Question

Compare the time periods of a simple pendulum at
places where g is 9.8 m and 4.36 m s- respectively.​

Answer 1

Explanation:

$t1 = 2\pi \sqrt{ \frac{l}{9.8} } \\ t2 = 2\pi \sqrt{ \frac{l}{4.36} } \\ \\ \frac{t1}{t2} = \sqrt{ \frac{4 .36}{9.8} }$

t1:t2 = √2.18 : √4.9