Question

The position of a particle moving along $x$ -axis is given by $x=x_{0}\left(cos\right)^{2}\left(\right.?t\left.\right)$ . 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)

$x_{0}\omega$

Answer 2

$x=x_{0}cos^{2}?t$ $\therefore x=x_{0}\left(\frac{cos 2 ?t + 1}{2}\right)$ $\Rightarrow x=\frac{x_{0}}{2}+\frac{x_{0}}{2}cos2?t$ $\therefore \frac{dx}{dt}=-\frac{x_{0}}{2}sin2?t \, . \, 2\omega$ $\Rightarrow \frac{dx}{dt}=-?x_{0}sin2?t$ So, the speed at mean position is $v_{0}=?x_{0}$