The time T of of oscillation of a simple pendulum of length l is given by `T=2pisqrt(l/g)`. Find the percentage error in T, corresponding to an error of `2%` in the value of l.

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

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

The period of oscillation T of a simple pendulum of length l is connected by the relation T = 2pi sqrt((i)/(g)) , where g is a constant. Find approximately the percentage error in the computed value of T correspounding to an error of 1% in the value of l.

The period of oscillation of a simple pendulum is T = 2pisqrt((L)/(g)). Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is

Consider a simple pendulum having a bob attached to a string that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l) , mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using the method of dimensions.