Suspended from an elevator's ceiling, a simple pendulum of length 1...

The time period of a simple pendulum in an accelerating elevator is given by $T = 2\pi \sqrt{\dfrac{l}{g_{\text{eff}}}}$.

Given $T = \dfrac{\pi}{3}$ s and $l = 1$ ft, substituting these values:

$\dfrac{\pi}{3} = 2\pi \sqrt{\dfrac{1}{g_{\text{eff}}}}$

Squaring both sides:

$\dfrac{1}{9} = \dfrac{4}{g_{\text{eff}}}$

$g_{\text{eff}} = 36 \text{ ft/s}^2$

The standard acceleration due to gravity is $g = 32 \text{ ft/s}^2$. Since $g_{\text{eff}} > g$, the elevator must be accelerating upwards.

For an elevator accelerating upwards with acceleration $a$, $g_{\text{eff}} = g + a$.

$36 = 32 + a$

$a = 4 \text{ ft/s}^2$

The acceleration of the elevator is $4 \text{ ft/s}^2$ upwards.