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Simple Harmonic Oscillator on a smooth surface

  1. Jul 10, 2014 #1
    I feel I understand what happens, and how to solve the equation of motion x(t) for a mass attached to a spring and released from rest horizontally on a smooth surface. We typically end up with

    [itex] x(t) = x_0 cos(ωt) [/itex]

    as the solution, with [itex] x_0 [/itex]as the amplitude of the oscillation.

    But I've been wondering what happens if the mass was released again from position [itex] x_0 [/itex]but with velocity [itex] v_0 [/itex] instead of being released from rest. Is x(t) the same or different to when the mass was released from rest?

    Thanks in advance.
  2. jcsd
  3. Jul 10, 2014 #2


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    Gold Member

    Have you tried doing this problem yourself using mathematics? Do you know where the original ##x(t)## comes from?

    If you've solved the differential equation, then the differential equation doesn't change, but the initial conditions do change. Released from rest means ##v(0)=\left.dx/dt\right|_{t=0}=0## and ##x(0)=x_0##. Now you have a different initial condition ##v(0)=v_0##. Mathematically this shouldn't be too hard to implement.

    Thinking about the problem physically, releasing the block with speed means giving it some initial kinetic energy. One would expect then that the amplitude of oscillations would be higher than if the block were released without speed since the turning points (where we define our amplitude of oscillations) are points where the kinetic energy has all been converted into potential energy. If you start off with more total energy, you expect higher amplitude oscillations.

    Nothing else really special happens.
  4. Jul 10, 2014 #3
    I tried doing the problem myself. What I did was use the trial function

    [itex] x(t) = x_0 e(jωt)[/itex]

    I found the first and second derivatives and substituted this back into Newton's law for motion of the particle

    [itex] x''(t) + ω^2 x(t) = 0 [/itex]

    where [itex] ω^2 [/itex] is equal to [itex] k/m [/itex], k being the stiffness constant of the spring and m being the mass of the particle.

    Then the real part of this is the solution, and from Euler's relation we end up with

    [itex] x(t) = x_0 cos(ωt) [/itex]

    Is this the correct way of thinking?

    So now, from initial conditions would the position at [itex] t = 0 [/itex]be given by

    [itex] x = x_0 cos(ωt) + v_0 t [/itex] just by integrating with respect to t and then re-arranging?

    Here's my reasoning.

    At time [itex] t = 0 [/itex] the velocity [itex] x'(t) = v_0 [/itex]

    Usually we'd have [itex] x(t) = x_0 cos(ωt) [/itex]. But differentiating this gives

    [itex] x'(t) = -x_0 ω sin(ωt) [/itex].

    But at [itex] t = 0 [/itex] this gives us an initial velocity of 0. We want the initial velocity to be [itex] v_0 [/itex] so we instead our velocity function is

    [itex] x'(t) = -x_0 ω sin(ωt) + v_0 [/itex],

    then integrating with respect to t gives

    [itex] x(t) = x_0 cos(ωt) + v_0 t [/itex]. Is that correct? I omitted the arbitrary constant on the end.
  5. Jul 10, 2014 #4
    You've only found one solution to the DE, once you've found the other linearly independant solution you will have two integration constants to adjust for [itex]x_0[/itex] and [itex]v_0[/itex].
    After solving for those two parameters you will end up with [itex] x(t) = x_0 cos(ωt)+(v_0/ω)sin(ωt) [/itex]
    Last edited: Jul 10, 2014
  6. Jul 10, 2014 #5
    The first part, where I found [itex] x(t) = x_0 cos(ωt) [/itex] was when initial velocity was 0.

    I don't know how to find this other solution non-linearly for when initial velocity is [itex] v_0 [/itex]. But the more I think about it, the less it makes sense for the solution to be sinusoidal + linear, because that would imply that after a very long time the effect of the cosine would become tiny and the particle would be displaced by a very large amount in the positive direction. So now I'm convinced that the solution is entirely sinusoidal but I don't know how to even start finding this solution.
  7. Jul 10, 2014 #6
    You're simply missing the other solution, when you made your original ansatz you should have added a Ae-iωt term which is also a solution to the DE.
  8. Jul 10, 2014 #7


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    There are three common ways of writing the general solution to the differential equation for the simple harmonic oscillator. All have two arbitrary constants, and are equivalent to each other. You can find the arbitrary constants for any one of them in terms of the arbitrary constants of either of the others:

    $$x(t) = A \cos \omega t + B \sin \omega t\\
    x(t) = C e^{i \omega t} + D e^{-i \omega t}\\
    x(t) = x_0 \cos (\omega t + \phi_0)$$

    (In the last one, you can use sin instead of cos, but with a different value for ##\phi_0##.)
  9. Jul 10, 2014 #8
    Hang on a second. I just realised why I need to add both the solutions to the DE.

    If I have a 2nd order linear differential equation, and there is a solution to it, like g(t), and another solution to it, h(t), then Ag(t) and Bh(t) are both solutions, where A, B are constants, and Ag(t) + Bh(t) is another solution! So, Ag(t) + Bh(t) encapsulates ALL solutions! I need to work on my differential equations. Thanks everyone.

    EDIT: Just came back to it and I understand it now, and got both the solutions. Thanks all.
    Last edited: Jul 10, 2014
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