The time period of a simple pendulum is given by `T = 2pi sqrt((l)/(g))`
where 'l' is the length of the pendulum and 'g' is the acceleration due to gravity at that place.
(a) Express 'g' in terms of T
(b) What is the length of the seconds pendulum ? (Take `g = 10 m s^(-2)`)

If the acceleration due to gravity on moon is one-sixth of that on the earth, what will be the length of a second pendulum there ? Take g = 9.8 ms^-2

Find the length of seconds pendulum at a place where g = pi^(2) m//s^(2) .

Find the length of seconds pendulum at a place where g = pi^(2) m//s^(2) .

If 'L' is length of simple pendulum and 'g' is acceleration due to gravity then the dimensional formula for (l/g)^(1/2) is same as that for

If 'L' is length of sample pendulum and 'g' is acceleration due to gravity then the dimensional formula for ((l)/(g))^(1/2) is same that for

The time period of a simple pendulum is given by the formula, T = 2pi sqrt(l//g) , where T = time period, l = length of pendulum and g = acceleration due to gravity. If the length of the pendulum is decreased to 1/4 of its initial value, then what happens to its frequency of oscillations ?

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

Time period T of a simple pendulum of length l is given by T = 2pisqrt((l)/(g)) . If the length is increased by 2% then an approximate change in the time period is