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 $2 \sin 	heta \cos 	he

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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 $2 \sin 	heta \cos 	heta = \sin(2	heta)$,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 equation is of the form $y = A \sin(\omega t + \phi)$,which represents $S.H.M.$Here,the angular frequency $\omega = 20\pi \ rad/s$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{20\pi} = \frac{1}{10} \ s = 0.1 \ s$.Therefore,the motion is $S.H.M.$ 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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