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Homework Help: Motion of a point is equal to dynamic harmonic oscilation

  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Point with mass is moving along the positive direction of x axis, its velocity is described by (A-Bx^2)^(1/2). Show that its equation of motion describes dynamic harmonic oscillation and find period (T) of this oscillation.

    2. Relevant equations
    A and B is known constants

    3. The attempt at a solution

    Hi, guys!
    My first instinct is to try to get something from F=m*a=-k*x.
    Further more m*(dv/dt)=-k*x, but immediately i see that its wrong because v=v(x) not v(t).

    So ,please, could someone give me some advice.
    Thank you!
  2. jcsd
  3. Apr 2, 2016 #2
    Hint short:

    If energy is conserved for a SHO along the lines of
    $$ stuff1 \times v^2 + stuff2 \times x^2 = constant$$

    Hint, long:

    $$ F = m {dv \over dt} = m {dv \over dx} {dx \over dt} = m v {dv \over dx} = -kx$$

    Integrate dx both sides, what do you get?
  4. Apr 2, 2016 #3
    Thank you for your hints, I had forgotten connections between derivatives.

    If I take the equation mv (dv/dx)=-kx
    I obtain m*v*dv=-k*x*dx
    By integrating m *(v^2)/2+C1=-k*(x^2)/2+C2
    I suppose I could now substitute v with given equation but if I do so, the equation becomes very messy and I do not think that it could be reduced to something useful.

    What do you suggest?
  5. Apr 2, 2016 #4
    Does your equation with the squares look like a conservation equation to you? :D (protip, you could haved used conservation of energy, but instead you derived it like a beast!)

    Anyways, if you solve for v from x from that equation, what do you get? What is the equivalent of
    $$\omega = \sqrt{k \over m}$$ when you try to match your equation to the given equation $$v = (A-Bx^2)^{1 \over 2}$$
    Last edited: Apr 2, 2016
  6. Apr 2, 2016 #5


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    There is a simpler alternative to showing it satisfies the usual SHM ODE: show it satisfies the solution to an SHM ODE.
  7. Apr 3, 2016 #6
    If I solve for v, I obtain [itex]v = ({\frac{\ 2(C1-C2)}{m}} - {\frac{\ k x^2}{m}})^{(1/2)} [/itex]
    which looks similar to given equation and if I assume that
    [itex]A≡{\frac{\ 2(C1-C2)}{m}} [/itex]
    it can be rewritten
    [itex]v = (A - ω^2x^2)^{(1/2)} [/itex]
    which means that
    [itex] B≡ ω^2 [/itex]
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