Question

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

• A. $v=\sqrt{v_0^2+{\omega }^2{x}^2}$
• B. $v=\sqrt{v_0^2-{\omega }^2{x}^2}$
• C. $v=3\sqrt{v_0^3+{\omega }^3{x}^3}$
• D. $v=v_0-({\omega }^3{x}^3{e}^{x^3}{)}^{1/3}$

Answer 1

Answer: (B)

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

As it is SHM so the equation of motion will be $F=-kx$ or $\frac{vdv}{dx}=-{\omega }^{2}x$

Now integrating the expression with boundary condition, ${\int }_{v_0}^{v}vdv=-{\omega }^2{\int }_0^{x}xdx$

or $\frac{1}{2}[v^2-v_0^2]=-\frac{{\omega }^2{x}^2}{2}$

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