The amplitude of an SHM particle is 4 , cm. A | vedclass

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(B) Let the potential energy be $U$ and kinetic energy be $K$. We are given that $U = K$.The potential energy of a particle in $SHM$ at a distance $x$

Vedclass · Accepted Answer

$2\sqrt{2} \, cm$

Vedclass · Answer

(B) Let the potential energy be $U$ and kinetic energy be $K$. We are given that $U = K$.The potential energy of a particle in $SHM$ at a distance $x$ from the mean position is $U = \frac{1}{2} k x^2$.The kinetic energy of the particle at distance $x$ is $K = \frac{1}{2} k (A^2 - x^2)$,where $A$ is the amplitude.Setting $U = K$,we get:$\frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2)$Canceling $\frac{1}{2} k$ from both sides:$x^2 = A^2 - x^2$$2x^2 = A^2$$x^2 = \frac{A^2}{2}$$x = \pm \frac{A}{\sqrt{2}}$Given $A = 4 \, cm$,we have:$x = \pm \frac{4}{\sqrt{2}} = \pm 2\sqrt{2} \, cm$.Thus,the potential energy and kinetic energy are equal at a distance of $2\sqrt{2} \, cm$ from the mean position.

The amplitude of an $SHM$ particle is $4 \, cm$. At what distance from the mean position will the potential energy and kinetic energy be equal?

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