Question

Compare the time period of a simple pendulum at places where
g= 9.8m/s^2 and 4.36m/s^2 respectively

Answer 1

Explanation:

formula T=2π√l/g

taking square

(T1/T2)²= g2/g1

= 4.36/9.8

=1/2 approx

taking square root

T1/T2=1/√2

Answer 2

Explanation:

Time period T inversly proportional to rot of acceleration. There fore T1/T2=

$\sqrt{4.36} \div \sqrt{9.8 } = 0.667$