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Simple harmonic motion

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    a particle of mass m moves in one dimension with an applied force Fx=-Fosinh([tex]\alpha[/tex]x) where [tex]\alpha[/tex] and Fo are constants. find the frequency of small oscillations about the equilibrium position. next, find the potential energy function and sketch it.


    2. Relevant equations



    3. The attempt at a solution

    Guess: x(t)=Asinh(wt)
    dx/dt=Awcosh(wt)
    d2x/dt2=Aw2sinh(wt)

    mAw2sinh(wt)=-Fosinh([tex]\alpha[/tex]x)
    w=[tex]\alpha[/tex]
    Fo=-Aw2m
    A=-Fo/mw2

    I have a feeling this is not the right way to go about this...Is it correct? If not, why? and how do i go about finding the frequency?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 23, 2009 #2

    Doc Al

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    Staff: Mentor

    For SHM, how must the force depend on displacement from equilibrium? Hint: Do a series expansion of F(x) to see how it behaves for small x.
     
  4. Sep 23, 2009 #3
    ok so after the series expansion F(x)=-F([tex]\alpha[/tex]x)+(([tex]\alpha[/tex]x)3)/3!+...)

    the force depends on the displacement directly. but how does this help me get the frequency?
     
  5. Sep 23, 2009 #4

    Doc Al

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    Staff: Mentor

    Compare that force function to that of a spring for small x. (Assuming you know how to find the frequency of an oscillating spring. If not, look it up.)
     
  6. Sep 23, 2009 #5
    the force for a spring is just F=-kx. so does F[tex]\alpha[/tex] =k?
     
  7. Sep 23, 2009 #6

    Doc Al

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    Staff: Mentor

    Yes. k = F0α
     
  8. Sep 23, 2009 #7
    ok, but to get the frequency i need to solve the differential equation that i attempted above. how do i go about that?
     
  9. Sep 24, 2009 #8

    Doc Al

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    Staff: Mentor

    The differential equation you need to solve is just the standard one for SHM:
    m d2x/dt2 = -kx
     
  10. Sep 24, 2009 #9
    The key point here, that I'm afraid you've missed, is that for sufficiently small values of [tex]x[/tex], the function is linear in [tex]x[/tex]

    From here, you can say that the force is of the form:
    [tex]F(x)=-(\alpha F_0)\cdot x[/tex]

    I think you've fallen into the pit-fall of this question. You're used to SHM being described by the regular trigonometric functions, [tex]sin(x), cos(x)[/tex], but here you were confronted with the hyperbolic sine function and were thrown off.

    Before we try and guess the solution to the differential equation, let's first find it.
    Looking at NSL for this mass, we can find the following differential equation:

    [tex]ma=-(\alpha F_0)\cdot x[/tex]

    [tex]m\ddot x=-(\alpha F_0)\cdot x[/tex]

    [tex]\ddot x=-(\frac{\alpha F_0}{m})\cdot x[/tex]

    Solving this differential equation is fairly simple. Just like the equation [tex]5=10-x[/tex] asks "which number, when subtracted from 10 gives 5?"
    This differential equation asks, "which function, when differentiated twice with respect to time, gives itself times a negative constant?"

    [tex]\omega^2\equiv \frac{\alpha F_0}{m}[/tex]

    The differential equation then turns to:

    [tex]\ddot x=-\omega^2 x[/tex]

    Now we'll guess!

    [tex]x(t)=?[/tex]

    But how do we know this is really a solution? How do we know if we guessed right?
    Just like how you would double-check your answer for [tex]5=10-x[/tex] by plugging in [tex]x=5[/tex] and seeing that you get a true statement, we'll do the same here.

    Differentiating our solution once with respect to time gives us:

    [tex]\frac{dx(t)}{dt}=?=\dot x(t)[/tex]

    Differentiating once more:

    [tex]\frac{d\dot x(t)}{dt}=?=\ddot x(t)[/tex]

    And then look to see if it checks out or not.

    Once you have [tex]x(t)[/tex], you can easily find its period. Before you get caught up in the math, remember that you don't care about the amplitude or phase shift of the periodic function, but only about its period.
     
    Last edited: Sep 24, 2009
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