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Exercises · 13.11

Q.Two particles each move uniformly on a circle lying in a plane, the centre of each circle at the origin of its own xx and yy axes. For each, obtain the simple harmonic motion of the xx-projection of the radius vector of the revolving particle P.
Motion (a): circle of radius 33 cm, period of revolution 22 s. At the instant t=0t = 0 the particle P is at the lowest point of the circle, on the negative yy-axis at coordinates (0, −3 cm)(0,\ -3\ \text{cm}), and it revolves in the clockwise sense.
Motion (b): circle of radius 22 m, period of revolution 44 s. At t=0t = 0 the particle P is at the left-most point of the circle, on the negative xx-axis at coordinates (−2 m, 0)(-2\ \text{m},\ 0), and it revolves in the anticlockwise sense.

Figure 13.20
Figure 13.20
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Each uniformly revolving particle casts an xx-projection that executes SHM with amplitude equal to the circle's radius and angular frequency ω=2π/T\omega = 2\pi/T. Reading off the starting point and the sense of rotation fixes the phase. Motion (a) gives x(t)=−3sin⁡(πt)x(t) = -3\sin(\pi t) cm; motion (b) gives x(t)=−2cos⁡(πt/2)x(t) = -2\cos(\pi t/2) m.

Concept - SHM as the projection of uniform circular motion

If a particle P moves uniformly on a circle of radius AA with angular speed ω\omega, its projection on the xx-axis is

x(t)=Acos⁡(ωt+ϕ),x(t) = A\cos(\omega t + \phi),

a simple harmonic motion of amplitude AA and angular frequency ω=2πT\omega = \dfrac{2\pi}{T}. The phase constant ϕ\phi is fixed by where P sits at t=0t = 0 and by the sense of revolution.

Motion (a) - radius 3 cm, T = 2 s, starts at the bottom, clockwise

Here A=3 cmA = 3\ \text{cm} and ω=2π2=π rad s−1\omega = \dfrac{2\pi}{2} = \pi\ \text{rad s}^{-1}.

At t=0t = 0, P is on the negative yy-axis, so its xx-coordinate is x(0)=0x(0) = 0. Turning clockwise from the bottom, P first swings toward the left (−x-x):

x(0)=0,x ⁣(T4)=x(0.5)=−3,x(1)=0,x(1.5)=+3 cm.x(0) = 0,\quad x\!\left(\tfrac{T}{4}\right) = x(0.5) = -3,\quad x(1) = 0,\quad x(1.5) = +3\ \text{cm}.

The function with x(0)=0x(0) = 0 that then becomes negative is

x(t)=−3sin⁡(πt) cm  =  3cos⁡ ⁣(πt+π2) cm.x(t) = -3\sin(\pi t)\ \text{cm} \;=\; 3\cos\!\left(\pi t + \tfrac{\pi}{2}\right)\ \text{cm}.

Motion (b) - radius 2 m, T = 4 s, starts at the left extreme …

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