A block of mass m rests on a platform. The pl | vedclass

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(B) For the block to remain in contact with the platform,the downward acceleration of the platform must not exceed the acceleration due to gravity $g$

Vedclass · Accepted Answer

$\frac{1}{2\pi}\sqrt{g/d}$

Vedclass · Answer

(B) For the block to remain in contact with the platform,the downward acceleration of the platform must not exceed the acceleration due to gravity $g$.The maximum downward acceleration of a platform undergoing $SHM$ is given by $a_{max} = \omega^2 d$,where $\omega$ is the angular frequency and $d$ is the amplitude.To ensure the block does not leave the platform,we set $a_{max} \le g$,which implies $\omega^2 d = g$.Solving for $\omega$,we get $\omega = \sqrt{g/d}$.Since angular frequency $\omega = 2\pi f$,where $f$ is the frequency,we have $2\pi f = \sqrt{g/d}$.Therefore,the maximum frequency is $f = \frac{1}{2\pi}\sqrt{\frac{g}{d}}$.

$A$ block of mass $m$ rests on a platform. The platform is given up and down $SHM$ with an amplitude $d$. What is the maximum frequency so that the block never leaves the platform?

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