An object undergoing SHM taken 0.5 s to travel from one point of zero velocity to the next such point. The distance between those point is 50 cm. The period, frequency and amplitude of the motion is:

The velocity of particle executing SHM is zero at extreme positions.

$A$ is the amplitude

The distance between two extremes $=2 \mathrm{~A}$

According to question $2 A=50 \mathrm{~cm}$

$\therefore A=25 \mathrm{~cm}$

Time taken to reach from one extreme to other

$t=T / 2$

According to question; $T / 2=0.5 \mathrm{sec}$

$\therefore \quad T=1 \mathrm{sec}$

Frequency is given as $f=1 / T=1 \mathrm{~Hz}$

(option-A)