A particle executes a simple harmonic motion with a periodic time $8$ second. At time $t = 0$ , it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is
( $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$ , $sin 9 0^{\circ} = cos 0^{\circ} = 1$ )

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The time period of the simple harmonic motion is $T = 8$ s.

The angular frequency is $ω = \frac{2 π}{T} = \frac{2 π}{8} = \frac{π}{4}$ rad/s.

Since the particle starts from the mean position at $t = 0$ , its displacement equation is given by:
$x (t) = A sin (ω t) = A sin (\frac{π}{4} t)$

The distance traveled by the particle in the 1st second ( $t = 0$ to $t = 1$ s) is:
$D_{1} = x (1) - x (0) = A sin (\frac{π}{4}) - 0 = \frac{A}{2}$

The distance traveled by the particle in the 2nd second ( $t = 1$ to $t = 2$ s) is:
$D_{2} = x (2) - x (1) = A sin (\frac{π}{2}) - A sin (\frac{π}{4}) = A - \frac{A}{2}$

The ratio of the distance traveled in the 2nd second to that in the 1st second is:
$\frac{D _{2}}{D _{1}} = \frac{A - \frac{A}{2}}{\frac{A}{2}} = \frac{1 - \frac{1}{2}}{\frac{1}{2}} = 2 - 1$

Answer: $2 - 1$

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