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(C) The kinetic energy $K$ of a particle executing simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$,where $k$

Vedclass · Accepted Answer

$5: 4$

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(C) The kinetic energy $K$ of a particle executing simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k (A^2 - x^2)$,where $k$ is the force constant and $A$ is the amplitude.At displacement $x_1 = \frac{A}{4}$,the kinetic energy is $K_1 = \frac{1}{2} k (A^2 - (\frac{A}{4})^2) = \frac{1}{2} k (A^2 - \frac{A^2}{16}) = \frac{1}{2} k (\frac{15A^2}{16})$.At displacement $x_2 = \frac{A}{2}$,the kinetic energy is $K_2 = \frac{1}{2} k (A^2 - (\frac{A}{2})^2) = \frac{1}{2} k (A^2 - \frac{A^2}{4}) = \frac{1}{2} k (\frac{3A^2}{4}) = \frac{1}{2} k (\frac{12A^2}{16})$.The ratio of the kinetic energies is $\frac{K_1}{K_2} = \frac{\frac{15A^2}{16}}{\frac{12A^2}{16}} = \frac{15}{12} = \frac{5}{4}$.

$A$ particle is executing simple harmonic motion with amplitude $A$. The ratio of the kinetic energies of the particle when it is at displacements of $\frac{A}{4}$ and $\frac{A}{2}$ from the mean position is

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