The velocity vector $v$ and displacement vector $x$ of a particle executing SHM are related as \[V\dfrac{{dv}}{{dx}} = - {w^2}x\] with the initial condition \[v = {v_0}\]at \[x = 0\] the velocity \[{v_s}\] when displacement is $x$, is
(A) \[v = \sqrt {v{}_0^2 + {w^2}{x^2}} \]
(B) \[v = \sqrt {v{}_0^2 - {w^2}{x^2}} \]
(c) \[v = \sqrt {v{}_0^3 + {w^3} + {x^3}} \]
(D) \[v = {v_0} - {({w^3}{x^3}{e^x}^{^3})^{\dfrac{1}{3}}}\]

Answer

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Hint: In mechanics and physics, simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. It is vibratory motion in a system in which the restoring force is proportional to the displacement from equilibrium.

Complete step by step solution:

As it is Simple harmonic motion so the equation of motion will be

\[F = - kx\]

Or the equation can be written as

\[\dfrac{{vdv}}{{dx}} = - {\omega ^2}x\]

Now integrating the expression with boundary condition,

\[\int\limits_{{v_0}}^v {vdv} = - {\omega ^2}\int\limits_0^x {xdx} \]

After integrating the above equation we get

$\left[ {\dfrac{{{v^2}}}{2}} \right]_{{v_0}}^v = - {\omega ^2}\left[ {\dfrac{{{x^2}}}{2}} \right]_0^x$

Now we have to apply the limit in the above equation so we get

$\left[ {\dfrac{{{v^2}}}{2} - \dfrac{{v_0^2}}{2}} \right] = - {\omega ^2}\left[ {\dfrac{{{x^2}}}{2}} \right]$

Now we have to do further calculation then we get

$\dfrac{1}{2}\left[ {{v^2} - v_0^2} \right] = \dfrac{{ - {\omega ^2}{x^2}}}{2}$

Now after simplifying the above equation we get

$v = \sqrt {v{}_0^2 - {\omega ^2}{x^2}} $

Hence the correct answer is option is (B).

Note:

In SHM, the total energy is a constant and the velocity is maximum at equilibrium position and the acceleration is maximum at extreme position. Also, kinetic energy is maximum and potential energy is minimum at the mean position. In the extreme positions, kinetic energy is minimum and the potential energy is maximum.

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