Q:

Tick-a-tick Tick-tick

Charlie has two clocks that hold reasonably good time. That is, each clock ticks at a constant rate, but its ticks are not necessarily exactly 1 second apart. The period between ticks is constant for each clock, with each period in the range $1 \pm 0.1$ seconds.
Charlie sleeps for 8 hours each night. If Charlie hears the clocks ticking out of phase when he goes to sleep one evening and in phase when he wakes up the next morning, what's the largest possible difference between the periods of the two clocks?
Assume that at the moment Charlie bought the clocks, they happened to tick at exactly the same time.
Clarifications

The period of the clock is the time between one tick and the next. The period of each clock must be between 0.9 seconds and 1.1 seconds. It could be 0.92 or 1.1 or even 1 second, so long as it is a constant between 0.9 and 1.1 seconds.
Out of phase means that the clocks are ticking exactly 0.5 seconds apart. They sound like tick-ka-tick over a space of about 1 second, with 'tick' coming from one clock and 'ka' coming from the other. Update: this may be taken to imply that Charlie falls asleep upon hearing a 'ka' exactly half a second after the other clock 'tick'ed.
In phase means that the clocks tick at exactly the same time. They sound like just one (loud) clock for that tick.
If one clock ticks at $1 + t_1$ seconds and the other ticks at $1 + t_2$ seconds, find $|t_1 - t_2|$.

Bonus: generalise the answer for clocks with periods in the range $1 \pm t$ seconds.
This is not a lateral-thinking puzzle. Reasoning for the answer must be provided. The correct answer with the simplest valid reasoning wins. No need for spoiler tags.

A:

NOTE: I have no idea if this answer still applies after the edit.

It is known that $(a \pm b) + (c \pm d) = (a+c) \pm (b+d)$
Therefore $2(a \pm b) = 2a \pm 2b$
For general rule $n(a \pm b) = na \pm nb$
Total number of seconds in $8$ hours $= 8 \times 60 \times 60 = 28800$ seconds
We have $a=1$ and $b=0.1$ and $n=28800$
Therefore $$n(a \pm b) = na \pm nb$$ $$\Rightarrow 28800(1 \pm 0.1) = 28800 \pm 2880$$
So, the maximum value is $= 28800+2880$
And , the minimum value is $=28800-2880$
So, the largest possible difference between the periods of the two clocks is 
Maximum value - Minimum value
$= (28800+2880) - (28800-2880)$
$=5760 $ seconds

