The amplitude of a particle executing SHM abo | vedclass

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(D) Given amplitude $A = 10 \, cm$. For option $A$: $K.E. = 0.64 \, K.E._{\max}$. Since $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and $K.E._{\max} =

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All of the above

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(D) Given amplitude $A = 10 \, cm$. For option $A$: $K.E. = 0.64 \, K.E._{\max}$. Since $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and $K.E._{\max} = \frac{1}{2} m \omega^2 A^2$,we have $A^2 - x^2 = 0.64 A^2$,which gives $x^2 = 0.36 A^2$,so $x = 0.6 A = 6 \, cm$. This is correct. For option $B$: When $x = 5 \, cm = A/2$,$P.E. = \frac{1}{2} k x^2 = \frac{1}{2} k (A/2)^2 = \frac{1}{4} P.E._{\max}$. Since $T.E. = K.E. + P.E. = P.E._{\max}$,then $K.E. = P.E._{\max} - 0.25 P.E._{\max} = 0.75 P.E._{\max}$. This is correct. For option $C$: In $SHM$,the total mechanical energy is conserved and is equal to the maximum $K.E.$ (at the mean position) or maximum $P.E.$ (at the extreme position). This is correct. Therefore,all statements are correct.

The amplitude of a particle executing $SHM$ about $O$ is $10 \, cm.$ Then:

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