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Finding The Position of A Mass As A Function of Time

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a mass m on the end of a spring of force constant k and constrained
    to move along the horizontal x axis. If we place the origin at the spring’s equilibrium
    position, the potential energy is U = 1/2kx^2 At time t = 0 the mass is sitting at the
    origin and is given a sudden kick to the right so that it moves out to a maximum
    displacement x_max = A and then continues to oscillate about the origin.

    (a) Write down the equation for conservation of energy and solve it to give the
    mass’s velocity [itex]\dot{x}[/itex] in terms of the position x and the total energy E.

    (b) Show that E = 1/2kA^2 and use this to eliminate E from your expression for [itex]\dot{x}[/itex]. Find the time for the mass to move from the origin out to a position x.

    (c) Solve the result of part (b) to give x as a function of t and show that the mass
    executes simple harmonic motion with period [itex]2 \pi \sqrt{m/k}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    All right. I am currently trying to evaluate the integral, which was instructed of me to find in part (b).

    Before I proceed any further, here is a little preliminary work:

    [itex]\dot{x}(x) = kA^2 - \frac{k}{m} x^2[/itex]

    And I used this fact to find the integral: [itex] \dot{x} = \frac{dx}{dt} [/itex] which becomes by separation of variables [itex]dt = \frac{dx}{\dot{x}}[/itex]

    Substituting in, and integrating both sides, this is the integral thus far:

    [itex]\int_0^t dt' = \int_0 ^x \frac{dx}{\sqrt{kA^2 - \frac{k}{m}x^2 }} [/itex]

    [itex]kA^2 = (k^{1/2}A)^2[/itex] and [itex]\frac{k}{m}x^2= (\omega x)^2[/itex], where

    [itex]\omega = \sqrt{\frac{k}{m}}[/itex]. This leads to trigonometric substitution:

    [itex]sin \theta = \frac{(k^{1/2}A)^2}{\sqrt{kA^2 -(\omega x)^2}} [/itex]

    The problem I have is finding how dx is equal to dθ. I tried using the fact that [itex]\cos \theta = \frac{( \omega x)^2}{(k^{1/k} A)^2}[/itex], then taking the arccos of this, and then taking the derivative of theta with respect to x. This didn't really seem helpful.

    How can I get rid of dx, so that I might have an integral involving only theta?
     
  2. jcsd
  3. Oct 15, 2013 #2

    Simon Bridge

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    (b) Show that E = 1/2kA^2 and use this to eliminate E from your expression for ##\dot{x}. Find the time for the mass to move from the origin out to a position x.

    Starting with:
    $$\frac{dx}{dt}=kA^2+\omega^2 x^2 : \omega^2=\frac{k}{m}$$ I get $$dt = \frac{dx}{kA^2+\omega x^2}$$
     
  4. Oct 15, 2013 #3
    Are you certain that that is correct? Here is my work from the beginning:

    [itex]E = T(\dot{x}) U(x) ~ \rightarrow E =\frac{1}{2}m \dot{x} + \frac{1}{2}kx^2[/itex]

    When [itex] x_max = A [/itex], then [itex]\dot{x} = 0[/itex]; thus, [itex]E = 0 + \frac{1}{2}kA^2[/itex]

    Taking the equation and solving for [itex]\dot{x}[/itex]:

    [itex] \frac{1}{2}kA^2 =\frac{1}{2}m \dot{x} + \frac{1}{2}kx^2 [/itex]

    [itex]\dot{x}^2 = kA^2 - \omega x^2 [/itex]

    [itex]\dot{x} = \sqrt{kA^2 - \omega x^2}[/itex]

    And then I applied the fact that [itex]\dot{x} = \frac{dx}{dt}[/itex], which, by separation of variables, can be written as [itex]dt = \frac{dx}{\dot{x}}[/itex]

    I am not certain as to how you were able to do this without introducing a square root.
     
  5. Oct 15, 2013 #4

    Simon Bridge

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    From post #1:
    ... that's what I started from. I take it that's a typo?

    ... makes more sense :) ... shouldn't that be ##\omega^2## though?

    $$\frac{dx}{dt}= \sqrt{kA^2 - \omega^2 x^2}$$

    Try substitution:
    $$x=\sqrt{\frac{kA^2}{\omega^2}}\sin\theta\\

    dx = \sqrt{\frac{kA^2}{\omega^2}}\cos\theta\; d\theta$$
     
  6. Oct 16, 2013 #5
    Yes, sorry for the typos. How where you able to get [itex]x = \sqrt{\frac{kA^2}{\omega^2}} \sin \theta[/itex]?
     
  7. Oct 16, 2013 #6

    Simon Bridge

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    Did you try the substitution?

    To understand the choice of substitution, consider:
    If the stuff under the surd was ##1-x^2## ...... what would be the substitution?
     
  8. Oct 17, 2013 #7
    Oh, I just realized that I drew my triangle incorrectly. I wrote the square of the lengths of the hypotenuse and one of the legs; that is, for the hypotenuse, I wrote[itex](k^{1/2} A)^2[/itex], when I should have written [itex]k^{1/2} A[/itex], and for the leg I wrote[itex](\omega A)^2[/itex], rather than [itex]\omega A[/itex].

    From the way I have my triangle drawn, [itex]\sec \theta = \frac{k^{1\2}A}{\sqrt{(k^{1/2}A)^2 - (\omega A)^2}} \implies \frac{\sec \theta}{k^{1/2} A} = \frac{1}{\sqrt{(k^{1/2}A)^2 - (\omega A)^2}}[/itex]

    For the differential of x, [itex]x = \frac{k^{1/2}A}{\omega} \sin \theta \implies dx = \frac{k^{1/2}A}{\omega} \cos \theta d \theta[/itex]

    Using these two expressions, I'll substitute them into the integral, for which I will get:

    [itex]t = \int \frac{\sec \theta}{k^{1/2} A} \frac{k^{1/2}A}{\omega} \cos \theta d \theta[/itex]

    Simplifying,

    [itex]t = \frac{1}{\omega} \int d \theta \implies t = \frac{\theta}{\omega}[/itex]

    Going back to the triangle, we find that theta is [itex]\theta = \sin^{-1} (\frac{\omega x}{k^{1/2}A})[/itex]

    Substituting this in, simplifying, and solving for x, I get

    [itex]x = \sqrt{m}A \sin (\omega t)[/itex].

    How do I get rid of that square root of the mass?
     
  9. Oct 17, 2013 #8

    Simon Bridge

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    Your dimensions don't stack up - check your algebra. (Dimensional analysis o the DE you constructed.)
    Note: I can no longer see what you have done.
     
    Last edited: Oct 17, 2013
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