The time period of a simple pendulum of infinite length is (R=radius of earth).

The time period of a simple pendulum of length 9.8 m is

Calculate the time period of a simple pendulum whose length is equal to radius of earth. Hint : L =R_(e) = 6.4 xx 10^(6) m , g = 9.8 ms^(-2) T ' = 2pi sqrt(( R_(e))/(2g))

What is the period of a simple pendulum whose length is 50 cm?

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g / 3 is

The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

How would the time period of a simple pendulum be affected if (a) the length is doubled (b) the amplitude is doubled ?

Which of the following is/are likely to affect the time period of a simple pendulum? (i) length (ii) amplitude of vibration (iii) mass of the bob

STATEMENT -1 : The time period of oscillation of a simple pendulum of constant length is more at a place inside a mine than on the surface of the earth. and STATEMENT -2 : The frequency of oscillations of a simple pendulum is more at a place inside a mine than on the surface of the earth .

The time period of a simple pendulum is 2 s. Find its frequency .

State how does the time period of a simple pendulum depend on length of pendulum.