Question

For particle P revolving round the centre O with radius of circular path r and angular velocity ω, as shown in below figure, the projection of OP on the x-axis at time t is

• A. x(t) = rcos (ωt )
• B. x(t) = rcos (ωt - $\frac{π}{6}$ ω)
• C. x(t) = rcos (ωt + $\frac{π}{6}$)
• D. x(t) = rsin (ωt + $\frac{π}{6}$)

Answer 1

The Correct Option is C

Let the particle \( P \) revolve around the center \( O \) with a uniform angular velocity \( \omega \). The angular displacement of the particle at time \( t \) is: \[ \theta = \omega t. \] From the figure, the angle of the particle from the \( x \)-axis at \( t = 0 \) is \( \frac{\pi}{6} \). Hence, at time \( t \), the total angle with respect to the \( x \)-axis is: \[ \theta_{\text{total}} = \omega t + \frac{\pi}{6}. \] The projection of \( OP \) on the \( x \)-axis is given by: \[ x(t) = r \cos\left(\theta_{\text{total}}\right). \] Substitute \( \theta_{\text{total}} = \omega t + \frac{\pi}{6} \): \[ x(t) = r \cos\left(\omega t + \frac{\pi}{6}\right). \] Final Answer: The projection of \( OP \) on the \( x \)-axis is: \[ \boxed{x(t) = r \cos\left(\omega t + \frac{\pi}{6}\right)}. \]