Question

Time period of a particle executing SHM is 8 sec. At t=0 it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is:

• A. $\frac{1}{\sqrt{2}}+1$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}+1$

Answer 1

The correct option is D $\sqrt{2}+1$

$x=A\sin \omega tx(t=1s)=A\sin \left(\frac{2\pi }{T}t\right)=A\sin \left(\frac{2\pi }{8}\right)=A\sin \left(\frac{\pi }{4}\right)=\frac{A}{\sqrt{2}}x(t=2s)=A\sin \left(\frac{2\pi }{T}t\right)=A\sin (\frac{2\pi }{8}\times 2)=A\sin \left(\frac{\pi }{2}\right)=A$

And the required ratio is:
$\frac{x(t=1s)}{x(t=2s)-x(t=1s)}=\frac{\frac{A}{\sqrt{2}}}{A-\frac{A}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$