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{(L)/(g ) }`

=>
`(T1)/(T2 )` =
`\sqrt{(g2)/(g1) }`

Given T1= 3s g1= 9.79 g2= 9.82

=> T2 = 3.00459 s