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Q.

A body dropped from a height h into the floor makes elastic collision with the floor. The frequency of oscillation of its periodic motion is

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a

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

b

$\frac{1}{2 π} \sqrt{\frac{g}{2 h}}$

c

$2 π \sqrt{\frac{g}{2 h}}$

d

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

answer is A.

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Detailed Solution

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$\begin{array}{l} s = ut + \frac{1}{2} {at}^{2} \\ h = \frac{1}{2} {gt}^{2} \\ t = \sqrt{\frac{2 h}{g}} \end{array}$
But t = T/2
$\frac{T}{2} = \sqrt{\frac{2 h}{g}}; T = 2 \sqrt{\frac{2 h}{g}}$
frequency $n = \frac{1}{T} = \frac{1}{2} \sqrt{\frac{g}{2 h}}$

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A body dropped from a height h into the floor makes elastic collision with the floor. The frequency of oscillation of its periodic motion is