# Oscillating springs in relativistic time frame

• Samuelriesterer
In summary: This should help.In summary, the problem statement, equations, and work done are: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. The oscillators are damped such that the amplitude decreases by 20% in 140 seconds. The coefficient in the damping force (b in F = -bv) and the time constant (τ) for the energy decay ( E= Eo e-t/τ) are 63.726% and 21.66 s, respectively. A second observer passes at 0.6c, traveling in the direction one of the oscillators moves.

#### Samuelriesterer

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, traveling 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

Samuelriesterer said:
<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.

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?

Samuelriesterer said:
I think I get it. So Delta x would shrink due to the length dilation
contraction. Dilation would be lengthening, as in time dilation.
Samuelriesterer said:
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.

Would it be:

##\vartriangle x’ = \frac{\vartriangle x}{\gamma}##
##m’ = \gamma \vartriangle x##
##k = \frac{m’ a}{\vartriangle x’}##

Samuelriesterer said:
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:
PeroK said:
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##.
Samuelriesterer said:
##m’ = \gamma \vartriangle x##
Shouldn't there be an m on the right, not an x?

haruspex said:
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.

Can anybody shed some light on finding the inertia and stretch in the second frame.

Samuelriesterer said:
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?

I have a lot of doubts about physics. But I didn't think you could transform non uniform acceleration..

## 1. What is the concept of oscillating springs in a relativistic time frame?

The concept of oscillating springs in a relativistic time frame involves the study of how the behavior of a spring changes when it is subjected to high speeds or strong gravitational fields. This is based on the principles of special and general relativity, which show that time and space are not constant and can be affected by factors like velocity and gravity.

## 2. How does the behavior of an oscillating spring differ in a relativistic time frame?

In a relativistic time frame, the behavior of an oscillating spring can be significantly different from its behavior in a non-relativistic frame. For example, the frequency, amplitude, and energy of the oscillations may change due to time dilation and length contraction. The spring may also experience a change in its rest length and stiffness due to the effects of gravity.

## 3. What factors affect the behavior of oscillating springs in a relativistic time frame?

The behavior of oscillating springs in a relativistic time frame is affected by several factors, including the speed of the spring, the strength of the gravitational field it is subjected to, and the direction of its motion. Other factors like the mass and material of the spring may also play a role in determining its behavior.

## 4. How can the study of oscillating springs in a relativistic time frame be useful?

The study of oscillating springs in a relativistic time frame can have various practical applications. For example, it can help in the design and development of accurate and precise clocks that are unaffected by changes in velocity or gravity. It can also aid in understanding the behavior of materials and structures in extreme conditions, such as those found in space or near black holes.

## 5. Are there any real-world examples of oscillating springs in a relativistic time frame?

Yes, there are several real-world examples of oscillating springs in a relativistic time frame. One famous example is the clock on board the International Space Station, which is designed to account for the effects of time dilation due to its high speed. Another example is the gravitational waves detected by the LIGO observatory, which are caused by the oscillations of spacetime itself due to the movement of massive objects like black holes.