Frequency f of a simple pendulum depends on its length l and acceleration g due to gravity according to the following equation `f=1/(2pi)sqrt(g/l)`
Graph between which of the following quantities is a straight line?

The time period T of a simple pendulum depends on length L and acceleration due to gravity g Establish a relation using dimensions.

The period of oscillation of a simple pendulum at a given place with acceleration due to gravity 'g' depends on

Consider a simple pendulum. The period of oscillation of the simple pendulum depends on its length and acceleration due to gravity. Then the expression for its time period is

The period of a conical pendulum in terms of its length (l) , semivertical angle ( theta ) and acceleration due to gravity (g) is

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

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

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 period of oscillation (T) of a simple pendulum depends on the probable quantities such as mass 'm' of a bob, length 'l' of the pendulum and acceleration due to gravity 'g' at the place. Derive an equation using dimensional analysis.