The displacement of a particle from its mean | vedclass

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(C) Given the displacement equation:$y = 0.2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)$Using the trigonometric identity $\sin(2A) = 2 \sin A

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Simple harmonic motion with period $0.1 \ s$

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(C) Given the displacement equation:$y = 0.2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)$Using the trigonometric identity $\sin(2A) = 2 \sin A \cos A$,we can rewrite the equation as:$y = 0.1 	imes [2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)]$$y = 0.1 \sin [2(10 \pi t + 1.5 \pi)]$$y = 0.1 \sin (20 \pi t + 3 \pi)$This is the standard form of a Simple Harmonic Motion $(SHM)$ equation,$y = A \sin(\omega t + \phi)$,where the angular frequency $\omega = 20 \pi \ rad/s$.The time period $T$ is given by:$T = \frac{2 \pi}{\omega} = \frac{2 \pi}{20 \pi} = \frac{1}{10} \ s = 0.1 \ s$.Thus,the motion is Simple Harmonic Motion with a period of $0.1 \ s$.

The displacement of a particle from its mean position (in metre) is given by
$y = 0.2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)$
The motion of the particle is

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The displacement of a particle from its mean position (in metre) is given by$y = 0.2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)$The motion of the particle is