So A is

    [itex]A = -V\sqrt{\frac{m}{K}}[/itex]

    So the general solution becomes

    [itex]x(t) = Vt - V\sqrt{\frac{m}{K}} sin (\sqrt{\frac{m}{K}} t)[/itex]

    So after I've figured out the time taken for the string to become slack, I figured that the total distance traveled until the mass gets to point B will be L+Vt=x which rearranges to get [itex]t = \frac{x-l}{V}[/itex]

    is this along the right lines for the second part?

    Thanks!
     
  7. Jul 22, 2011 #6
    You inverted the k and the m in the sine. It's convenient to write [itex]\omega=\sqrt{\frac{k}{m}}[/itex], and write:[itex]x(t)=V t - \frac{V}{\omega} \sin(\omega t)[/itex]

    Do you think the answer should depends on the length of the string?
     
  8. Jul 23, 2011 #7
    Ok, so our answer for [itex]x(t) = Vt - \frac{V}{\omega } sin (wt)[/itex].

    So assuming that at the instant the string looses tension the following is true

    [itex]Vt-x=0[/itex]

    we can replace Vt with x in the equation for x(t) and after some cancelations figure out that

    [itex]sin(wt) = 0[/itex]

    So [itex]wt = n\pi[/itex]

    Assuming that when the string loses tension for the first time n=1 then

    [itex]t = \frac{\pi }{\omega } = \pi\sqrt{\frac{m}{K}}[/itex]

    To find the distance traveled I assume that all we need to do is substitute in t into the equation above. In which case it does depend on L because the mass has to travel the length of the string to catch up with point B?
     
  9. Jul 24, 2011 #8
    So I've been working on this problem a bit more... I've been trying to find the total distance adding the distance taken for the string to become slack x(t), when
    [itex]t = \pi\sqrt{\frac{K}{m}}[/itex] to the distance travelled by the point B [itex]x = Vt_{2}[/itex] and then adding the length of the string to that giving me....

    [itex]x = V\pi\sqrt{\frac{K}{m}} + L + Vt_2[/itex]

    From the answer that we are given:

    [itex] x = V\pi\sqrt{\frac{K}{m}} + 2L [/itex]

    it's clear that [itex]Vt_2[/itex] should equal L, but I'm unsure how to show it.

    Thanks a million for your help so far!
     
  10. Jul 25, 2011 #9
    [itex]x[/itex] is the displacement of particle B. Why is there a need to add [itex]L[/itex] to it?

    You are right that string loses tension at [itex]t_1 = \frac{\pi}{\omega}[/itex]. So the distance traveled is [itex]x(t_1) = V t_1 = \frac{\pi V}{\omega}[/itex].
     
  11. Jul 25, 2011 #10
    I think we need to add L because the question requires that we find the displacement of the mass when it catches up with the end of the string, which is L distance away from the other end of the string where the mass is attached. Since we've got the time taken for the string to slacken again, the corresponding distance can be calculated, and this distance added to L will give us the displacement at point B. I hope my logic is correct.
     
  12. Jul 25, 2011 #11
    Oops, didn't read the question right. Then you just need to find the speed and distance when the string lost tension and just add the catch up time.
     
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