The time period of a simple pendulum is given by the formula, `T = 2pi sqrt(l//g)`, where T = time period, l = length of pendulum and g = acceleration due to gravity.
If the length of the pendulum is decreased to 1/4 of its initial value, then what happens to its frequency of oscillations ?

Explanation

The time period of a simple pendulum is
` T= 2 pi sqrt(l //g )`
When length is decreased to `1//4^(th)` of its initial value then `l^1 = l//4`
` therefore ` New time period, (`T^1`)
` = 2 pi sqrt((l//4)/(g)) = (2 pi)/(2) sqrt((l)/(g)) = 1/2 (T)`
` therefore ` Time taken to perform one oscillation is reduced to half of its initial value.
` therefore `Frequency of oscillation becomes double that of the initial value.