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Homework Help: Work done by a harmonic oscillator

  1. Oct 13, 2007 #1
    In the case of an undamped oscillator, the work done by the system is written as ( assume initial position is 0 ) :

    [tex] W = - \Delta U = - K \frac{x^2}{2} [/tex]

    But to verify this , we must assume that the force acting on the oscillator is constant , which is not true as F = f(x) according to hook's law.

    To find an expression for the work done by the system I start with :

    [tex] dW = Cos(F,x) \ d(Fx) = d(Fx) = -K d(x^2) [/tex]

    then it follows that

    [tex] W = - K x^2 [/tex]

    Ofcourse this eq must be wrong , but I wonder why. Why should the force of the spring be constant ?
  2. jcsd
  3. Oct 13, 2007 #2

    Doc Al

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    Staff: Mentor

    The force of the spring isn't constant:

    [tex]dW = Fdx = kx dx[/tex]

    Integrate that and you'll get the correct answer.
  4. Oct 14, 2007 #3
    When you write dW=Fdx you already assume the force is constant . Otherwise , why would F be outside the differential ? We should write dW = d(Fx) keeping F inside the differential as it is not a constant. dW = Fdx is correct only when F = constant. True ?

    Notice , when calculating the work done by an electron in a battery , we write dW = d(Fx) = d(qEx) = Vdq assuming the potential difference V = const while q is a variable. ( In this case F is not constant ). Had we writen directly dW = Fdx , we would not obtain the correct expression. It is not necessary that dW = Fdx always holds true, it depends on the situation.

    I hope you got my point.
  5. Oct 14, 2007 #4

    Doc Al

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    Staff: Mentor

    No, not true. The differential element of work is dW = Fdx, not dW = d(Fx). (What does dF even mean? A small change in F?) Work is Force times distance; we need to add up (integrate) all the contributions of Force*dx over the variable x. Whether F is constant or not is a different story: if F is constant, then you can take it out of the integral. In this case F is a function of x, so dW = Fdx = kx dx:

    [tex]W = \int F(x) dx \neq F(x) \int dx[/tex]

    [tex]W = \int F(x) dx = \int k x dx = 1/2 kx^2[/tex]

    Again, incorrect. dW = F dx is always true--it's the basic definition of work. In the case where force is constant (and in the same direction as displacement):

    [tex]W = \int F dx = F \int dx = Fx[/tex]
  6. Oct 14, 2007 #5
    Thanks , I think it makes much more sense now.
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