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)

$x=x_{0}co{s}^2\omega t$
$\therefore x=x_0\left(\frac{cos2\omega t+1}{2}\right)$
$\Rightarrow x=\frac{x_0}{2}+\frac{x_0}{2}cos2\omega t$
$\therefore \frac{dx}{dt}=-\frac{x_0}{2}sin2\omega t.2\omega$
$\Rightarrow \frac{dx}{dt}=-\omega x_{0}sin2\omega t$
So, the speed at mean position is $v_0=\omega x_0$