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(B) The speed of a particle executing simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$, where $a$ is the amplitude, $\omega$ is the an

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$\frac{5}{8 \pi}$

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(B) The speed of a particle executing simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$, where $a$ is the amplitude, $\omega$ is the angular frequency, and $x$ is the displacement from the equilibrium position.Squaring both sides, we get $v^2 = \omega^2 (a^2 - x^2)$.For the two given cases:$v_1^2 = \omega^2 (a^2 - x_1^2) \implies 13^2 = \omega^2 (a^2 - 3^2) \implies 169 = \omega^2 (a^2 - 9) \quad (1)$$v_2^2 = \omega^2 (a^2 - x_2^2) \implies 12^2 = \omega^2 (a^2 - 5^2) \implies 144 = \omega^2 (a^2 - 25) \quad (2)$Subtracting equation $(2)$ from $(1)$:$169 - 144 = \omega^2 (a^2 - 9 - a^2 + 25)$$25 = \omega^2 (16)$$\omega^2 = \frac{25}{16} \implies \omega = \frac{5}{4} \ rad/s$.The frequency $f$ is related to angular frequency by $\omega = 2 \pi f$.Therefore, $f = \frac{\omega}{2 \pi} = \frac{5/4}{2 \pi} = \frac{5}{8 \pi} \ Hz$.

The velocity of a particle executing a simple harmonic motion is $13 \ m/s$, when its distance from the equilibrium position $(Q)$ is $3 \ m$ and its velocity is $12 \ m/s$, when it is $5 \ m$ away from $Q$. The frequency of the simple harmonic motion is

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