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

Question

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

Accepted Answer

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

Answer

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

Answer

$ v = \sqrt[3]{v_0^3 + \omega^3 x^3} $

Answer

$ v = v_0 - (\omega^3 x^3 e^{x^3})^{1/3} $

AIIMS 2015 Physics Simple Harmonic Motion MCQ Question

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