Answer the following . The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angle of oscillation, a more involved analysis shows that T is greater than `2 (pi) sqrt ((l/g))`. Think of a qualitative argument to appreciate this result.

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

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 :

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

Answer the following . Time period of a particle in S.H.M. depends on the force constant k and mass m of the particle : T = 2 (pi) (( sqrt m /k)). A simple pendulum executes SHM approximately . Why then is the time period of a pendulum independent of the mass of the pendulum ?

A student measures the value of g with the help of a simple pendulum using the formula g = (4pi^(2)L)/(T^(2)) . He measures length L with a meter scale having least count 1 mm and finds it 98.0 cm . The time period is measured with the help of a watch of least count 0.1s . The time of 20 oscillations is found to be 40.4 s. The error Deltag in the measurment of g is (" in" m//s^(2)) .