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(C) Given: Amplitude $A = 0.06 \,m$,Kinetic Energy $K.E. = 10 \,J$,Potential Energy $P.E. = 8 \,J$. Total Energy $T.E. = K.E. + P.E. = 10 \,J + 8 \,J

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$0.04$

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(C) Given: Amplitude $A = 0.06 \,m$,Kinetic Energy $K.E. = 10 \,J$,Potential Energy $P.E. = 8 \,J$. Total Energy $T.E. = K.E. + P.E. = 10 \,J + 8 \,J = 18 \,J$. The formula for Potential Energy is $P.E. = \frac{1}{2} kx^2$ and Total Energy is $T.E. = \frac{1}{2} kA^2$. Taking the ratio: $\frac{P.E.}{T.E.} = \frac{\frac{1}{2} kx^2}{\frac{1}{2} kA^2} = \frac{x^2}{A^2}$. Substituting the values: $\frac{8}{18} = \frac{x^2}{(0.06)^2}$. $\frac{4}{9} = \frac{x^2}{(0.06)^2}$. Taking the square root on both sides: $\frac{2}{3} = \frac{x}{0.06}$. $x = \frac{2}{3} 	imes 0.06 = 2 	imes 0.02 = 0.04 \,m$.

An object executes $SHM$ along the $x$-axis with an amplitude of $0.06 \,m$. At a certain distance '$x$' metres from the mean position,it has a kinetic energy of $10 \,J$ and a potential energy of $8 \,J$. The distance '$x$' will be: (in $\,m$)

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