$A$ simple pendulum of frequency $f$ has a metal bob. If the bob is charged negatively and is allowed to oscillate above a large positively charged plate,the frequency will be:

The acceleration due to gravity on the surface of the moon is $1.7 \; m s^{-2}$. What is the time period of a simple pendulum on the surface of the moon if its time period (in $s$) on the surface of the earth is $3.5 \; s$? ($g$ on the surface of the earth is $9.8 \; m s^{-2}$)

What is the length of a simple pendulum,which ticks seconds (in $m$)?

$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.

$Assertion:$ The time period of a simple pendulum on a satellite orbiting the Earth is infinity.
$Reason:$ The time period of a simple pendulum is inversely proportional to $\sqrt{g}$.

Two pendulums having lengths $1.0 \ m$ and $1.21 \ m$ start oscillating simultaneously in the same phase. They will again come in the same phase after how many oscillations of the smaller pendulum?