Question

The position vector of a particle $\overrightarrow{R}$ as a function of time is given by $\overrightarrow{R}=4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}$ where $R$ is in meters, $t$ is in seconds and $\hat{i}\text{and}\hat{j}$ denotes unit vectors along $x-\text{and}y-\text{directions}$, respectively. Which one of the following statements is wrong for the motion of particle?

• A. Magnitude of acceleration vector is $\frac{v^2}{R}$, where $v$ is the velocity of particle.
• B. Magnitude of the velocity of particle is $8\pi \text{m/s}$.
• C. Path of the particle is a circle of radius $4\text{m}$.
• D. Acceleration vector is along $-\overrightarrow{R}$.

Answer 1

The correct option is B Magnitude of the velocity of particle is $8\pi \text{m/s}$.

Given:
$\overrightarrow{R}=4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}$ .... (i)

The velocity of the particle is

$\overrightarrow{v}=\frac{d\overrightarrow{R}}{dt}$
$=\frac{d}{dt}\left[4\sin \right(2\pi t)\hat{i}+4\cos (2\pi t\left)\hat{j}\right]$

$\overrightarrow{v}=8\pi \cos \left(2\pi t\right)\hat{i}-8\pi \sin \left(2\pi t\right)\hat{j}$ .... (ii)

Its magnitude is

$\left|\overrightarrow{v}\right|=\sqrt{{\left(8\pi \cos \right(2\pi t\left)\right)}^2+{(-8\pi \sin (2\pi t\left)\right)}^2}$

$=\sqrt{128{\pi }^2}=8\pi \sqrt{2}$ $\text{m/s}$
$\because {\sin }^2\theta +{\cos }^2\theta =1$

The acceleration of particle is

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$
$=\frac{d}{dt}\left[8\pi \cos \cos \right(2\pi t)\hat{i}-8\sin \sin (2\pi t\left)\hat{j}\right]$

$\overrightarrow{a}=-16{\pi }^2\sin \sin \left(2\pi t\right)\hat{i}-16{\pi }^2\cos \cos \left(2\pi t\right)\hat{j}$ .... (iii)

From equation (i) and (iii)

$\overrightarrow{a}=-4{\pi }^2\overrightarrow{R}$

$\therefore$ Acceleration is along $-\overrightarrow{R}$

Magnitude of acceleration : $\left|\overrightarrow{a}\right|=|-{\left(2\pi \right)}^2\overrightarrow{R}|=4{\pi }^2\left|\overrightarrow{R}\right|$
$=16{\pi }^2\frac{{\left(8\pi \right)}^2}{4}=\frac{v^2}{R}$

$R_x=x=4\sin \left(2\pi t\right)$

$R_y=y=4\cos \left(2\pi t\right)$

$R_x^2+R_y^2=x^2+y^2=4^2$

$\Rightarrow$ The path of the particle is a circle of radius $4$.
Hence, option $\left(A\right)$ is correct option.