Answer
See answer.
Work Step by Step
The frequency of a simple pendulum is given by equation 11–11b, $f=\frac{1}{2 \pi}\sqrt{\frac{g}{\mathcal{l}}}$.
If the pendulum is accelerated vertically upward/downward, this is equivalent to increasing/decreasing the acceleration due to gravity.
a. An upward acceleration effectively increases g.
$$f_{new}= \frac{1}{2 \pi}\sqrt{\frac{g+a}{\mathcal{l}}}=\frac{1}{2 \pi}\sqrt{\frac{g+0.35g}{\mathcal{l}}}=\sqrt{1.35}\frac{1}{2 \pi}\sqrt{\frac{g}{\mathcal{l}}}=1.16f$$
b. A downward (i.e., negative) acceleration effectively decreases g.
$$f_{new}= \frac{1}{2 \pi}\sqrt{\frac{g+a}{\mathcal{l}}}=\frac{1}{2 \pi}\sqrt{\frac{g-0.35g}{\mathcal{l}}}=\sqrt{0.65}\frac{1}{2 \pi}\sqrt{\frac{g}{\mathcal{l}}}=0.81f$$
