The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in $1^{\text {st }}$ second to the $2^{\text {nd }}$ second is .............. $s$
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(c)
$T=8 \,s$
$ \omega=\frac{2 \pi}{8}=\frac{\pi}{4}$
$x_1=A \sin \frac{\pi}{4}=\frac{A}{\sqrt{2}}$
$x_2=A \sin \frac{\pi}{4} \times 2-A \sin \frac{\pi}{4}=A-\frac{A}{\sqrt{2}}=\frac{A}{\sqrt{2}}(\sqrt{2}-1)$
$\frac{x_1}{x_2}=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$
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