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Q. A simple harmonic motion is represented by: $y = 5(sin3 \pi t + \sqrt{3} cos3 \pi t) cm$ The amplitude and time period of the motion are:

JEE MainJEE Main 2019Oscillations

Solution:

$y \, = \, 5 \, [sin(3 \, \pi \, t) + \sqrt{3} \, cos(3\pi t)]$
$=10 sin \bigg(3\pi t \, + \, \frac{\pi}{3}\bigg)$
Amplitude = 10 cm
$T \, = \, \frac{2\pi}{w} = \frac{2\pi}{3\pi} = \frac{2}{3}sec$

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