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

Answer:

T=2π√l/g.... initially.

now T=2π√l/4g

now... after solving T=√l

so,, frequency,f= 1/√l.

thank you