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

(a) $ \displaystyle \frac{1}{\sqrt{2} + 1} $

(b) √2

(c) $ \displaystyle \frac{1}{\sqrt{2} } $

(d) √2 +1

Ans: (d)

Sol: $\large y_1 = a sin\frac{2\pi}{8}\times 1$ …(i)

$\large y_1 + y_2 = a sin\frac{2\pi}{8}\times 2$ …(ii)

Dividing (ii) by (i)

$\large \frac{y_1 + y_2}{y_1} = \frac{sin\frac{2\pi}{4}}{sin\frac{2\pi}{8}}$

$\large 1 + \frac{y_2}{y_1} = \sqrt{2}$

$\large \frac{y_2}{y_1} = \sqrt{2} – 1 $

$\large \frac{y_1}{y_2} = \frac{1}{\sqrt{2} – 1} $

$\large \frac{y_1}{y_2} = \frac{1}{\sqrt{2} – 1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1} $

$\large \frac{y_1}{y_2} = \sqrt{2}+1 $