A particle executes a simple harmonic motion | vedclass

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(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $x(t) = A \sin(\omega t)$, where $A$ is the a

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$\sqrt{2} - 1$

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(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $x(t) = A \sin(\omega t)$, where $A$ is the amplitude and $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} 	ext{ rad/s}$.At $t = 0$, $x(0) = 0$.At $t = 1 	ext{ s}$, $x(1) = A \sin(\frac{\pi}{4} 	imes 1) = A \sin(45^\circ) = \frac{A}{\sqrt{2}}$.Distance traveled in the $1^{st}$ second is $d_1 = |x(1) - x(0)| = \frac{A}{\sqrt{2}}$.At $t = 2 	ext{ s}$, $x(2) = A \sin(\frac{\pi}{4} 	imes 2) = A \sin(90^\circ) = A$.Distance traveled in the $2^{nd}$ second is $d_2 = |x(2) - x(1)| = |A - \frac{A}{\sqrt{2}}| = A(1 - \frac{1}{\sqrt{2}}) = A(\frac{\sqrt{2} - 1}{\sqrt{2}})$.The ratio of distance traveled in the $2^{nd}$ second to the $1^{st}$ second is $\frac{d_2}{d_1} = \frac{A(\frac{\sqrt{2} - 1}{\sqrt{2}})}{\frac{A}{\sqrt{2}}} = \sqrt{2} - 1$.

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