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(B) Let the amplitude be $A = 6 \ cm$ and the displacement from the mean position be $x$.The potential energy $(U)$ of a particle in simple harmonic m

Vedclass · Accepted Answer

$4$

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(B) Let the amplitude be $A = 6 \ cm$ and the displacement from the mean position be $x$.The potential energy $(U)$ of a particle in simple harmonic motion is given by $U = \frac{1}{2} k x^2$.The kinetic energy $(K)$ of the particle is given by $K = \frac{1}{2} k (A^2 - x^2)$.Given the ratio of potential energy to kinetic energy is $U/K = 4/5$.Substituting the expressions, we get $\frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{4}{5}$.This simplifies to $\frac{x^2}{A^2 - x^2} = \frac{4}{5}$.Cross-multiplying gives $5x^2 = 4(A^2 - x^2) = 4A^2 - 4x^2$.Rearranging terms, $9x^2 = 4A^2$, which implies $x^2 = \frac{4}{9} A^2$.Taking the square root, $x = \frac{2}{3} A$.Given $A = 6 \ cm$, we find $x = \frac{2}{3} 	imes 6 \ cm = 4 \ cm$.

The amplitude of a particle executing simple harmonic motion is $6 \ cm$. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4:5$ is (in $cm$)

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