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Finding the period of small vertical oscillations

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    I need to find the period of small vertical oscillations about equilibrium position of a string whose motion can be described by the following equation:
    d2x/dt2 = (-g/h)*x

    Answer: 2π√(h/g)

    2. Relevant equations
    I know that the time period is given by the formula
    T = 2πω where ω = √({d2V/dx2} / m) where V is the potential

    3. The attempt at a solution
    I tried solving the differential equation but I got stuck by doing the following:
    Transform differential equation into (by taking dx/dt on both sides):
    d2x/dt2 dx/dt = (-g/h)*x dx/dt
    Denote dx/dt = m, then we can write:
    ∫m*dm/dt dt = ∫(-g/h)*x * dx/dt dt
    (m)2/2 = -gx2/2h
    which is:
    (dx/dt)2/2 = -gx2/2h
    which gives that dx/dt is equal to √(-gx2/h) which is impossible since there is a negative sign inside.

    Thank you in advance!
     
  2. jcsd
  3. Nov 5, 2016 #2

    haruspex

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    What about the constant of integration?
     
  4. Nov 5, 2016 #3
    The equation of SHM is a = - ω2 x
    d2x/dt2 Is the acceleration...can you see what is ω2
    To find out what is t ?

    T = 2Pi/ω ....not 2πω
     
  5. Nov 5, 2016 #4
    Ohhh, thank you so much! So ω is √(g/h) and then T = 2π/ω = 2π√(h/g)
    The thing is that we did not learn about this formula before and so I was wondering if there is a way to deduce T by solving the equation?
     
  6. Nov 5, 2016 #5
    The main thing you need to know is that acceleration is d2x/dt2
    SHM is motion where acceleration is proportional to displacement...ie acc = - k x ....( - sign because restoring force, and therefore acceleration, is directed towards the equilibrium position)
    In your case k = g/h
     
  7. Nov 5, 2016 #6

    haruspex

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    Yes, see post #2.
     
  8. Nov 7, 2016 #7
    Hi, thank you for your reply!
    I totally forgot about the constant of integration. However, I still cannot get the needed result by just simply solving the equation.
    So, with the constant of integration, I get:
    dx/dt=v=√(c-gx2/h)
    then I solve this differential equation again and get:
    T = (√h) (arcsin ((x√g)/√ch) (in this case constant of integration is zero as at time t=0 the displacement from equilibrium position is zero)
    But then I am not sure what I have to do with c.
     
  9. Nov 7, 2016 #8

    haruspex

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    That does not look right. There should be a g outside the arcsin.
     
  10. Nov 7, 2016 #9
    Ooops, I am really sorry - I made a slip. For the solution I got T = √(h/g) arcsin ({x√g}/{√ch}) but then I am not sure where does 2π in the solution come from :(
     
  11. Nov 7, 2016 #10

    haruspex

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    It might be more obvious if you turn it around to express x as a function of time. What is the smallest increment of time after which the function repeats?
     
  12. Nov 8, 2016 #11
    Ohhh, I see, thank you so much! It makes sense! :)
     
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