# How does the period of an oscillating mass-spring system comply with the relativistic time dilation as viewed by a moving observer?

Consider a simple mass-spring system. The spring is attached to the ceiling in the lab frame of reference and a block of mass $$m'$$ is suspended from the spring. The mass starts to oscillate with simple harmonic motion of period $$T'$$ in the lab frame of reference. If the spring constant is measured to be $$k'$$ by the lab observer, how can we calculate its period $$T$$, which must be dilated by the traditional Lorentz factor, as viewed by a moving observer? Assume that the moving observer travels at a considerable fraction of light speed towards the lab observer and perpendicular to the alignment of the spring.

Remember that since the definition of the relativistic mass is not widely accepted, it is not legitimate to insert the relativistic mass increase ($$m=\gamma m'$$) straightforwardly into the period equation as follows:

$$T=2\pi\sqrt\frac{m}{k}=2\pi\sqrt\frac{\gamma m'}{\alpha k'}=2\pi\gamma\sqrt\frac{m'}{k'}=\gamma T'\space,$$

where $$\alpha=1/\gamma$$.

• Doppler effect. – m4r35n357 Aug 4 '20 at 13:16
• If you are a moving observer looking at any oscillating system, you see it's frequency (calculated in its own frame of reference) modified by the Doppler Effect. Like watching a clock. – m4r35n357 Aug 4 '20 at 13:24
• Are you seriously saying the pendulum is moving close to the speed of light? I assumed you meant the observer (otherwise why have one?). – m4r35n357 Aug 4 '20 at 13:31
• So are you saying the pendulum and the observer are moving close to the speed of light? If so, I cannot help any further. – m4r35n357 Aug 4 '20 at 13:42
• @m4r35n357 well, 1/2 of the Doppler effect. – JEB Aug 4 '20 at 13:55

## 2 Answers

Consider coordinates $$\mathbf x = (ct,x,y)$$. In the lab frame, $$\mathbf x = \pmatrix{ct\\0\\A\cos(\omega t)}$$

In a frame moving with speed $$v$$ in the $$+\hat x$$ direction, we find via Lorentz transformation that $$\mathbf x' = \pmatrix{\gamma ct\\ -\gamma v t \\ A\cos(\omega t)}=\pmatrix{ct'\\-vt'\\A\cos\left(\frac{\omega t'}{\gamma}\right)}$$

where $$t' = \gamma t$$. We therefore see that the frequency is reduced by a factor of $$\gamma$$, so the period is increased by the same factor.

There is no gravity in this scenario.

Let's consider two springs attached to the opposite sides of a ceiling fan with two massive blades in the lab frame of reference, and two block of masses that are suspended from the springs. The masses start to oscillate with simple harmonic motion of period T' in the lab frame of reference.

Then the ceiling fan is turned on. (This contraption is designed so that g-forces do not have any effect on the oscillation period)

If the spring constants are measured to be k' by the lab observer, when the fan is off, it will be measured to be k when the fan is on. k = k' / gamma.

If the rest mass of the system consisting of the two masses is measured to be m' by the lab observer, when the fan is off, it will be measured to be m when the fan is on. m = gamma * m'.

Now we can calculate the period of the oscillation when the fan is on:

$$T=2\pi\sqrt\frac{m}{k}=2\pi\sqrt\frac{\gamma m'}{\alpha k'}=2\pi\gamma\sqrt\frac{m'}{k'}=\gamma T'\space,$$

where $$\alpha=1/\gamma$$

Next, one of the blades breaks off, and the blade-spring-mass system keeps moving and oscillating at unchanged rate. We have calculated the oscillation period of a different system, or a 'wrong' system, but we know that the 'right' system has the same oscillation period as the 'wrong system'.