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Question

(C) The total energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$.Potential energy $U$ at displacement $x$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The total energy $E$ of a particle in simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$.Potential energy $U$ at displacement $x$ is given by $U = \frac{1}{2} m \omega^2 x^2$.According to the problem, the potential energy is half of the total energy, so $U = \frac{E}{2}$.Substituting the expressions, we get $\frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \left( \frac{1}{2} m \omega^2 A^2 \right)$.Simplifying the equation, we get $x^2 = \frac{A^2}{2}$.Taking the square root on both sides, we find $x = \pm \frac{A}{\sqrt{2}}$.

$A$ particle is executing linear simple harmonic motion of amplitude $A$. At what displacement is the energy of the particle half potential and half kinetic?

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