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(D) The equation for the displacement of a particle starting from the mean position in $SHM$ is given by $x(t) = A \sin(\omega t)$.Given: $T = 8 \ s$,

Vedclass · Accepted Answer

$1: (\sqrt{2} - 1)$

Vedclass · Answer

(D) The equation for the displacement of a particle starting from the mean position in $SHM$ is given by $x(t) = A \sin(\omega t)$.Given: $T = 8 \ s$,so $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$. Amplitude $A = 4\sqrt{2} \ m$.Thus,$x(t) = 4\sqrt{2} \sin(\frac{\pi}{4} t)$.Distance in the $1^{	ext{st}}$ second ($t=0$ to $t=1$): $x(1) = 4\sqrt{2} \sin(\frac{\pi}{4}) = 4\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 4 \ m$.Distance in the $2^{	ext{nd}}$ second ($t=1$ to $t=2$): $x(2) = 4\sqrt{2} \sin(\frac{2\pi}{4}) = 4\sqrt{2} \sin(\frac{\pi}{2}) = 4\sqrt{2} 	imes 1 = 4\sqrt{2} \ m$.The distance travelled in the $2^{	ext{nd}}$ second is $d_2 = x(2) - x(1) = 4\sqrt{2} - 4 = 4(\sqrt{2} - 1) \ m$.The ratio of distance in the $1^{	ext{st}}$ second to the $2^{	ext{nd}}$ second is $\frac{d_1}{d_2} = \frac{4}{4(\sqrt{2} - 1)} = \frac{1}{\sqrt{2} - 1}$.

Similar Questions

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