Question

The velocity v and displacement x of a particle executing simple harmonic motion are related as

$v\frac{dv}{dx}=-{\omega }^{2}x$

At $x=0,v=v_0$. Find the velocity v when the displacement becomes x.

• A. $v=\sqrt{v_0^2-{\omega }^2{x}^2}$
• B. $\sqrt{v_0^2+{\omega }^2{x}^2}$
• C. $v=\sqrt{v_0-{\omega }_0^{2}x}$
• D. None of these

Answer 1

The correct option is A

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

We have, $v\frac{dv}{dx}=-{\omega }^{2}x$

or, $vdv=-{\omega }^{2}xdx$

or, $\underset{v_0}{\overset{v}{\int }}vdv=\underset{0}{\overset{x}{\int }}-{\omega }^{2}xdx$ .......(i)

When summation is made on $-{\omega }^{2}xdx$ the quantity to be varied is x. When summation is made on $vdv$ the quantity to be varied is $v$. As x varies from 0 to x the velocity varies from $v_0$ to v. Therefore, on the left the limits of integration are from $v_0$ to v and on the right they are from 0 to x. Simplifying (i),

$[\frac{1}{2}{v}^2{]}_{v_0}^v=-{\omega }^2[\frac{x^2}{2}{]}_0^x$

or, $\frac{1}{2}(v^2-v_0^2)=-{\omega }^2\frac{x^2}{2}$

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

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