Question

The position vector of a particle from origin is given by $\overrightarrow{r}=A(\overrightarrow{i}\cos \omega t+\overrightarrow{j}\sin \omega t)$. The motion of the particle is

• A. simple harmonic
• B. on a straight line
• C. on a circle
• D. with constant acceleration

Answer 1

The correct option is A simple harmonic

Given,

$\overrightarrow{r}=A(\hat{i}cos\omega t+\hat{j}sin\omega t)$. . . . . .(1)

velocity, $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=A(-\hat{i}\omega sin\omega t+\hat{j}\omega cos\omega t)$

$\overrightarrow{v}=A\omega (-\hat{i}sin\omega t+\hat{j}cos\omega t)$

acceleration, $\overrightarrow{a}=\frac{d^2\overrightarrow{r}}{d{t}^2}=-A{\omega }^2(\hat{i}cos\omega t+\hat{j}sin\omega t)$

$\overrightarrow{a}=-A\omega \overrightarrow{r}$

$\overrightarrow{a}\propto (-\overrightarrow{r})$

The acceleration is directly proportional to the negative of displacement.

The motion is simple harmonic.

The correct option is A.