The displacement of a particle in simple harm | vedclass

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Question

(A) The kinetic energy $(K.E.)$ and potential energy $(P.E.)$ in $SHM$ are given by:$K.E. = \frac{1}{2} m \omega^2 (A^2 - y^2)$$P.E. = \frac{1}{2} m \

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$\sqrt{\frac{\pi}{3}}$

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(A) The kinetic energy $(K.E.)$ and potential energy $(P.E.)$ in $SHM$ are given by:$K.E. = \frac{1}{2} m \omega^2 (A^2 - y^2)$$P.E. = \frac{1}{2} m \omega^2 y^2$Given that $K.E. = 8 	imes P.E.$Substituting the expressions:$\frac{1}{2} m \omega^2 (A^2 - y^2) = 8 	imes \frac{1}{2} m \omega^2 y^2$$A^2 - y^2 = 8 y^2$$A^2 = 9 y^2$$y = \pm \frac{A}{3}$From the given equation $y = \sqrt{3 \pi} \sin \left(\frac{100}{\pi} t + \frac{\pi}{4}ight)$, the amplitude $A = \sqrt{3 \pi}$.Therefore, the displacement $y = \frac{\sqrt{3 \pi}}{3} = \frac{\sqrt{3} \sqrt{\pi}}{\sqrt{3} \sqrt{3}} = \sqrt{\frac{\pi}{3}}$.

The displacement of a particle in simple harmonic motion $(SHM)$ is given by $y = \sqrt{3 \pi} \sin \left(\frac{100}{\pi} t + \frac{\pi}{4}\right)$. What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

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