Question

The displacement vector of a particle of mass \(m\) is given by \(\overrightarrow{r}(t) = \hat{i} A \cos \omega t + \hat{j} B \sin \omega t\)

A Show that the trajectory is an ellipse.

B Show that \(\overrightarrow{F} = - m \omega^2 \overrightarrow{r}\)

Answer 1

Generally, displacement vector is given by,

\(r = x \hat{i} + y \hat{j}\)

\(x = A \cos \omega t, y = B \sin \omega t\)

\(\dfrac{x}{A} = \cos \omega t\) ...\((i)\)

\(\dfrac{y}{B} = \sin \omega t\) ...\((ii)\)

Squaring & adding \((i)\), \((ii)\)

\(\dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1\)

Which is the equation of ellipse.

\(\text{Final Answer}: \dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1\)


\(\text{Find velocity of the particle.}\)

Given,

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

Differentiate equation \((i)\) with respect to \(t\)

\(\dfrac{dr}{dt} = v = - \hat{i} \omega A \sin \omega t + \hat{j} \omega B \cos \omega t\) \((ii)\)

\(\text{Find acceleration of the particle.}\)

Differentiate equation \((ii)\) with respect to \(t\)

\(a = \dfrac{dv}{dt} = - \hat{i} \omega^2 A \cos \omega t - \hat{j} \omega^2 B \sin \omega t\)

\(a = - \omega^2 ( \hat{i} A \cos \omega t + \hat{j} B \sin \omega t) = - \omega^2 \overrightarrow{r}\)

We know, \(F = ma\)

\(\therefore\) \(\overrightarrow{F} = -m \omega^2\overrightarrow{r}\)