A large horizontal surface moves up and down | vedclass

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

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) For the mass to remain in contact with the surface,the downward acceleration of the surface must not exceed the acceleration due to gravity $g$.The maximum acceleration of a particle in $S.H.M.$ is given by $a_{max} = \omega^2 A$.To maintain contact,we require $a_{max} \leq g$.Substituting $\omega = 2 \pi f$,we get $(2 \pi f)^2 A \leq g$.Given $A = 1 \, cm = 0.01 \, m$ and taking $g \approx 10 \, m/s^2$:$4 \pi^2 f^2 (0.01) = 10$$f^2 = \frac{10}{4 \pi^2 	imes 0.01} = \frac{10}{0.04 \pi^2} = \frac{250}{\pi^2}$$f = \sqrt{\frac{250}{\pi^2}} = \frac{\sqrt{250}}{\pi} \approx \frac{15.81}{3.14} \approx 5.03 \, Hz$.Thus,the maximum frequency is approximately $5 \, Hz$.

$A$ large horizontal surface moves up and down in $S.H.M.$ with an amplitude of $1 \, cm$. If a mass of $10 \, kg$ placed on the surface is to remain continuously in contact with it,the maximum frequency of $S.H.M.$ will be .... $Hz$.

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