Question

The period of a particle in SHM is $8$ seconds. At $t$ $=$ $0$ it is at the mean position. The ratio of the distance traveled by it in the 1st and the 2nd seconds is

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

Answer 1

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

In first one second angle moved is $\frac{\pi }{4}$
$\therefore$, distance is $A/\sqrt{2}$
In second second angle moved $\frac{\pi }{4}$ more
$\therefore$ distance in second second equals $A-\frac{A}{\sqrt{2}}$
$\therefore \frac{d_1}{d_2}=\frac{\frac{A}{\sqrt{2}}}{A-\frac{A}{\sqrt{2}}}$
$=\sqrt{2}+1$