A large horizontal surface moves up and down in $SHM$ with an amplitude of $1 \,cm$. If a mass of $10\, kg$ (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of $S.H.M.$ will be ... $Hz$
AIIMS 1995, Medium
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(c) For body to remain in contact ${a_{\max }} = g$
$\therefore {\omega ^2}A = g$
$\therefore {\omega ^2}A = g$
$\Rightarrow 4{\pi ^2}{n^2}A = g$
$ \Rightarrow {n^2} = \frac{g}{{4{\pi ^2}A}} = \frac{{10}}{{4{{(3.14)}^2}0.01}} = 25$
$ \Rightarrow n = 5\;Hz$
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