Question

The co-ordinates of a particle moving in a plane are given by $x\left(t\right)=a\cos \left(pt\right)$ and $y\left(t\right)=b\sin \left(pt\right)$ where $a,b(b<a)$ 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 $t=\frac{\pi }{2p}$
• C. The acceleration of the particle is always directed towards origin.
• D. The distance travelled by the particle in time internal $t=0$ to $t=\frac{\pi }{2p}$ is $a$.

Answer 1

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

The acceleration of the particle is always directed towards origin.

Given:
$x\left(t\right)=a\cos \left(pt\right)$

$\Rightarrow \cos \left(pt\right)=\frac{x}{a}.......\left(1\right)$

$y\left(t\right)=b\sin \left(pt\right)$

$\Rightarrow \sin \left(pt\right)=\frac{y}{b}.........\left(2\right)$

From Eq.(1) and (2),

${\sin }^2\left(pt\right)+{\cos }^2\left(pt\right)=1$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

This is the equation of an ellipse.

Also,
$\overrightarrow{r}\left(t\right)=a\cos \left(pt\right)\hat{i}+b\sin \left(pt\right)\hat{j}......\left(3\right)$

So, $\overrightarrow{v}\left(t\right)=-ap\sin \left(pt\right)\hat{i}+bp\cos \left(pt\right)\hat{j}....\left(4\right)$

and
$\overrightarrow{a}\left(t\right)=-a{p}^2\cos \left(pt\right)\hat{i}-b{p}^2\sin \left(pt\right)\hat{j}....\left(5\right)$

Now, at $t=\frac{\pi }{2p}$,
from eq. (4) and eq. (5)

$\overrightarrow{v}=-ap\hat{i};\overrightarrow{a}=-b{p}^2\hat{j}$

Thus, $\overrightarrow{v}$ and $\overrightarrow{a}$ are perpendicular to each other.

From (3) and (5),

$\overrightarrow{a}\left(t\right)=-p^2\overrightarrow{r}\left(t\right)$
[radial acceleration]

Therefore, acceleration of the particle is always directed towards the origin.



Now, at $t=0$, from eq.(3)
$\overrightarrow{r_1}=a\hat{i}$

At $t=\frac{\pi }{2p};{\overrightarrow{r}}_2=b\hat{j}$

Hence, displacement of the particle is equal to length $\text{AB}=\sqrt{a^2+b^2}$.

Therefore, option $\left(a\right),\left(b\right)$ and $\left(c\right)$ are the correct answer.

Why this question ? This is a very beautiful question which check your basic understanding of kinematics and it's graphical interpretation.