JEE Main 2023 · 8 April · evening · Single correct · +4 / −1
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)=cos(ω t-π/6 ω)
- (b) x(t)=cos(ω t)
- (c) x(t)=r cos (ω t+π/6)
- (d) x(t)=r sin (ω t+π/6)
Answer and solution
Answer (c) x(t)=r cos (ω t+π/6)
Solution
Physics sir · 3 steps
Find the angle OP makes with the x-axis at time t, then take r times its cosine.
- Starting angle. At t = 0, P is already 30° above the x-axis. In radians that is π/6, so the initial phase is π/6.
- Angle at time t. P moves anticlockwise, away from the x-axis, at angular velocity ω. In time t it sweeps an extra angle ω t, so OP makes an angle ω t + π/6 with the x-axis.
x(t) = r cos(ω t + π/6)
- So. The projection of a radius r on the x-axis is r times the cosine of its angle with that axis. This gives option (c). The period T = 6 s only fixes ω = 2π/6 = π/3 rad s⁻¹ and does not change the form.
The idea to keep. For a point on a circle, the x-projection is r cos(angle) and the y-projection is r sin(angle), where the angle is ω t plus the starting angle from the x-axis.
Answer from the NTA final answer key. Solution by PrepWiser.