The displacement of a particle varies according to the relation $x = 3 \sin 100 \, t + 8 \cos ^2 50\,t $. Which of the following is/are correct about this motion .
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$.x=3 \sin 100 t+8 \cos ^{2} 50 t=3 \sin 100 t+4(1+\cos 100 t) $
using $ 2 \cos ^{2} A=1+\cos 2 A$
$x=3 \sin 100 t+4 \cos 100 t+4$
$\frac{d x}{d t}=300 \cos 100 t-400 \sin 100 t$
$\frac{d^{2}(x-4)}{d t^{2}}=-30000 \sin 100 t+40000 \cos 100 t=-100(3 \sin 100 t+4 \cos 100 t)=$
$-100(x-4)$
so it is a $SHM.$
Amplitude of the motion, $A=\sqrt{3^{2}+4^{2}}=5$ units
for max distance, $\frac{d x}{d t}=300 \cos 100 t-400 \sin 100 t=0 \Rightarrow \tan 100 t=\frac{3}{4}$
$\therefore \sin 100 t=\frac{3}{5}, \cos 100 t=\frac{4}{5}$
now, $x_{\max }=3 \times \frac{3}{5}+4 \times \frac{4}{5}+4=5+4=9$ units
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