since T = 2π√(L/g), g = (4π^2L)/T^2
the ratio of
g1/g2 = (4π^2L1)/(T1^2) / (4π^2L2)/T2^2
= (4L)/(6T)^2 / (3L)/(2T)^2
= (4/36) / (3/4)
= 1/9 * 4/3
= 4/27
On planet e a pendulum has a length of 4L has a period of 6T, on planet F a pendulum has a length of 3L has a period of 2T what is the ratio of gravitational acceleration of these two planets
2 answers
g=4π^2L/T^2
So,
gF/gE=(4π^2L/T^2)/(4π^2L/T^2)=(4π^2(3L)/(2T)^2)/(4π^2(4L)/(6T)^2)
gF/gE=(4π^2(3L)/(2T)^2)/(4π^2(4L)/(6T)^2)=(3L/4T^2)/(4L/36T^2)
gF/gE=(3L/4T^2)/(4L/36T^2).
gF/gE=(3/4)*9=27/4=6.75
So,
gF/gE=(4π^2L/T^2)/(4π^2L/T^2)=(4π^2(3L)/(2T)^2)/(4π^2(4L)/(6T)^2)
gF/gE=(4π^2(3L)/(2T)^2)/(4π^2(4L)/(6T)^2)=(3L/4T^2)/(4L/36T^2)
gF/gE=(3L/4T^2)/(4L/36T^2).
gF/gE=(3/4)*9=27/4=6.75