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Worked Examples · Example 13.4

Q.The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the xx-projection of the radius vector of the rotating particle P in each case. Motion (a): a circle of radius AA, period of revolution T=4T = 4 s. At t=0t = 0 the particle P is at an angle of 45°45° above the positive xx-axis, and it revolves in the anticlockwise sense. Motion (b): a circle of radius BB, period of revolution T=30T = 30 s. At t=0t = 0 the particle P is at the topmost point of the circle, on the positive yy-axis, and it revolves in the clockwise sense.

Figure — two circular motions
Figure
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Write the position vector's angle with the xx-axis as a function of time, then take

its xx-projection x(t)=Rcos⁡(angle)x(t) = R\cos(\text{angle}). Case (a): x(t)=Acos⁡(π2t+π4)x(t) = A\cos\left(\frac{\pi}{2}t + \frac{\pi}{4}\right), SHM of period 4 s. Case (b): x(t)=Bsin⁡(π15t)x(t) = B\sin\left(\frac{\pi}{15}t\right), SHM of period 30 s.

Why This Approach Works

When a particle P moves uniformly on a circle of radius RR with angular speed

ω\omega, its position vector makes an angle θ(t)\theta(t) with the +x+x-axis that

changes linearly with time. The xx-projection of P, x(t)=Rcos⁡θ(t)x(t) = R\cos\theta(t), then

automatically executes SHM — this is exactly the reference-circle connection between

uniform circular motion and SHM developed in this section. The only work needed per

case is to write down θ(t)\theta(t) correctly from the given initial angle, period, and

sense of rotation (clockwise subtracts from the initial angle; anticlockwise adds to

it).

Step-by-Step Solution

Case (a): radius AA, T=4T = 4 s, initial angle 45°=π/445° = \pi/4, anticlockwise.

The angular speed is ω=2πT=2π4=π2 rad/s\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\ \text{rad/s}.

Since the motion is anticlockwise, the angle with the xx-axis increases with time:

θ(t)=π2t+π4\theta(t) = \frac{\pi}{2}t + \frac{\pi}{4}

The xx-projection is therefore

x(t)=Acos⁡(π2t+π4)x(t) = A\cos\left(\frac{\pi}{2}t + \frac{\pi}{4}\right)

This is SHM of amplitude AA, angular frequency π/2\pi/2 rad/s (period 4 s), and

initial phase π/4\pi/4.

Case (b): radius BB, T=30T = 30 s, initial angle 90°=π/290°=\pi/2 (on the +y+y-axis), clockwise.

The angular speed is ω=2πT=2π30=π15 rad/s\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{30} = \dfrac{\pi}{15}\ \text{rad/s}. …

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