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Oscillating springs in relativistic time frame

  1. Jan 26, 2015 #1
    1. The problem statement, equations, and work done:

    [A]
    Two identical spring and mass oscillators are set in motion in perpendicular directions. The masses are each 4.0 kg and the spring constants are 196 N/m.

    [1] Calculate the angular frequency and the period of the oscillators.

    ##\omega = \sqrt{\frac{k}{m}} = .143 rad/sec##

    ##T = \frac{2\pi}{\omega} = 43.938 s##

    [2] The oscillators are damped such that the amplitude decreases by 20% in 140 seconds. Determine the coefficient in the damping force (b in F = -bv) and the time constant (τ) for the energy decay ( E= Eo e-t/τ)

    ##A_{\% decrease\hspace{1 mm}per\hspace{1 mm}cycle} = \frac{A_{total\hspace{1 mm}decrease} \hspace{1 mm}X\hspace{1 mm}time}{T} = \frac{.20 X 140 s}{43.938 s} = 63.726 \%##

    ##A = A_0 - A_{\%decrease\hspace{1 mm}per\hspace{1 mm}cycle} = A_0 e^{\frac{-bT}{2m}} →##

    ##1 - .63726 = 1\hspace{1 mm}X\hspace{1 mm}e^{\frac{-b(43.938 s)}{2(4 kg)}} →##

    ##b = .184636 kg/s##

    ##\tau = \frac{m}{b} = \frac{.18636 kg/s}{4 kg} = 21.66 s##

    [3] A second observer passes at 0.6c, travelling in the direction one of the oscillators moves. Write the force law for the spring in the frame where one end is stationary (as in parts 1 and 2). Then transform this to the second observer's frame by finding the inertia and stretch in the second frame. Do this twice, once for each oscillator.

    ##F_{stationary\hspace{1 mm}reference} = ma = -k\Delta x##

    <This is where I am confused. How to I convert this to second observer's frame by finding the inertia and stretch?>

    [4] Using your force law in [3], derive the expressions for the angular frequency and the period of the oscillators in the second frame.

    [5] Now compare the periods you calculated to the result of a time dilation calculation
     
  2. jcsd
  3. Jan 26, 2015 #2

    haruspex

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    I'm no expert on relativity, but I would guess the inertia would be increased according to the relativistic mass formula, and stretch reduced according to the distance contraction.
     
  4. Jan 27, 2015 #3
    I think I get it. So Delta x would shrink due to the length dilation and k, which represents the inertia?, would expand; but what equation represents this k expansion?
     
  5. Jan 27, 2015 #4

    haruspex

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    contraction. Dilation would be lengthening, as in time dilation.
    No, inertia means mass. m increases according to the usual equation. Use the spring equation to find how these x and m changes affect k.
     
  6. Jan 28, 2015 #5
    Would it be:

    ##\vartriangle x’ = \frac{\vartriangle x}{\gamma}##
    ##m’ = \gamma \vartriangle x##
    ##k = \frac{m’ a}{\vartriangle x’}##
     
  7. Jan 28, 2015 #6

    PeroK

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    Aren't you supposed to use the transformations of forces for different IRF's?

    ##F'_x = ...##

    ##F'_y = ...##

    As well as the length contraction formula.
     
    Last edited: Jan 28, 2015
  8. Jan 28, 2015 #7

    haruspex

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    Probably. It says 'the force law' which suggests it's the ##F=k\Delta x## equation that's to be transformed. I was looking at ##ma = k\Delta x##.
    Shouldn't there be an m on the right, not an x?
     
  9. Jan 28, 2015 #8
    Oh yes, that was a mistake. An m on the right.
     
  10. Jan 30, 2015 #9
    Can anybody shed some light on finding the inertia and stretch in the second frame.
     
  11. Jan 30, 2015 #10

    haruspex

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    You wrote ##k = \frac{m’ a}{\vartriangle x’}## (I guess you meant ##k' = \frac{m’ a}{\vartriangle x’}##), but you also need to transform a.
    You already posted an equation for transforming the extension. Do you have doubts about it?
     
  12. Jan 30, 2015 #11
    I have a lot of doubts about physics. But I didn't think you could transform non uniform acceleration..
     
  13. Jan 30, 2015 #12

    haruspex

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