For a particle executing S.H.M.,its potential | vedclass

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(C) Potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$ and kinetic energy is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Given that t

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$\frac{2 \sqrt{2} A}{3}$

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(C) Potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$ and kinetic energy is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Given that the potential energy is $8$ times the kinetic energy at displacement $x$:$U = 8K$Substituting the expressions:$\frac{1}{2} m \omega^2 x^2 = 8 	imes \frac{1}{2} m \omega^2 (A^2 - x^2)$$x^2 = 8(A^2 - x^2)$$x^2 = 8A^2 - 8x^2$$9x^2 = 8A^2$$x^2 = \frac{8A^2}{9}$Taking the square root on both sides:$x = \frac{\sqrt{8} A}{3} = \frac{2\sqrt{2} A}{3}$

For a particle executing $S.H.M.$,its potential energy is $8$ times its kinetic energy at a certain displacement $x$ from the mean position. If $A$ is the amplitude of $S.H.M.$,the value of $x$ is:

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