The period of oscillation of simple pendulum increases by 20% where its length is increased by 44 cm. Find its
initial period of oscillation.

The period of oscillation of simple pendulum increases by 20% where its length is increased by 44 cm. Find its initial length

The period of oscillations of a simple pendulum increses by 10%, when its length is increased by 21 cm. Find its initial length and initial period.

The period of a simple pendulum of infinite length is

The period of a simple pendulum is found to increase by 50% , when the length of the pendulum is increased by 0.6 m. Calculate the initial length and initial period of oscillation at a place where g = 9.8 frac(m)(s^2)

The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is C

If the length of simple pendulum is increased by 300%, then the time period will be increased by

If length of a simple pendulum is increased by 44%, then what is the gain in the time period of pendulum?

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 length of simple pendulum executing SHM is increased by 69% The percentage increase in the time period of the pendulum is