The velocity vector v and displacement vector x of a particle executing SHM are related as Vdvdx-w2x with the initial condition v-v0 at x-0- The velocity vs when displacement is x- is

The correct option is B v=√( v _ 0 ^ 2 w ^ 2 x ^ 2 ) Hence the correct option is B. #160; ...

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Standard IX

Physics

Fundamental and Derived Units

The velocity ...

Question

The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $V \frac{d v}{d x} = - w^{2} x$ with the initial condition $v = v_{0}$ at $x = 0$ . The velocity $v_{s}$ when displacement is $x$ , is

v = \sqrt{v_{0}^{2} + w^{2} x^{2}}

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v = \sqrt{v_{0}^{2} - w^{2} x^{2}}

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v = \sqrt{v_{0}^{3} + w^{3} + x^{3}}

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v = v_{0} - {(w^{3} x^{3} e^{x^{3}})}^{1 / 3}

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Solution

The correct option is B $v = \sqrt{v_{0}^{2} - w^{2} x^{2}}$

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The velocity vector v and displacement vector x of a particle executing SHM are related as Vdvdx=−w2x with the initial condition v=v0 at x=0. The velocity vs when displacement is x, is

The correct option is B v=√v20−w2x2Hence the correct option is B.

Hence the correct option is B.

The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as $V \frac{d v}{d x} = - w^{2} x$ with the initial condition $v = v_{0}$ at $x = 0$ . The velocity $v_{s}$ when displacement is $x$ , is

The correct option is B $v = \sqrt{v_{0}^{2} - w^{2} x^{2}}$