Multicorrect for JEE from Irodov - SHM

The answer to this question should be available in most standard course books of physics.

It is assumed that the reader is familiar with Simple Harmonic Motion . Since the question asks only for possibility, we consider two cases:

Case I: One dimensional motion Just because displacement (and hence, acceleration) and velocity may be just along one line since the start of motion, it is possible that the trajectory is a straight line.

Case II: Two dimensional motion Firstly, we justify that the path cannot be open. Since the velocity is finite, a finite force acting for a finite time, can bring the velocity to zero (in the corresponding direction). The force in our case, increases with distance from the origin, and so, it can bring down to zero any finite velocity. (There is a more better way of justifying this at Motion in Two Dimensional Harmonic Potential .) Thus, a hyperbola is not possible.

Next, we write the acceleration in component form: $a_x\hat{i}+a_y\hat{j}=-w^2(x\hat{i}+y\hat{j}$ , or simply $a_x=-w^2x$ and $a_y=-w^2y$ . The solution to this is (can be) $x=asin^2(wt+\alpha), y=bsin^2(wt+\beta)$ . In a special case (when $\beta-\alpha=\pi/2$ ), we can write $x=asin^2\theta, y=bcos^2\theta$ . Eliminating $\theta$ , we get $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (equation of an ellipse), which when $a=b$ reduces to the equation of a circle.

The equation of an ellipse can also be obtained even when $\beta-\alpha=\pi/2$ is not true, but themath involved is 'bad'. It can be found at Motion in Two Dimensional Harmonic Potential .

Thus, 1101 is the answer

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