A coin is placed on a horizontal platform whi | vedclass

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(B) Let $m$ be the mass of the coin and $N$ be the normal reaction force exerted by the platform on the coin.The equation of motion for the coin in th

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for an amplitude of $\frac{g}{\omega^2}$

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(B) Let $m$ be the mass of the coin and $N$ be the normal reaction force exerted by the platform on the coin.The equation of motion for the coin in the vertical direction is given by:$mg - N = m a$where $a$ is the acceleration of the platform. For simple harmonic motion,the acceleration is $a = -\omega^2 x$,where $x$ is the displacement from the mean position.Substituting this into the equation of motion:$mg - N = m(-\omega^2 x)$$N = m(g + \omega^2 x)$However,when the platform moves upwards,the acceleration is directed downwards,so $a = -\omega^2 x$ (where $x$ is positive upwards). When the platform is at the highest point,$x = A$ (amplitude),and the acceleration is $a = -\omega^2 A$.The equation becomes $mg - N = m(-\omega^2 A)$,which implies $N = m(g + \omega^2 A)$. This does not lead to $N=0$.When the platform moves downwards,the acceleration is directed upwards. At the highest point of the oscillation,the platform's acceleration is directed downwards. The coin loses contact when the normal force $N$ becomes zero.This happens when the downward acceleration of the platform exceeds the acceleration due to gravity $g$.The maximum downward acceleration is $\omega^2 A$. Thus,the coin leaves contact when $\omega^2 A \ge g$,or $A \ge \frac{g}{\omega^2}$.Therefore,the coin will leave contact for the first time when the amplitude reaches $\frac{g}{\omega^2}$ at the highest position of the platform.

$A$ coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega$. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

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