A particle is executing SHM with an amplitude | vedclass

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Question

(A) The total energy $(E)$ of a particle executing $SHM$ is given by $E = \frac{1}{2} kA^2$,where $k$ is the force constant and $A$ is the amplitude.T

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

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(A) The total energy $(E)$ of a particle executing $SHM$ is given by $E = \frac{1}{2} kA^2$,where $k$ is the force constant and $A$ is the amplitude.The potential energy $(U)$ at a displacement $x$ is given by $U = \frac{1}{2} kx^2$.According to the problem,the potential energy is half of the total energy:$U = \frac{1}{2} E$Substituting the expressions for $U$ and $E$:$\frac{1}{2} kx^2 = \frac{1}{2} \left( \frac{1}{2} kA^2 \right)$Canceling $\frac{1}{2} k$ from both sides:$x^2 = \frac{1}{2} A^2$Taking the square root of both sides:$x = \frac{A}{\sqrt{2}}$

$A$ particle is executing $SHM$ with an amplitude $A$. What is the displacement of the particle when its potential energy is half of its total energy?

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