Period of oscillation

Let L be the original ength of the pendulum.
Period T = 2.pi.square-root[L/g]
where g is the acceleration due to gravity, and pi = 3.14159.....

After 3/4 of the length of the suspension string is fixed, the length of the pendulumn now becomes L/4.
Let the new periof be T', hence,
T' = 2.pi.square-root[(L/4)/g]
or T' = 2.pi.square-root[L/4g]

Dividing: T'/T = square-root[(L/4g).(g/L)] = square-root[1/4] = 1/2
hence, T' = T/2

Since it takes half a period to swing from the equilibrium position to the extreme left and then return to the equilibrium position, thus time needed
= T'/2 = T/4