Question

A pendulum has a period of 1.8 s.

The original pendulum is taken to a planet where g = 16 m/s2.

What is its period on that planet?

Answer 1

A pendulum has a period of 1.8 s. The original pendulum is taken to a planet where $g = 16 \, \text{m/s}^2$. What is its period on that planet?

The period of a pendulum can be approximated by:

$$T = 2 \pi \sqrt {\frac {L}{g}}$$

where $L$ is the length of the pendulum and $g$ is the local acceleration of gravity.

On the Earth $g = g_{0} = 9.8\frac{m}{s^{2}}$, therefore period equals:

$$T _ {0} = 2 \pi \sqrt {\frac {L}{g _ {0}}}$$

And on the planet where $g = 16\frac{m}{s^2}$:

$$T = 2 \pi \sqrt {\frac {L}{g}}$$

Therefore:

$$\frac {T}{T _ {0}} = \sqrt {\frac {g _ {0}}{g}}$$ $$T = T _ {0} \sqrt {\frac {g _ {0}}{g}} = 1.8 \, s \sqrt {\frac {9.8}{16}} \cong 1.4 \, s$$

Answer: $1.4 \, s$