The displacement of a particle varies accordi | vedclass

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Question

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

Vedclass · Accepted Answer

The amplitude of the $S.H.M.$ component is $3 \ units$

Vedclass · Answer

(B) Given the displacement equation: $x = 3 \sin(100t) + 8 \cos^2(50t)$.Using the trigonometric identity $\cos^2(	heta) = \frac{1 + \cos(2	heta)}{2}$,we can rewrite the equation as:$x = 3 \sin(100t) + 8 \left( \frac{1 + \cos(100t)}{2} ight)$$x = 3 \sin(100t) + 4 + 4 \cos(100t)$$x = 4 + [3 \sin(100t) + 4 \cos(100t)]$.Let $x' = x - 4 = 3 \sin(100t) + 4 \cos(100t)$.This represents an $S.H.M.$ about the mean position $x = 4$.The amplitude of this $S.H.M.$ is $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ units$.Since the particle oscillates between $4 - 5$ and $4 + 5$,the range is $[-1, 9]$.The maximum displacement from the origin is $9 \ units$.Comparing with the options,statement $A$ is correct (it is $S.H.M.$),statement $B$ is $INCORRECT$ (amplitude is $5$,not $3$),statement $C$ is $INCORRECT$ (amplitude is $5$,not $\sqrt{73}$),and statement $D$ is correct.

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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