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    Okay, this is a nasty practise question I've been asked to do. I have tried diagrams, but I can't come up with a very mathematical way of showing what i need to. If someone can show me how, I will give them some reputation.

    Here goes:

    Qa) A particle has unit mass. It moves along the x-axis and is subject to a restoring force -n^2x. It is also subject to constant frictional resistance, with magnitude n^2a, but friction only acts when x is positive. Show that the differential equation for x(t) will be:

    a + n^2.x = -n^2a, when x > 0
    a + n^2.x = 0 , for x equal to or less than 0

    NB - particle is released from rest when x = -b < 0. Show it next comes to rest when x = (a^2 + b^2)^1/2 - a.

    I appreciate this problem is extremely complex and not easy, but any and all help is welcome, as I have no idea how to solve this or even begin really aside fromd drawing inane diagrams.

    Please help!
    Rob
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    (Original post by RobbieC)
    Okay, this is a nasty practise question I've been asked to do. I have tried diagrams, but I can't come up with a very mathematical way of showing what i need to. If someone can show me how, I will give them some reputation.

    Here goes:

    Qa) A particle has unit mass. It moves along the x-axis and is subject to a restoring force -n^2x. It is also subject to constant frictional resistance, with magnitude n^2a, but friction only acts when x is positive. Show that the differential equation for x(t) will be:

    a + n^2.x = -n^2a, when x > 0
    a + n^2.x = 0 , for x equal to or less than 0

    NB - particle is released from rest when x = -b < 0. Show it next comes to rest when x = (a^2 + b^2)^1/2 - a.

    I appreciate this problem is extremely complex and not easy, but any and all help is welcome, as I have no idea how to solve this or even begin really aside fromd drawing inane diagrams.

    Please help!
    Rob
    Looks like damped harmonic motion.
    Haven't done it yet, sorry! It's M4 stuff I think.
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    It is hideous, and I have no idea what to do... That's fair enough, its a uni standard question, though Id argue not for a first year who is meant to only do number cruncher maths. Ugh.

    But anyone who fancies giving it a go, there is reppage available.
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    What you've written aren't differential equations - there are no derivatives in there. Do you want to check that?
    Surely here you'll be applying F=m\ddot{x} twice, once for the situation where >0 and once where x<0?
    By the way, did you have any luck with the problems in the other topic you made?
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    force = mass * acceleration
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    If you ignore the friction, the particle is undergoing SHM so you have

    \ddot{x} = -n^{2}x

    Nice and simple and the behaviour of the particle when x<0. Now bring in the frictional force when x>0. The friction is of magnitude n^{2}\ddot{x}. But since it is friction it is negative, so you'll get

    \ddot{x} = -n^{2}x - n^{2}\ddot{x}

    Rearrange to

    \ddot{x} + n^{2}x = -n^{2}a

    Initial conditions are x(0) = -b&lt;0 and \dot{x}(0)=0

    Since -b<0, use the SHM equation \ddot{x} = -n^{2}x

    You should know from your work on SHM that if an SHM system has an amplitude of b and a frequency "n" than at the centre x=0, it moves with a speed bn.

    Therefore, once the particle is released, it accelerates towards x=0, and when it gets to x=0 it is moving with speed \dot{x}=bn.

    Now consider the governing equation for x>0

    \ddot{x} + n^{2}x = -n^{2}a

    This is SHM but with an inhomogeneous term. The general solution is

    x = A \sin (nt) + B \cos (nt) - a giving

    \dot{x} = An \cos (nt) - Bn \sin (nt)

    Put in initial conditions, which are at t=0 x=0 and \dot{x}=bn gives

    x = b\sin(nt) + a\cos(nt) - a
    and so
    \dot{x} = bn\cos(nt) - an\sin(nt)

    \dot{x} = 0 \quad \Rightarrow \quad bn\cos(nt) - an\sin(nt) = 0 or b\cos(nt) = a\sin(nt)

    With a bit of thought (or drawing of triangles and thinking about trig) you'll see that this gives you the answer you require,

    x = \sqrt{a^{2}+b^{2}}-a
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    That is very helpful AlphaN. Thank you very much, ill think it all through later when Im back from badminton, but cheers. Rep you as soon as my rep power returns.

    And Gaz, I am yet to find out whether or not I completed the problems correctly, as hand in date has gone past. I will actually find out tomorrow. In the end I found something in my notes id missed, which helped with a couple of questions, the others I am just hoping my effort was sufficient to gain marks.
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    Hmm cant do the final stage AN. I just cant get rid of the trig and still have the answer you speak of.

    What do you mean, draw triangles??!
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    bump, in case you happen to read back over this
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    Argh!! I can't get the trig out of the answer to get what I want
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    ... because \tan(nt) = b/a, so \sin(nt) = b/\sqrt{a^2+b^2} and \cos(nt) = a/\sqrt{a^2+b^2}
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    That doesnt show what im supposed to though. Thought at least there's a square root there...
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    Consider what Shiny says :
    (Original post by shiny)
    ... because \tan(nt) = b/a, so \sin(nt) = b/\sqrt{a^2+b^2} and \cos(nt) = a/\sqrt{a^2+b^2}
    and my general solution

    x = b\sin(nt)+a\cos(nt) - a

    Put those values into the general solution gives

    x = \frac{b^{2}}{\sqrt{a^{2}+b^{2}}} + \frac{a^{2}}{\sqrt{a^{2}+b^{2}}} - a = \frac{b^{2}+a^{2}}{\sqrt{a^{2}+b  ^{2}}} - a = \sqrt{a^{2}+b^{2}} - a

    Therefore it does give the answer you want.
 
 
 
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