Clock on a pendulum If a clock is swinging on a pendulum (assume it's on some massive object with gravity) what happens when that system is moving at relativistic speeds? To an inertial observer, on the back-swing (i.e. opposite the direction of movement of the system) the pendulum has a lower ("absolute") velocity than on the forward swing. The clock should experience less time dilation on the back-swing, and that should register as a different number of 'ticks' on the clock, but that can't be because then the 'moving' observer would notice that too!
Is the length (duration) of each swing different to cancel out that effect, and what causes the change in duration: (is it because the weight/bob is relativistically more massive on the forward swing?)
I don't know too much, so please keep the explanation simple!
Please note, gravity should not play much of a role here. I could ask the same question with an oscillating tuning fork to avoid most gravitational complications.
 A: Yes, although the observer travelling with the pendulum measures the same speed forwards and back, however in the inertial observer's frame, the speed of the pendulum relative to the moving frame  varies between the forward swing and the back swing. You can verify this using the velocity addition formula used to calculate relative velocities (shown below using natural units, that is all velocities are expressed as a fraction of 1, where 1 is the speed of light).
$$u={v+u' \over 1+vu'}$$
where $v$ is the velocity of the system containing the pendulum, $u'$ is the velocity of the pendulum as measured in that system, and $u$ is the absolute velocity of the pendulum as measured in the inertial observers frame. The speed of the pendulum relative to the pivot as measured by the inertial observer is $|u - v|$ and you can calculate that forward swing $+u'$ yields a slower relative speed than $-u'$ on the back-swing.
Since the pendulum is swinging faster on the back-swing, the duration of its swing will be shorter and that will compensate for the fact that the clock is ticking faster (due to less time-dilation) and the number of ticks will add up exactly to the number of ticks on the forward-swing, which has a longer duration but more time dilation.
We can express this relationship algebraically, noting that the duration of a half-period in the inertial frame is $\Delta t = {L \over |u - v|}$ where $L$ is the distance traveled in a half-swing; and the formula for time elapsed in the clock's frame (i.e. the number of ticks) is $\Delta t' = \Delta t \sqrt {1 - u^2}$. Manipulating and simplifying the formula for the case where $u$ is positive (the forward swing) and negative (the back swing) will demonstrate that $\Delta t$ is the same for both even though they travel at different speeds.
(I have not yet explored the underlying mechanism for why the speed varies between the forward and back swing).
A: I see two main issues with your scenario:
1) A pendulum is not an inertial reference frame, so special relativity does not really work here.
2) If you took the problem to be a bunch of instantaneous inertial reference frames, as one does in the classic relativistic rocket problem, you will find that your assumption of how time is measured is incorrect: the two observers will not observe a different number of ticks on the front and back swings, because the number of ticks is dilated from the perspective of a stationary observer as well. So it might take six ticks for the pendulum to go forward, and six ticks to go back, but to the stationary observer, the six forward-moving ticks will happen more slowly than the ones on the backswing. An observer sitting on the pendulum would see them all happen at the same rate, and an observer sitting on the (constant-velocity) pivot point of the pendulum would see symmetric time dilation for the front and back swings.
A pendulum basically is a clock, so if we ignore the problem of inertial reference frames, any relativistic effects that affect the pendulum will similarly affect the clock.
