Question

The coordinates of a particle moving in a plane are given by x(t) = a cos (pt) and y(t) = b sin (pt) where a, b (<a) and p are positive constants of appropriate dimensions. Then

• A. the path of the particle is an ellipse
• B. the velocity and acceleration of the particle are normal to each other at $t=\pi /\left(2p\right)$
• C. the acceleration of the particle is always directed towards a focus
• D. the distance travelled by the particle in time interval $t=0tot=\pi /\left(2p\right)$ is a

Answer 1

Answer: (A), (B), (C)

the acceleration of the particle is always directed towards a focus

$x=acospt\Rightarrow cos\left(pt\right)=\frac{x}{a}$ ...(i)
$y=bsinpt\Rightarrow sin\left(pt\right)=\frac{y}{b}$ ...(ii)

Squaring and adding (i) and (ii), we get
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore$ Path of the particle is in ellipse.

Hence option (a) is correct.

From the given equations
We can find,
$\frac{dx}{dt}=v_x=-apsinpt$

$\frac{d^2}{x}=a_x=-a{p}^{2}cospt$
$\frac{dy}{dt}=v_y=bpcospt$
and
$\frac{d^{2}y}{dt}=a_y=-b{p}^{2}sinpt$

At time $t=\frac{\pi }{2p}orpt=\frac{\pi }{2}$
$a_{x}and{v}_y$ become zero (because cos $\frac{\pi }{2}=0)only{v}_{x}and{a}_{y}areleft.$
or we can say that velocity is along negative x-axis and acceleration along y-axis.

Hence, at $t=\frac{\pi }{2p},$ velocity and acceleration force the particle are normal to each other. So option (b) is also correct. At t = t, position of the particle $\overrightarrow{r}\left(t\right)=x\hat{i}+y\hat{j}=acospt\hat{i}+bsinpt\hat{j}$ and acceleration of the particle is $\overrightarrow{a}\left(t\right)=a_x\hat{i}+a_y\hat{j}=p^2[acospt\hat{i}+bsinpt\hat{j}]$
$=p^2[x\hat{i}+y\hat{j}=-p^2\overrightarrow{r}(t\left)\right]$

Therefore acceleration of the particle is always directed towards origin.

Hence option (c) is also correct.

At t = 0, particle is at (a, 0) and at $t=\frac{\pi }{2p},$ particle is at (0, b). Therefore, the distance covered is one fourth of the elliptical path not a.

Hence option (d) is wrong.