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 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, 1.99 sec Find the Most Probable value

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

Find the principal solution of sec^-1 (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

If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity is

Find the ratio of the first quantity to the second quantity in the reduced form: 2 min 6 sec, 3 min 54 sec.

Prove the following : sec^2 theta + cosec^2 theta = sec^2 theta times cosec^ theta

(sec2A+1)sec^2A=