The velocity vector v and displacement vector | vedclass

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(B) Given the differential equation for $SHM$: $\frac{v dv}{dx} = -\omega^2 x$.To find the velocity $v$ at displacement $x$,we integrate both sides wi

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$v = \sqrt{v_0^2 - \omega^2 x^2}$

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(B) Given the differential equation for $SHM$: $\frac{v dv}{dx} = -\omega^2 x$.To find the velocity $v$ at displacement $x$,we integrate both sides with respect to their variables:$\int_{v_0}^{v} v dv = \int_{0}^{x} -\omega^2 x dx$.Performing the integration:$\left[ \frac{v^2}{2} \right]_{v_0}^{v} = -\omega^2 \left[ \frac{x^2}{2} \right]_{0}^{x}$.$\frac{1}{2}(v^2 - v_0^2) = -\frac{\omega^2 x^2}{2}$.Multiplying by $2$ on both sides:$v^2 - v_0^2 = -\omega^2 x^2$.Rearranging for $v$:$v^2 = v_0^2 - \omega^2 x^2$.Taking the square root:$v = \sqrt{v_0^2 - \omega^2 x^2}$.

The velocity vector $v$ and displacement vector $x$ of a particle executing $SHM$ are related as $\frac{v dv}{dx} = -\omega^2 x$ with the initial condition $v = v_0$ at $x = 0$. The velocity $v$,when displacement is $x$,is

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