Oscillating springs in relativistic time frame

  • #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
 

Answers and Replies

  • #2
haruspex
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<This is where I am confused. How to I convert this to second observer's frame by finding the inertia and stretch?>
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.
 
  • #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?
 
  • #4
haruspex
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I think I get it. So Delta x would shrink due to the length dilation
contraction. Dilation would be lengthening, as in time dilation.
k, which represents the inertia?, would expand
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.
 
  • #5
Would it be:

##\vartriangle x’ = \frac{\vartriangle x}{\gamma}##
##m’ = \gamma \vartriangle x##
##k = \frac{m’ a}{\vartriangle x’}##
 
  • #6
PeroK
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Would it be:

##\vartriangle x’ = \frac{\vartriangle x}{\gamma}##
##m’ = \gamma \vartriangle x##
##k = \frac{m’ a}{\vartriangle x’}##
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:
  • #7
haruspex
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Aren't you supposed to use the transformations of forces
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##.
##m’ = \gamma \vartriangle x##
Shouldn't there be an m on the right, not an x?
 
  • #8
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?
Oh yes, that was a mistake. An m on the right.
 
  • #9
Can anybody shed some light on finding the inertia and stretch in the second frame.
 
  • #10
haruspex
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Can anybody shed some light on finding the inertia and stretch in the second frame.
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?
 
  • #11
I have a lot of doubts about physics. But I didn't think you could transform non uniform acceleration..
 

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