Time period of a particle executing SHM is 8 | vedclass

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(D) The displacement of a particle in $SHM$ starting from the mean position is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.Give

Vedclass · Accepted Answer

$\sqrt{2} + 1$

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(D) The displacement of a particle in $SHM$ starting from the mean position is given by $x(t) = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.Given $T = 8 \, s$,we have $\omega = \frac{2\pi}{8} = \frac{\pi}{4} \, rad/s$.Distance covered in the $1^{st}$ second $(d_1)$ is the displacement at $t = 1 \, s$:$d_1 = x(1) = A \sin\left(\frac{\pi}{4} 	imes 1ight) = A \left(\frac{1}{\sqrt{2}}ight) = \frac{A}{\sqrt{2}}$.Distance covered in the $2^{nd}$ second $(d_2)$ is the displacement at $t = 2 \, s$ minus the displacement at $t = 1 \, s$:$d_2 = x(2) - x(1) = A \sin\left(\frac{\pi}{4} 	imes 2ight) - \frac{A}{\sqrt{2}} = A \sin\left(\frac{\pi}{2}ight) - \frac{A}{\sqrt{2}} = A - \frac{A}{\sqrt{2}} = A \left(1 - \frac{1}{\sqrt{2}}ight) = A \left(\frac{\sqrt{2}-1}{\sqrt{2}}ight)$.The ratio of the distance covered in the $1^{st}$ second to the $2^{nd}$ second is:$\frac{d_1}{d_2} = \frac{A/\sqrt{2}}{A(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.Rationalizing the denominator:$\frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$.

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

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