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A particle moving under a conservative force

  1. Jun 25, 2014 #1
    1. The problem statement, all variables and given/known data

    From, Classical mechanics 5th edition, Tom W.B. Kibble, Frank H. Berkshire
    Chapter 2, problem 30

    A particle moving under a conservative force oscillates between x11 and x2. Show that the period of oscillation is

    τ = 2[itex]\int[/itex][itex]^{x_{2}}_{x^{1}}[/itex][itex]\sqrt{\frac{m}{2(V(x_{2})-V(x))}}[/itex]dx

    2. Relevant equations

    m[itex]\ddot{x}[/itex] + F(x) = 0

    F(x) = -[itex]\frac{d}{dx}[/itex]V(x)


    3. The attempt at a solution

    m[itex]\ddot{x}[/itex] + F(x) = 0

    → m[itex]\ddot{x}[/itex] -[itex]\frac{d}{dx}[/itex]V(x) = 0

    → [itex]\int[/itex][itex]^{x_{2}}_{x_{1}}[/itex]m[itex]\ddot{x}[/itex]dx = V(x2)-V(x1)

    Im not sure if i've started right and if I have I don't know how to go forward with the [itex]\ddot{x}[/itex]
     
  2. jcsd
  3. Jun 25, 2014 #2

    CAF123

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    Gold Member

    Rewrite $$m\ddot{x} + \frac{dV}{dx} = 0 \,\,\text{as}\,\,\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + V(x)\right) = 0$$ and carry on from there.
     
  4. Jun 27, 2014 #3
    When I try that I just end up with

    m[itex]\int[/itex][itex]\dot{x}[/itex][itex]\frac{d\dot{x}}{dx}[/itex]dx + V(x) = 0

    by parts on the integral just sends me in a circle?
     
  5. Jun 27, 2014 #4

    CAF123

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    Gold Member

    If $$\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 + V(x) \right) = 0\,\,\,\text{then}\,\,\,\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 + V(x) = \text{const}$$ Use what you know about the particle at the boundaries of its oscillations (i.e at ##x_1## and ##x_2##) to obtain the constant.

    Once you have this, you can separate variables to find T.
     
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