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(D) Given the displacement relation: $x = 3 \sin 100t + 8 \cos^2 50t$.Using the trigonometric identity $2 \cos^2 A = 1 + \cos 2A$,we can rewrite the e

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$(B)$ and $(C)$ both.

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(D) Given the displacement relation: $x = 3 \sin 100t + 8 \cos^2 50t$.Using the trigonometric identity $2 \cos^2 A = 1 + \cos 2A$,we can rewrite the expression as:$x = 3 \sin 100t + 4(1 + \cos 100t)$$x = 3 \sin 100t + 4 \cos 100t + 4$Let $y = x - 4$,then $y = 3 \sin 100t + 4 \cos 100t$.This is a standard form of $S.H.M.$ with angular frequency $\omega = 100 	ext{ rad/s}$.The amplitude $A$ is given by $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 	ext{ units}$.Since the motion is $S.H.M.$ about the mean position $x = 4$,the displacement $y$ oscillates between $-5$ and $+5$.Thus,$x = y + 4$ oscillates between $4 - 5 = -1$ and $4 + 5 = 9$.The maximum displacement from the origin is $9 	ext{ units}$.Therefore,both statements $(B)$ and $(C)$ are correct.

The displacement of a particle varies according to the relation $x = 3 \sin 100t + 8 \cos^2 50t$. Which of the following is/are correct about this motion?

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