Answer to Question #274951 in Physics for Dhea

Question

A pendulum on planet X, where the value of g is unknown, oscillates with a period of 2 seconds. What is the period of this pendulum on earth where g= 9.8meter per second square?

Answer 1

On planet X $T_1=2\pi\sqrt{l/g_1}$

On Earth $T_2=2\pi\sqrt{l/g_2}$

We get $\frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}\to T_2=T_1\cdot\sqrt{\frac{g_1}{g_2}}=2\cdot\sqrt{\frac{g_1}{9.8}}=$

$=0.639\cdot\sqrt{g_1}$ . You need to know either the acceleration or the length of the pendulum in order to calculate the period.