$17$ Multiple Choice $10$ points

$F_5^2=v_6^2+{20}_f(x-y_0)$

$$

${?}^{-mp}$

hase $=R_{1}n$

$v_t=mgy$

$=\frac{\text{d}\text{e}\text{l}\text{U}}{del{y}^{\text{'}}},{}_1=\frac{\text{d}\text{e}\text{l}\text{U}}{\text{d}\text{e}\text{l}\text{z}}$

A frictionless pendulum clock on the surface of the earth has a period of $1\mathrm{.}00$ s $.$ On a distant planet, the length of the pendulum must be shortened slightly to have a period of $1\mathrm{.}00$ s$.$ What is true about the acceleration due to gravity on the distant planet?

The gravitational acceleration on the planet is slightly greater than $g.$

The gravitational acceleration on the planet is slightly less than $g.$

We cannot tell because we do not know the mass of the pendulum.

The gravitational acceleration on the planet is equal to $g.$