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

To solve this problem, we need to understand the motion of the particle P as it revolves around the center O in a circular path with radius r and angular velocity \omega. We are focusing on the projection of the line OP on the x-axis at a given time t.

The general formula for the projection of a revolving vector on an axis is based on the angle it makes with the reference axis. Here, the angle is given by the angular displacement \theta = \omega t.

From the diagram provided, the particle P starts its motion at an initial angle of \frac{\pi}{6} (30 degrees), going counterclockwise.

The x-coordinate of the particle can therefore be determined by considering both the angular velocity and the initial phase:

• The displacement \theta at time t is given by \omega t + \frac{\pi}{6}, since it starts from \frac{\pi}{6}.
• The projection of OP on the x-axis is thus r \cos(\omega t + \frac{\pi}{6}).

Hence, the correct expression for the projection of OP on the x-axis at time t is:

x(t) = r \cos(\omega t + \frac{\pi}{6})

Thus, the correct option is:

x(t) = rcos (ωt + $\frac{π}{6}$)

This solution accounts for the initial phase angle introduced in the problem, providing a precise mathematical representation of the motion.