A large horizontal surface moves up and down | vedclass

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(C) For the mass to remain in contact with the surface,the downward acceleration of the surface must not exceed the acceleration due to gravity $g$. A

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) For the mass to remain in contact with the surface,the downward acceleration of the surface must not exceed the acceleration due to gravity $g$. At the highest point of the $SHM$,the acceleration is directed downwards and its magnitude is $a_{max} = \omega^2 A$. To maintain contact,we require $a_{max} \leq g$. Thus,$\omega^2 A = g$. Substituting $\omega = 2\pi n$,where $n$ is the frequency: $(2\pi n)^2 A = g$ $4\pi^2 n^2 A = g$ $n^2 = \frac{g}{4\pi^2 A}$. Given $g = 10 \, m/s^2$,$A = 1 \, cm = 0.01 \, m$,and $\pi^2 \approx 10$: $n^2 = \frac{10}{4 	imes 10 	imes 0.01} = \frac{10}{0.4} = 25$. $n = \sqrt{25} = 5 \, Hz$.

$A$ 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 $SHM$ will be ... $Hz$.

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