A metal ball released from height 'h' makes perfectly elastic collision with ground. The frequency of…

A metal ball released from height 'h' makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is $\quad(\mathrm{g}=$ acceleration due to gravity $)$
  1. $\frac{1}{2 \pi} \sqrt{\frac{g}{2 h}}$
  2. $\frac{1}{2} \sqrt{\frac{g}{2 h}}$
  3. $\frac{1}{2} \sqrt{\frac{2 h}{g}}$
  4. $\frac{1}{2 \pi} \sqrt{\frac{2 h}{g}}$

Solution

The correct option is $\mathbf{C} \frac{1}{2} \sqrt{\frac{g}{2 h}}$ $\begin{aligned} & \mathrm{h}=\frac{1}{2} \mathrm{gt} \\ & \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \\ & \mathrm{t}_{\text {total }}=2 \mathrm{t}-2 \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \\ & \mathrm{f}=\frac{1}{\mathrm{t}_{\text {total }}}=\frac{1}{2} \sqrt{\frac{\mathrm{~g}}{2 \mathrm{~h}}} \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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