JEE Main 2015 · 10 Apr · Shift 1 · Physics Q1

← JEE Main 2015 · 10 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 1 of 90

$x$ and $y$ displacements of a particle are given as $x(t)=a\sin\omega t$ and $y(t)=a\sin2\omega t$ . Its trajectory will look like :

A tilted ellipse

Trajectory in a square box with dashed x and y axes: an ellipse tilted along the line y = x.

An inverted parabola-like arch

Trajectory in a square box with dashed x and y axes: an arch rising from the bottom corners to a peak at the top centre.

A figure-of-eight with lobes on the left and right of the origin

Trajectory in a square box with dashed x and y axes: two lobes, one on each side of the y-axis, crossing at the origin.

A figure-of-eight with lobes above and below the origin

Trajectory in a square box with dashed x and y axes: two lobes, one above and one below the x-axis, crossing at the origin.

Answer and worked solution

Official answer: C

$y=a\sin2\omega t=2a\sin\omega t\cos\omega t=\dfrac{2x}a\sqrt{a^2-x^2}$ , so $y^2=\dfrac{4x^2(a^2-x^2)}{a^2}$ .

$y=0$ at $x=0$ and $x=\pm a$ , with $|y|$ largest ( $=a$ ) at $x=\pm\frac a{\sqrt2}$ . That is a figure-of-eight with one lobe on each side of the $y$ -axis, crossing at the origin: graph (3).

Topics:Motion in a plane Superposition of SHMs

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 1