MHT CET · Physics · Center of Mass Momentum and Collision
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 \()\)
- A \(\frac{1}{2 \pi} \sqrt{\frac{g}{2 h}}\)
- B \(\frac{1}{2} \sqrt{\frac{g}{2 h}}\)
- C \(\frac{1}{2} \sqrt{\frac{2 h}{g}}\)
- D \(\frac{1}{2 \pi} \sqrt{\frac{2 h}{g}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \sqrt{\frac{g}{2 h}}\)
Step-by-step Solution
Detailed explanation
\(t = \sqrt{\frac{2h}{g}}\) \(T = 2t = 2 \sqrt{\frac{2h}{g}}\)
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