An astronaut sets up a pendulum on the Moon,where the acceleration due to gravity is 1/6 that of Earth.If she finds that she must push the pendulum with a frequency of 0.10 Hz to maintain a swinging motion,what is the length of the pendulum?
A)4.1 m
B)1.6 m
C)99 m
D)7.8 m

Question

Quizplus · Accepted Answer

The frequency of a pendulum is given by $f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$, where $g$ is the acceleration due to gravity and $L$ is the length of the pendulum. On the Moon, $g = \frac{1}{6}g_{Earth}$, and given $f = 0.10$ Hz, we can rearrange the formula to solve for $L$: $L = \frac{g}{(2\pi f)^2}$. Substituting $g_{Moon} = \frac{1}{6}g_{Earth} = \frac{1}{6} \times 9.8 \, m/s^2$ and $f = 0.10$ Hz, we find $L \approx 4.1$ m.

An Astronaut Sets Up a Pendulum on the Moon,where the Acceleration