The time period of a particle in simple harmo | vedclass

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Question

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

Vedclass · Accepted Answer

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

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(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $y(t) = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.At $t=1 \ s$,the distance from the mean position is $y_1 = A \sin(\frac{\pi}{4} 	imes 1) = A 	imes \frac{1}{\sqrt{2}}$.Since the particle starts from the mean position,the distance travelled in the first second is $d_1 = y_1 = \frac{A}{\sqrt{2}}$.At $t=2 \ s$,the displacement is $y_2 = A \sin(\frac{\pi}{4} 	imes 2) = A \sin(\frac{\pi}{2}) = A$.The distance travelled in the second second is $d_2 = y_2 - y_1 = A - \frac{A}{\sqrt{2}} = A(1 - \frac{1}{\sqrt{2}})$.The ratio of the distances is $\frac{d_1}{d_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.

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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