Question

The position of a particle moving along $x$ -axis is given by $\mathrm{x}=\mathrm{x}_0 \cos ^2(\omega \mathrm{t})$. . Its speed when it is at mean position is

• A. $2x_{0}\omega $
• B. $x_{0}\omega ^{2}$
• C. $\frac{x_{0} \omega }{2}$
• D. $x_{0}\omega $

Answer 1

Answer: (D)

$x=x_{0}cos^{2}ωt$
$\therefore x=x_{0}\left(\frac{cos 2 ωt + 1}{2}\right)$
$\Rightarrow x=\frac{x_{0}}{2}+\frac{x_{0}}{2}cos2ωt$
$\therefore \frac{dx}{dt}=-\frac{x_{0}}{2}sin2ωt \, . \, 2\omega $
$\Rightarrow \frac{dx}{dt}=-ωx_{0}sin2ωt$
So, the speed at mean position is $v_{0}=ωx_{0}$