Question

On Planet X, a ball on a massless, rigid rod oscillates as a simple pendulum with a period of 2.0s. If the pendulum is taken to the moon of Planet X, where the free-fall acceleration g is half as big, the period will be

Answer 1

Answer:

Time period at the moon will be 2.828 sec

Explanation:

Let initially the length of the pendulum is l and acceleration due to gravity is g

And time period is given T = 2 sec

So time period of the pendulum is equal to $T=2\pi \sqrt{\frac{l}{g}}$-------------eqn 1

Now on the planet acceleration due to gravity $g_{new}=\frac{g}{2}$

So time period of the pendulum at this planet $T_{new}=2\pi \sqrt{\frac{l}{g_{new}}}$--------eqn 2

Now dividing eqn 1 by eqn 2

$\frac{T}{T_{new}}=\sqrt{\frac{\frac{g}{2}}{g}}$

$T_{new}=\sqrt{2}T=1.414\times 2=2.828sec$

So time period at the moon will be 2.828 sec