The time period of a particle in simple harmo | vedclass

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Question

(C) The equation for the position of a particle in simple harmonic motion starting from the mean position is $y(t) = A \sin(\omega t)$,where $\omega =

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

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(C) The equation for the position of a particle in simple harmonic motion starting from the mean position is $y(t) = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.The distance travelled in the first second ($t=0$ to $t=1$) is $y_1 = A \sin(\frac{\pi}{4} 	imes 1) = A 	imes \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}}$.The position at $t=2$ seconds is $y(2) = A \sin(\frac{\pi}{4} 	imes 2) = A \sin(\frac{\pi}{2}) = A$.The distance travelled in the second second ($t=1$ to $t=2$) is $y_2 = y(2) - y(1) = A - \frac{A}{\sqrt{2}} = A(1 - \frac{1}{\sqrt{2}})$.The ratio of the distances is $\frac{y_1}{y_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.

The time period of a particle in simple harmonic motion is $8 \ s$. At $t=0$,it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is

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