Question

The velocity vector $v$ and displacement $x$ of a particle executing SHM are related as $\dfrac{vdv}{dx}=-\omega^{2}x$, with initial condition $V=v_{0}$ at $x=0$. The velocity $v$, when displacement is $x$ is?
\[\begin{align}
  & A.v=\sqrt{v_{0}^{2}+{{\omega }^{2}}{{x}^{2}}} \\
 & B.v=\sqrt{v_{0}^{2}-{{\omega }^{2}}{{x}^{2}}} \\
 & C.v=\sqrt[3]{v_{0}^{3}+{{\omega }^{3}}{{x}^{3}}} \\
 & D.v={{v}_{0}}-{{({{\omega }^{3}}{{x}^{3}}{{e}^{{{x}^{3}}}})}^{\dfrac{1}{3}}} \\
\end{align}\]

Answer 1

Hint: We know that the general equation of the SHM displacement of particle is given as, $x=A\sin(\omega t)$, here, they have given a relationship between the velocity vector $v$ and displacement $x$ of a particle, using this equation, we can find the velocity vector $v$ at some displacement is $x$, by integrating the given equation.
Formula: $v=\dfrac{dx}{dt}$

Complete answer:
We know that the SHM or simple harmonic motion is the motion caused by the restoring force; it is directly proportional to the displacement of the object from its mean position. And is always directed towards the mean.
Given that $\dfrac{vdv}{dx}=-\omega^{2}x$. But we know that the velocity of the particle, whose displacement is known is defined as the rate of change of displacement with respect to time, and is mathematically given as $v=\dfrac{dx}{dt}$.
Now to find the velocity $v$ at some displacement is $x$, using the definition of velocity, we need to rearrange and integrate the given equations using appropriate limits. Rearranging the equation, we get,
$vdv=-\omega^{2}xdx$
 Here the limit of velocity varies from $v_{0}$ to $v$ and similarly the limit of the displacement $x$ varies from $x_{0}$ to $x$. Mathematically, it can be represented as,
$\implies\int_{v_{0}}^{v}vdv=\int_{x_{0}}^{x}-\omega^{2}xdx$
On integration and applying the limits, we get the following equations,
\[\begin{align}
  & \Rightarrow \left. \dfrac{{{v}^{2}}}{2} \right|_{{{v}_{0}}}^{v}=-{{\omega }^{2}}\left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{x} \\
 & \Rightarrow ({{v}^{2}}-v_{0}^{2})=-{{\omega }^{2}}{{x}^{2}} \\
 & \Rightarrow {{v}^{2}}=-{{\omega }^{2}}{{x}^{2}}+v_{0}^{2} \\
 & \therefore v=\sqrt{v_{0}^{2}-{{\omega }^{2}}{{x}^{2}}} \\
\end{align}\]

Thus, the correct answer is option \[B.v=\sqrt{v_{0}^{2}-{{\omega }^{2}}{{x}^{2}}}\]

Note:
Remember SHM motions are sinusoidal in nature. Assume, the particle is at mean when, $t=0$, $v=v_{0}$ and $x=0$. This makes the further steps easier. Since this question involves more of mathematics than physics, it is important to know some basic integration, to solve this sum. Also note that this is a very easy sum, provided one knows integration.