A simple pendulum makes 20 oscillations in one second on the surface of the earth. Determine the time period of the simple pendulum on the surface of a planet where the acceleration due to gravity is one fourth of the acceleration due to gravity on the surface of the earth .

Explanation

` T_(1) = 1/20 s T_(1) = 2pi sqrt (l/(g_(1))`
` Rightarrow 1/20 = 2pi sqrt(l/(g_(1))`
` T_(2) = 2pi sqrt(l/(g_(2))`
` ((2))/((1)) Rightarrow T_(2)/(1/20) = (2pi sqrt(l//g_(2)))/(2pisqrt(l//g_(1)))`
`20T_(2)= sqrt(l/(g_(2))xxg_(1)/l)=sqrtg_(1)/(g_(2))`
` T_(2) 1/20 sqrt (g_(1))/(g_(2))`
But ` g_(2) = sqrt(g_(1)/4)`
`= 1/20 sqrt(g_(1)/(1/4g_(1)))= 1/20 sqrt4 = 2/20 = 1/10 s 0.1 s `