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(D) Given the equation relating velocity $v$ and displacement $x$ for a particle in simple harmonic motion $(SHM)$: $4v^2 = 25 - x^2$ Dividing by $4$,

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$4\pi \ s$

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(D) Given the equation relating velocity $v$ and displacement $x$ for a particle in simple harmonic motion $(SHM)$: $4v^2 = 25 - x^2$ Dividing by $4$, we get: $v^2 = \frac{1}{4}(25 - x^2) = \frac{1}{4}(5^2 - x^2)$ Taking the square root on both sides: $v = \frac{1}{2}\sqrt{5^2 - x^2} \quad ... (i)$ We know that the standard equation for velocity in $SHM$ is: $v = \omega\sqrt{A^2 - x^2} \quad ... (ii)$ Comparing equations $(i)$ and $(ii)$, we find: Angular frequency $\omega = \frac{1}{2} \ rad/s$ Amplitude $A = 5 \ m$ The time period $T$ is given by the formula: $T = \frac{2\pi}{\omega}$ Substituting the value of $\omega$: $T = \frac{2\pi}{1/2} = 4\pi \ s$

$A$ particle is exhibiting simple harmonic motion and has its displacement $x$ and velocity $v$ related as $4v^2 = 25 - x^2$. The time period of the $SHM$ is

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