Ncert Exercises & Solutions

Q. 38 The displacement vector of a particle of mass m is given by $\vec{r (t)} = \hat{i} A \cos ωt + \hat{j} B \sin ωt$ .

(a) Show that the trajectory is an ellipse.
(b) Show that $\vec{F} = - {mω}^{2} \vec{r (t)}$ .

Hint: Acceleration of the particle

a (t) = \frac{dv (t)}{dt} .

Step 1: Find the relation between x and y by eliminating t.

(a)

The displacement vector of the particle of mass m is given by

r (t) = \hat{i} Acos ωt + \hat{j} Bsin ωt

∴

Displacement along the x-axis is,

\begin{array}{l} x = Acos ωt \\ or \frac{x}{A} = \cos ωt . . . (i) \end{array}

Displacement along the y-axis is,

\begin{array}{l} and, y = Bsin ωt \\ or \frac{y}{B} = \sin ωt \end{array}

Squaring and then adding Eqs. (i) and (ii), we get

\frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = \cos^{2} ωt + \sin^{2} ωt = 1

This is an equation of the ellipse.
Therefore, the trajectory of the particle is an ellipse.

Step 2: Find velocity by using

\vec{v (t)} = \frac{d \vec{r (t)}}{dt} .

(b)

Velocity Of the particle

\begin{array}{l} \vec{v (t)} = \frac{d \vec{r (t)}}{dt} = \hat{i} \frac{d}{dt} (Acos ωt) + \hat{j} \frac{d}{dt} (Bsin ωt) \\ = \hat{i} [A (- \sin ωt) . ω] + \hat{j} [B (\cos ωt) . ω] \\ = - \hat{i} Aωsin ωt + \hat{j} Bωcos ωt \end{array}

Step 2: Find acceleration by using

\vec{a (t)} = \frac{d \vec{v (t)}}{d t} .

Acceleration of the particle

\vec{a (t)} = \frac{d \vec{v (t)}}{d t} .

\begin{array}{l} a = - \hat{i} Aω \frac{d}{d} (\sin ωt) + \hat{j} Bω \frac{d}{dt} (\cos ωt) \\ = - \hat{i} Aω [\cos ωt] \cdot ω + \hat{j} Bω [- \sin ωt], ω \\ = - \hat{i} {Aω}^{2} \cos ωt - \hat{j} {Bω}^{2} \sin ωt \\ = - ω^{2} [\hat{i} Acos ωt + \hat{j} Bsin ωt] \\ = - ω^{2} r \end{array}

Step 3: Find the force acting on the particle.

∴

Force acting on the particle =

\vec{F} = m \vec{a (t)} = - {mω}^{2} \vec{r (t)} .