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The position vector of a particle R⃗\vec{R} as a function of time tt is given by: R⃗=4sin⁡(2πt)i^+4cos⁡(2πt)j^\vec{R} = 4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}, where RR is in meters, tt is in seconds and i^,j^\hat{i}, \hat{j} denote unit vectors along x and y-directions, respectively. Which one of the following statements is wrong for the motion of the particle?

A

Acceleration is along −R⃗-\vec{R}.

B

Magnitude of the acceleration vector is v2R\frac{v^2}{R}, where vv is the velocity of the particle.

C

Magnitude of the velocity of the particle is 8 m/s.

D

Path of the particle is a circle of radius 4 m.

Step-by-Step Solution

  1. Analyze the Path: The position coordinates are x=4sin⁡(2πt)x = 4\sin(2\pi t) and y=4cos⁡(2πt)y = 4\cos(2\pi t). Squaring and adding gives x2+y2=16(sin⁡2(2πt)+cos⁡2(2πt))=16x^2 + y^2 = 16(\sin^2(2\pi t) + \cos^2(2\pi t)) = 16. This represents a circle of radius R=4R = 4 m. (Statement 4 is correct).
  2. Calculate Velocity: Velocity is the time derivative of the position vector: v⃗=dR⃗dt\vec{v} = \frac{d\vec{R}}{dt}. v⃗=ddt[4sin⁡(2πt)i^+4cos⁡(2πt)j^]\vec{v} = \frac{d}{dt}[4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}] v⃗=4(2π)cos⁡(2πt)i^−4(2π)sin⁡(2πt)j^=8πcos⁡(2πt)i^−8πsin⁡(2πt)j^\vec{v} = 4(2\pi)\cos(2\pi t)\hat{i} - 4(2\pi)\sin(2\pi t)\hat{j} = 8\pi\cos(2\pi t)\hat{i} - 8\pi\sin(2\pi t)\hat{j}
  3. Magnitude of Velocity (Speed): ∣v⃗∣=(8π)2(cos⁡2(2πt)+sin⁡2(2πt))=8π≈25.1 m/s|\vec{v}| = \sqrt{(8\pi)^2 (\cos^2(2\pi t) + \sin^2(2\pi t))} = 8\pi \approx 25.1 \text{ m/s} The statement claiming the speed is 8 m/s is incorrect. (Statement 3 is wrong).
  4. Calculate Acceleration: Acceleration is the time derivative of velocity: a⃗=dv⃗dt\vec{a} = \frac{d\vec{v}}{dt}. a⃗=ddt[8πcos⁡(2πt)i^−8πsin⁡(2πt)j^]\vec{a} = \frac{d}{dt}[8\pi\cos(2\pi t)\hat{i} - 8\pi\sin(2\pi t)\hat{j}] a⃗=−16π2sin⁡(2πt)i^−16π2cos⁡(2πt)j^\vec{a} = -16\pi^2\sin(2\pi t)\hat{i} - 16\pi^2\cos(2\pi t)\hat{j} a⃗=−(2π)2[4sin⁡(2πt)i^+4cos⁡(2πt)j^]=−(2π)2R⃗\vec{a} = -(2\pi)^2 [4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}] = -(2\pi)^2\vec{R} The acceleration is directed along −R⃗-\vec{R} (towards the center). (Statement 1 is correct).
  5. Centripetal Acceleration Magnitude: In uniform circular motion, acceleration magnitude is ac=v2Ra_c = \frac{v^2}{R} . Let's check: v2R=(8π)24=64π24=16π2\frac{v^2}{R} = \frac{(8\pi)^2}{4} = \frac{64\pi^2}{4} = 16\pi^2. The magnitude of a⃗\vec{a} calculated above is also ∣−(2π)2R⃗∣=4π2(4)=16π2|-(2\pi)^2\vec{R}| = 4\pi^2(4) = 16\pi^2. They match. (Statement 2 is correct).
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