For any S.H.M.,the amplitude is 6, cm. If the | vedclass

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(B) The potential energy $U$ of a particle in $S.H.M.$ at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $E$ of the pa

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$4.2$

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(B) The potential energy $U$ of a particle in $S.H.M.$ at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy $E$ of the particle is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.According to the problem,the instantaneous potential energy is half the total energy,so $U = \frac{1}{2} E$.Substituting the expressions,we get $\frac{1}{2} m \omega^2 y^2 = \frac{1}{2} (\frac{1}{2} m \omega^2 a^2)$.Simplifying this,we get $y^2 = \frac{a^2}{2}$,which implies $y = \frac{a}{\sqrt{2}}$.Given $a = 6\, cm$,we have $y = \frac{6}{\sqrt{2}} = 3 	imes \sqrt{2} \approx 3 	imes 1.414 = 4.242\, cm$.Rounding to the nearest provided option,the distance is $4.2\, cm$.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.

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