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Q.

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 1st second to the 2nd second is

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a

$\frac{1}{\sqrt{2}}$

b

$(\sqrt{2} - 1)$

c

$\sqrt{2}$

d

$(\sqrt{2} + 1)$

answer is C.

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Detailed Solution

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$X = A \sin (ω t) X_{i n 1 \sec} = X_{1 s t \sec} = A \sin (\frac{2 π}{8} \times 1) A = \sin \frac{π}{4} = \frac{A}{\sqrt{2}} X_{2 second} = A \sin (\frac{2 π}{8} \times 2) X_{2 n d second} = X_{2} \sec - X_{1} \sec = A - \frac{A}{\sqrt{2}} \frac{X_{1 s t sec}}{X_{2 n d sec}} = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$

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