Physics - SHM Question with Solution | TestHub

Time taken by the ball to fall from height $h$ is given by $h=\frac{1}{2}g{t}^2$, which gives $t=\sqrt{\frac{2h}{g}}$.

Since the collision is perfectly elastic, the ball rebounds with the same speed and reaches the same height $h$. The time taken to ascend is also $t=\sqrt{\frac{2h}{g}}$.

The total time period of the periodic motion is $T=2t=2\sqrt{\frac{2h}{g}}$.

The frequency of the periodic vibratory motion is $f=\frac{1}{T}=\frac{1}{2}\sqrt{\frac{g}{2h}}$.

Answer: $\frac{1}{2}\sqrt{\frac{g}{2h}}$