x and y displacements of a particle are given as x(t) = a sin ω t and y(t) = a sin2 ω t. Its trajectory will look like: (a) (b) (c) (d)

Answer: (c). y = asin 2ω t = 2asin ω tcos ω t = 2x√(1 - x²/a²). y = 0 at x = 0 and x = ± a; |y| is maximum ( = a) at x = ± a/√(2). So the path is a figure of eight with one loop on each side of the y-axis, crossing at the origin. So the answer is (c).

x and y displacements of a particle are given as x(t) = a sin ω t and… | JEE Main 2015 (10 Apr, Shift 1)

Q16JEE Main 2015 · 10 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition

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:

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Answer: (c)

$y = a\sin 2\omega t = 2a\sin\omega t\cos\omega t = 2x\sqrt{1 - x^2/a^2}$ .
$y = 0$ at $x = 0$ and $x = \pm a$ ; $|y|$ is maximum ( $= a$ ) at $x = \pm a/\sqrt{2}$ .
So the path is a figure of eight with one loop on each side of the y-axis, crossing at the origin.
So the answer is (c).

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Damped, Forced Oscillations & Superposition question from JEE Main 2015 (10 Apr, Shift 1)

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