Question

(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79m/s2 is moved to a location where the acceleration due to gravity is 9.82m/s2. What is its new period

Answer 1

I assume this is the motion of the simple pendulum

T = 2π × $\sqrt{\frac{L}{g } }$

=> $\frac{T1}{T2 }$ = $\sqrt{\frac{g2}{g1} }$

Given T1= 3s g1= 9.79 g2= 9.82

=> T2 = 3.00459 s