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NCERT Exemplar · Q8

Q.A particle P moves uniformly around a circle of radius BB centred at the origin O, with the xx-axis horizontal (positive to the right) and the yy-axis vertical (positive upward). At t=0t=0 the particle is at the topmost point of the circle, on the positive yy-axis, at coordinates (0, B)(0,\,B). It revolves in the clockwise sense with a period of 30 s30\ \text{s}. The simple harmonic motion of the xx-projection (the foot of the perpendicular onto the xx-axis) of the radius vector of the rotating particle P is

(a) x(t)=Bsin⁡(2πt30)x(t)=B\sin\left(\dfrac{2\pi t}{30}\right)
(b) x(t)=Bcos⁡(πt15)x(t)=B\cos\left(\dfrac{\pi t}{15}\right)
(c) x(t)=Bsin⁡(πt15+π2)x(t)=B\sin\left(\dfrac{\pi t}{15}+\dfrac{\pi}{2}\right)
(d) x(t)=Bcos⁡(πt15+π2)x(t)=B\cos\left(\dfrac{\pi t}{15}+\dfrac{\pi}{2}\right)
Chhattisgarh CgbseMCQ· 1mImportance★★★★★est
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The xx-coordinate of a particle in uniform circular motion is x=Bcos⁡θx=B\cos\theta, where θ\theta is the angle of the radius vector from the +x+x-axis. With ω=2π/T=π/15 rad s−1\omega=2\pi/T=\pi/15\,\text{rad s}^{-1}, a start at the top (θ=90∘\theta=90^\circ) turning clockwise gives θ(t)=90∘−ωt\theta(t)=90^\circ-\omega t, so x(t)=Bsin⁡(ωt)=Bsin⁡(2πt/30)x(t)=B\sin(\omega t)=B\sin(2\pi t/30).

Concept: SHM as a projection of circular motion

For a particle moving on a circle of radius BB, the projection of its position on the xx-axis executes SHM:

x(t)=Bcos⁡θ(t),θ(t)=θ0±ωt,x(t)=B\cos\theta(t),\qquad \theta(t)=\theta_0\pm\omega t,

with ++ for anticlockwise and −- for clockwise revolution.

Step 1 — angular frequency

ω=2πT=2π30=π15 rad s−1.\omega=\frac{2\pi}{T}=\frac{2\pi}{30}=\frac{\pi}{15}\ \text{rad s}^{-1}.

Step 2 — initial angle

At t=0t=0 the particle is on the +y+y-axis, so θ0=90∘=π/2\theta_0=90^\circ=\pi/2.

Step 3 — apply the clockwise sense

Clockwise means θ\theta decreases: θ(t)=π2−ωt\theta(t)=\dfrac{\pi}{2}-\omega t. Hence

x(t)=Bcos⁡ ⁣(π2−ωt)=Bsin⁡(ωt)=Bsin⁡ ⁣(πt15)=Bsin⁡ ⁣(2πt30).x(t)=B\cos\!\left(\frac{\pi}{2}-\omega t\right)=B\sin(\omega t)=B\sin\!\left(\frac{\pi t}{15}\right)=B\sin\!\left(\frac{2\pi t}{30}\right). …

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