The ratio of kinetic energy to the potential | vedclass

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(A) The kinetic energy $(KE)$ of a particle executing $SHM$ is given by:$KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$Given the distance $y = \frac{A

Vedclass · Accepted Answer

$3:1$

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(A) The kinetic energy $(KE)$ of a particle executing $SHM$ is given by:$KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$Given the distance $y = \frac{A}{2}$,where $A$ is the amplitude:$KE = \frac{1}{2} m \omega^{2} (A^{2} - (\frac{A}{2})^{2}) = \frac{1}{2} m \omega^{2} (A^{2} - \frac{A^{2}}{4}) = \frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})$The potential energy $(PE)$ of a particle executing $SHM$ is given by:$PE = \frac{1}{2} m \omega^{2} y^{2}$Substituting $y = \frac{A}{2}$:$PE = \frac{1}{2} m \omega^{2} (\frac{A}{2})^{2} = \frac{1}{2} m \omega^{2} (\frac{A^{2}}{4})$The ratio of kinetic energy to potential energy is:$\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^{2} (\frac{3A^{2}}{4})}{\frac{1}{2} m \omega^{2} (\frac{A^{2}}{4})} = \frac{3/4}{1/4} = 3$Therefore,the ratio is $3:1$.

The ratio of kinetic energy to the potential energy of a particle executing $SHM$ at a distance equal to half its amplitude,the distance being measured from its equilibrium position,is:

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