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(B) In simple harmonic motion, the force is given by $F = -kx = -m\omega^2 x$. The maximum force is $F_{max} = m\omega^2 A$. Given that the force at a

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(B) In simple harmonic motion, the force is given by $F = -kx = -m\omega^2 x$. The maximum force is $F_{max} = m\omega^2 A$. Given that the force at a position $x$ is $86.6 \%$ of $F_{max}$, we have $F = 0.866 F_{max} = \frac{\sqrt{3}}{2} F_{max}$. Since $F = m\omega^2 x$ and $F_{max} = m\omega^2 A$, we get $x = \frac{\sqrt{3}}{2} A$. The velocity of a particle in $SHM$ at position $x$ is $v = \omega \sqrt{A^2 - x^2}$. Substituting $x = \frac{\sqrt{3}}{2} A$, we get $v = \omega \sqrt{A^2 - (\frac{\sqrt{3}}{2} A)^2} = \omega \sqrt{A^2 - \frac{3}{4} A^2} = \omega \sqrt{\frac{1}{4} A^2} = \frac{1}{2} \omega A$. The maximum velocity is $v_{max} = \omega A$. Therefore, the ratio of velocity at that point to the maximum velocity is $\frac{v}{v_{max}} = \frac{\frac{1}{2} \omega A}{\omega A} = \frac{1}{2}$.

$A$ particle is executing simple harmonic motion. If the force acting on the particle at a position is $86.6 \%$ of the maximum force on it, then the ratio of its velocity at that point and its maximum velocity is

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