Question

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 ?

Answer 1

The time period of a simple pendulum is
`T = 2pi 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))`
`= 2pi sqrt((l//4)/(g)) = (2pi)/(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.