If displacement x and velocity v are related | vedclass

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(C) The given equation is $4v^2 = 16 - x^2$.Dividing both sides by $16$,we get $\frac{4v^2}{16} + \frac{x^2}{16} = 1$,which simplifies to $\frac{v^2}{

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$4\pi$

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(C) The given equation is $4v^2 = 16 - x^2$.Dividing both sides by $16$,we get $\frac{4v^2}{16} + \frac{x^2}{16} = 1$,which simplifies to $\frac{v^2}{4} + \frac{x^2}{16} = 1$.Comparing this with the standard $SHM$ velocity-displacement relation $\frac{v^2}{(A\omega)^2} + \frac{x^2}{A^2} = 1$,we identify:$A^2 = 16 \Rightarrow A = 4$ and $(A\omega)^2 = 4 \Rightarrow A\omega = 2$.Substituting $A = 4$ into $A\omega = 2$,we get $4\omega = 2$,so $\omega = 0.5 	ext{ rad/s}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{0.5} = 4\pi 	ext{ s}$.

If displacement $x$ and velocity $v$ are related as $4v^2 = 16 - x^2$ in a $SHM$,then the time period of the given $SHM$ is (consider $SI$ units).

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