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How to derive the equations of oscillation

  1. Apr 2, 2013 #1
    I am new to this site.
    I have a problem with the derivations of second order equations for SHM.
    F= -kx
    m(second time derivative of x)+k(first time derivative of x)=0
    As my text book says above equation implies that x(t)=Acos(ωt+∅)
    But I can't understand why. From where did they get those ω,∅ and cosine function.
    Please help
  2. jcsd
  3. Apr 2, 2013 #2
    You derived the second-order ODE right: [itex]m\ddot{x} + kx = 0[/itex]. This yields the equation [itex]\ddot{x}= - \left ( \frac{k}{m} \right ) x[/itex]. Only functions whose second derivative have the same form as the original function are valid solutions. Exponentials and trigonometric functions fit this description (Euler's identity shows how they are related). Thus, [itex]x(t) = A \cos{(ωt-θ)}[/itex] is a valid solution. Substitute it into the ODE and see for yourself.
  4. Apr 2, 2013 #3
    I didn't see the other part to your post. The ω and θ just come from the general form of a sinusoidal function. You can solve for them by substituting x(t) into your ODE and solving for them. [itex]ω = \sqrt{k/m}[/itex] and θ is the phase angle, which is a way of accounting for the fact that SHM might not start from a rest position.
  5. Apr 2, 2013 #4


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    Staff: Mentor

    This is an example of a linear differential equation. There are standard methods for solving them, which you can find e.g. on Wikipedia, which even gives the simple harmonic oscillator as an example:


    Or you can look in any introductory textbook on differential equations, which will have more details about the method.
  6. Apr 3, 2013 #5


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    This is the annoying thing about differential equations. :biggrin: You often have to start, knowing what sort of answer you are likely to get. In this case, you need the answer to be in the form of a function for which the second time derivative is linearly related to that function. We know that differentiating Sin(x) twice gives you -Sin(x) so that a Sin function can fit as a solution (and so can a Cos function) Because of the integration involved in solving the equation, there are other constants that come into the solution and their actual values will depend upon the 'initial conditions.

    But Maths is always a bit like that, isn't it? Why multiply both sides by x? Why take Logs? Why subtract those two simultaneous equations? I'm sure you've already been there and that you already know some of the tricks.
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