Question

The time period of a particle executing $SHM$ is $8s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in $1^{st}$ second to the $2^{nd}$ second is

• A. $\left(\frac{1}{\sqrt{2}-1}\right)$
• B. $\sqrt{2}$
• C. $(\sqrt{2}+1)$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $\left(\frac{1}{\sqrt{2}-1}\right)$

$\begin{array}{c}{Y}^{\prime }=A\sin w=A\sin \frac{2\pi }{8}=\frac{A}{\sqrt{2}}\\ Dis\tan cewhent=2\sec is\\ Y^2=A\sin w=A\sin \frac{\pi }{2}=A\\ Dis\tan cecoveredin2\sec is\frac{A-A}{\sqrt{2}}\\ \frac{Y^1}{Y^2}=\frac{\left(\frac{A}{\sqrt{2}}\right)}{\left(A-\frac{A}{\sqrt{2}}\right)}\\ =\frac{1}{\sqrt{2}-1}\end{array}$