The time period of a particle executing S.H.M | vedclass

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(C) The time period $T = 8 \,s$. The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \,rad/s$.Since the particle starts fr

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$(\sqrt{2}+1)$

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(C) The time period $T = 8 \,s$. The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \,rad/s$.Since the particle starts from the mean position,the displacement equation is $x(t) = A \sin(\omega t)$.Distance covered in the $1^{	ext{st}}$ second ($t=0$ to $t=1$): $d_1 = x(1) - x(0) = A \sin(\frac{\pi}{4} 	imes 1) - 0 = A \frac{1}{\sqrt{2}}$.Distance covered in the $2^{	ext{nd}}$ second ($t=1$ to $t=2$): $d_2 = x(2) - x(1) = A \sin(\frac{\pi}{4} 	imes 2) - A \sin(\frac{\pi}{4} 	imes 1) = A \sin(\frac{\pi}{2}) - A \frac{1}{\sqrt{2}} = A(1 - \frac{1}{\sqrt{2}})$.Ratio $\frac{d_1}{d_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.Rationalizing the denominator: $\frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \sqrt{2}+1$.

The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in the $1^{\text{st}}$ second to the $2^{\text{nd}}$ second is:

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