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Velocity function of a variable force

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle with a mass of 3kg is at rest at x = 3m, and then a force = (12 N/m)x is applied to it. What is the acceleration of the particle when it reaches x = 5m? Determine the v(t) for the particle.


    2. Relevant equations
    [tex]
    \sum F = ma
    [/tex]

    3. The attempt at a solution

    The first part was simple. I just solved for acceleration and got the same answer as the book, 20 m/s^2. However there was no answer for the second part so I want to know if I solved for v(t) properly.

    [tex]
    a = (4 \frac {1}{s^2}) x
    [/tex]

    [tex]
    a = \frac {d^2x} {dt^2}
    [/tex]

    [tex]
    \frac {d^2x} {dt^2} - (4 \frac {1}{s^2}) x = 0
    [/tex]

    Which is just a 2nd order differential equation, whose general solution is:
    [tex]
    x = C_1 e^(\lambda_1 t) + C_2 e^(\lambda_2 t)
    [/tex]

    Where lambda is the solution to the auxiliary equation,
    [tex]
    \lambda^2 - 4\lambda = 0
    [/tex]

    Therefore
    [tex]
    x = C_1 e^(2 t) + C_2 e^(-2 t)
    [/tex]

    and v(t) is
    [tex]
    v = 2 C_1 e^(2 t) - 2 C_2 e^(-2 t)
    [/tex]

    The problem stipulates that at x=3m, v=0, so when if I set t=0 at x=3m, then I can solve for C1 and C2
    [tex]
    3 = C_1 + C_2
    [/tex]

    [tex]
    0 = 2 C_1 - 2 C_2
    [/tex]

    [tex]
    C_1 = \frac {3}{2} and C_2 = \frac {-3}{2}
    [/tex]

    So v(t) becomes
    [tex]
    v = (3 \frac {m}{s}) ( e^(2 t) + e^(-2 t) )
    [/tex]

    Then I wanted to make it look nicer so I rewrote it as:
    [tex]
    v = (6 \frac {m}{s}) \cosh {(2 \frac {1}{s} t)}
    [/tex]

    Is this the proper solution to the question?
    If not could someone please explain to me how to get the correct one. Thank you for the help.
     
  2. jcsd
  3. Jul 6, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi Piamedes,

    Since the problem involves v and x, I would write

    [tex]
    \frac {d^2x} {dt^2}
    [/tex]

    in a form that does not have t in it:

    [tex]
    \frac {d^2x} {dt^2} =v \frac{dv}{dx}
    [/tex]

    Then you can integrate over v and x directly. (But you might also want to drop the (1/s^2) that is inside your equation.)
     
  4. Jul 6, 2008 #3
    I tried doing that at first, but since the problem required v as a function of time when I integrated again to solve for x as a function of t it came out to
    [tex]
    \ln [x + \sqrt{x^2 - 9}] = 2 t
    [/tex]

    So when I tried solving for x it I lost the +/- portion of the square. I went back and used the general solution because it seemed easier than trying to figure out a mistake in my page of algebra work.

    Does doing it different ways yield different answers?
     
  5. Jul 6, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

    I'm not sure I understand what happened, but different way should definitely give the same answer.

    Looking over your original post, I think your values for C1 and C2 are incorrect. Your second equation for them is:

    [tex]
    0=2 C_1 - 2 C_2
    [/tex]
    so they must be equal.
     
  6. Jul 7, 2008 #5
    oops. I solved it all the way through to the end and messed up some algebra.
    Its should be
    [tex]
    C_1 = C_2 = \frac {3}{2}
    [/tex]

    Which means that
    [tex]
    x = (\frac {3}{2} \frac {m}{s}) (e^(2 t) + e^(-2 t) ) = (\frac {3}{2} \frac {m}{s}) \cosh {(2t)}
    [/tex]

    [tex]
    x = (3 \frac{m}{s}) (e^(2 t) - e^(-2 t) ) = (3 \frac {m}{s}) \sinh {(2t)}
    [/tex]

    [tex]
    a = (6 \frac {m}{s}) (e^(2 t) + e^(-2 t) ) = (6 \frac {m}{s}) \cosh {(2t)}
    [/tex]

    And if I check from the original equation
    [tex]
    a = 4x
    [/tex]

    [tex]
    a = 4 ( (\frac {3}{2} \frac {m}{s}) \cosh {(2t)} ) = 4x
    [/tex]

    By showing that my solution fits the given parameters, is that proof that the answer is correct or am I still missing something.
    Thanks for all the help so far.
     
  7. Jul 8, 2008 #6

    alphysicist

    User Avatar
    Homework Helper

    You have a small error here. Notice that the exponential form gives x=3 at t=0, while the cosh from doesn't.
     
  8. Jul 8, 2008 #7
    Thanks for catching that error. I must have forgotten to factor out the 1/2 when I transformed it into cosh.

    It should be
    [tex]
    x = (3 \frac {m}{s}) \cosh {(2t)}
    [/tex]

    Thanks for the help.
     
  9. Jul 8, 2008 #8

    alphysicist

    User Avatar
    Homework Helper

    Sure, glad to help! (And the way you put all the details of your work in your posts definitely made it easy to read through.)
     
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