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(B) Given: Time period $T = 3 \,s$, Displacement $x = 0.6 \,A$ (where $A$ is the amplitude).The kinetic energy $(K.E.)$ of a particle in simple harmon

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$16: 9$

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(B) Given: Time period $T = 3 \,s$, Displacement $x = 0.6 \,A$ (where $A$ is the amplitude).The kinetic energy $(K.E.)$ of a particle in simple harmonic motion is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(P.E.)$ of the particle is given by $P.E. = \frac{1}{2} m \omega^2 x^2$.The ratio of kinetic energy to potential energy is $\frac{K.E.}{P.E.} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2} = \left(\frac{A}{x}\right)^2 - 1$.Substituting $x = 0.6 \,A = \frac{6}{10} \,A = \frac{3}{5} \,A$, we get $\frac{A}{x} = \frac{5}{3}$.Therefore, the ratio is $\left(\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{25 - 9}{9} = \frac{16}{9}$.

$A$ particle is executing simple harmonic motion with a time period of $3 \,s$. At a position where the displacement of the particle is $60 \%$ of its amplitude, the ratio of the kinetic and potential energies of the particle is

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