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 1st second to the 2nd 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 $