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Question

(C) Given the equation: $x = 3 \sin 100t + 8 \cos^2 50t$.Using the trigonometric identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$x = 3 \

Vedclass · Accepted Answer

The amplitude of the resultant $S.H.M.$ is $\sqrt{73} 	ext{ units}$

Vedclass · Answer

(C) Given the equation: $x = 3 \sin 100t + 8 \cos^2 50t$.Using the trigonometric identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$x = 3 \sin 100t + 8 \left( \frac{1 + \cos 100t}{2} ight)$$x = 3 \sin 100t + 4 + 4 \cos 100t$$x = 3 \sin 100t + 4 \cos 100t + 4$Let $3 = A \cos \phi$ and $4 = A \sin \phi$,then $A = \sqrt{3^2 + 4^2} = 5$.The equation becomes $x = 5 \sin(100t + \phi) + 4$.This represents $S.H.M.$ about the mean position $x = 4$ with an amplitude of $5 	ext{ units}$.The maximum displacement from the origin is $x_{max} = 4 + 5 = 9 	ext{ units}$.Option $C$ is incorrect because the amplitude is $5$,not $\sqrt{73}$.

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

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