What do they mean by a pendulum losing seconds? In many pendulum related question, a pendulum is taken do a different place where it loses seconds. For example:

A second's pendulum is taken to a mountain and it loses 20 seconds per day. What
  is the height of the mountain? (the radius of the earth is 6400km)

My problem: If the pendulum has a time period of 2 seconds, how can it lose 20 seconds? And why would it lose seconds?
I know this question is really elementary but I am really confused
 A: This is my interpretation:
If the pendulum has a period $\tau$ (2 seconds in this case) then the number of oscillations per day,$N$, is the number of seconds in a day divided by $\tau$:
$$ N = \frac{86400}{\tau} $$
If the pendulum loses $T$ seconds per day (20 seconds in this case) then the number of seconds lost per oscillation, $\Delta\tau$, is:
$$ \Delta\tau = \frac{T}{N} = \frac{T\tau}{86400} $$
And the period of the pendulum is therefore reduced to:
$$ \tau' = \tau - \Delta\tau = \tau\left(1 - \frac{T}{86400}\right) $$
Since the period of the pendulum is related to the gravitational acceleration you can now calculate $g$ and therefore the change in distance from the centre of the Earth.
A: My interpretation of the question;  specifically the "lost seconds per day"
A pendulum is used to regulate a clock;  in its original location, the clock keeps perfect time, as judged against some standard:  WWV time signal, chronometer, etc...
The clock is packed up, transported to a new location, set up and set running.  The clock is set to the correct time (say, midnight or $00:00:00.00$) by the same standard as above. When the standard says that the next midnight has arrived, the pendulum clock reads only $23:59:40$.
The pendulum has allowed the clock to mark off only $86,380$ seconds.  The pendulum is swinging slower than it did originally;  the period of the pendulum has increased, by a factor of $\frac{86400}{86380}$.
That said, other answers have detailed how to use this information to find the height of the mountain...
