Successive measurements of the period of oscillation of a simple pendulum come out to be 2.63 s, 2.56 s, 2.71 s and 2.81 s. Calculate Mean absolute error and Percentage error

Obtain the expression for the period of a simple pendulum performing S.H.M.

If the period of oscillation of mass m suspended from a spring is 2 s, then the period of mass 4 m will be

In ohm's experiment, the values of unknown resistances were found to be 6.12 Omega , 6.15 Omega , calculate the mean absolute error, relative error and percentage error in these measurements.

The time period ,for the oscillation , of a simple pendulum were recorded, 5 readings were taken they were 2.00 sec., 2.02 sec., 1.96 sec., 2.03 sec, 1.101 sec Find the Percentage error.

The time period ,for the oscillation , of a simple pendulum were recorded, 5 readings were taken they were 2.00 sec., 2.02 sec., 1.96 sec., 2.03 sec, 1.100 sec Find the Final absolute error

The path length of oscillation of simple pendulum of length 1 m is 16 cm. Its maximum velocity is (take, g=pi^(2)m//s^(2) )

The period of oscillation of a simple pendulum is given by T=2pisqrt((l)/(g)) where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measured by a stop watch of least count 0.1 s. The percentage error is g is

The speed of light in a medium is 2.25 xx 10^(8) m//s . Calculate absolute refractive index of the medium.

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