Oscillations

JEE Main level, single correct, medium

22 of 194

Question 22 of 194

For particle PP revolving round the centre OO with radius of circular path rr and regular velocity ω\omega, as shown in below figure, the projection of OPOP on the xx-axis at time tt is

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Answer: (B) x(t)=rcos⁡(ωt+π6)x(t)=r\cos \left(\omega t+\dfrac{\pi }{6}\right)

As can be seen from the diagram, the initial phase of the particle is given by
ϕ=30∘=π6\begin{matrix}\phi & = & 30^{\circ} \\ & = & \dfrac{\pi }{6}\end{matrix}
The equation of motion of the particle at its initial position at time
t=0t=0 can be written as
x(t)=rcos⁡θ...(1)x\left(t\right)=r\cos \theta ...(1)
where,
θ\theta is the angular displacement of the particle at time t=tt=t.
Since, the position of the particle at a later instant of time always lags behind its initial position at time
t=0t=0, the equation of motion of the particle at any instant of time is given by
x(t)=rcos⁡(θ+ϕ)...(2)x\left(t\right)=r\cos \left(\theta +\phi \right)...\left(2\right)
Substitute the values of the known parameters into equation (2) to obtain the required expression

x(t)=rcos⁡(ωt+π6)x(t)=r\cos \left(\omega t+\dfrac{\pi }{6}\right)

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