A particle oscillates in XY plane according to equation r = A sin ω t i ^ + A cos 2 ω t j ^ . The trajectory of particle is :-

x = A sin ω t . y = A cos 2 ω t = A ( 1 − 2 sin 2 ω t ) = A − A 2 x 2 . The trajectory is a downward-opening parabola - (A)

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PhysicsSHMSuperposition of SHMMedium2 minQB

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A particle oscillates in XY plane according to equation $r = A sin ω t \hat{i} + A cos 2 ω t \hat{j}$ . The trajectory of particle is :-

Answer:

Solution:

$x = A sin ω t$ .

$y = A cos 2 ω t = A (1 - 2 sin^{2} ω t) = A - \frac{2 x ^{2}}{A}$ .

The trajectory is a downward-opening parabola - (A)

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Superposition of SHM

⏱ 2mℹ️ Source: QB

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