Answer
$a = 2.08~m/s^2$
Work Step by Step
Let $F_T$ be the tension in the pendulum. Let $m$ be the mass of the bob. The vertical component of $F_T$ is equal in magnitude to the bob's weight.
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
We can find the acceleration:
$ma = F_T~sin~\theta$
$ma = \frac{mg~sin~\theta}{cos~\theta}$
$a = g~tan~\theta$
$a = (9.80~m/s^2)~tan~12^{\circ}$
$a = 2.08~m/s^2$
