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Question

(C) The total energy of a particle in $SHM$ is given by $E = \frac{1}{2} m \omega^2 a^2$.The kinetic energy $(K)$ of the particle at displacement $y$

Vedclass · Accepted Answer

$y=\frac{a}{2}$

Vedclass · Answer

(C) The total energy of a particle in $SHM$ is given by $E = \frac{1}{2} m \omega^2 a^2$.The kinetic energy $(K)$ of the particle at displacement $y$ is given by $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.Given that $K = \frac{3E}{4}$,we substitute the expressions for $K$ and $E$:$\frac{1}{2} m \omega^2 (a^2 - y^2) = \frac{3}{4} \left( \frac{1}{2} m \omega^2 a^2 \right)$.Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides,we get:$a^2 - y^2 = \frac{3}{4} a^2$.Rearranging the terms to solve for $y^2$:$y^2 = a^2 - \frac{3}{4} a^2 = \frac{1}{4} a^2$.Taking the square root of both sides,we find:$y = \frac{a}{2}$.

$A$ particle starts executing simple harmonic motion $(SHM)$ of amplitude $a$ and total energy $E$. At any instant,its kinetic energy is $\frac{3E}{4}$. Then its displacement $y$ is given by:

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