JEE Main 2014 · 9 Apr · Shift 1 · Physics Q14

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PhysicsSingle correct+4 / -1Question 14 of 90

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by ; $x=a_1\cos\omega t$ and $y=a_2\cos2\omega t$ traces a curve given by :

An upward-opening parabola with its vertex at (0, −a₂), passing through (±a₁, a₂)

Within the box |x| ≤ a₁, |y| ≤ a₂: a parabola opening upward, lowest at (0, −a₂) and reaching (±a₁, a₂).

A parabola opening to the left with its vertex at (−a₁, 0), passing through (a₁, ±a₂)

Within the box: a parabola opening to the left, vertex at (−a₁, 0), reaching (a₁, ±a₂).

A downward-opening parabola with its vertex at (0, a₂), passing through (±a₁, −a₂)

Within the box: a parabola opening downward, highest at (0, a₂) and reaching (±a₁, −a₂).

A straight line through the origin from (−a₁, −a₂) to (a₁, a₂)

Within the box: a straight line through the origin along the diagonal from (−a₁, −a₂) to (a₁, a₂).

Answer and worked solution

Official answer: A

$y=a_2\cos2\omega t=a_2\left(2\cos^2\omega t-1\right)=a_2\left(\dfrac{2x^2}{a_1^2}-1\right)$ .

This is a parabola opening upward with its vertex at $(0,-a_2)$ , reaching $y=a_2$ at $x=\pm a_1$ . That is graph (1).

Topics:Superposition of SHMs

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 14