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MHT CET 2026 (12 May, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2026 (12 May, Shift 2)

Q5MHT CET 2026 · 12 MAY · SHIFT 2Medium

Periodic Motion & SHM Basics

A ball released from height $(h)$ which makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is ( $g$ = acceleration due to gravity)

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Answer: (b)

Time to fall height $h$ : $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}}$ .
Elastic bounce: it rises back to $h$ in the same time, so $T = 2\sqrt{\frac{2h}{g}}$ .
$f = \frac{1}{T} = \frac{1}{2}\sqrt{\frac{g}{2h}}$ .
So the answer is (b).

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