The equation of a particle executing SHM is g | vedclass

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Question

(B) The given equation is $x(t) = 3 \cos(\frac{\pi}{2}t)$.The angular frequency is $\omega = \frac{\pi}{2}$ rad/s.The time period is $T = \frac{2\pi}{

Vedclass · Accepted Answer

$27 - \frac{3}{\sqrt{2}}$ cm

Vedclass · Answer

(B) The given equation is $x(t) = 3 \cos(\frac{\pi}{2}t)$.The angular frequency is $\omega = \frac{\pi}{2}$ rad/s.The time period is $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/2} = 4$ s.In one time period ($T = 4$ s),the particle travels a distance of $4A$,where $A = 3$ cm.So,distance in one period = $4 	imes 3 = 12$ cm.In $8.5$ seconds,the number of periods is $n = \frac{8.5}{4} = 2.125$ periods.Distance in $2$ full periods = $2 	imes 12 = 24$ cm.Remaining time is $0.5$ s.At $t = 8$ s,the particle is at $x(8) = 3 \cos(\frac{\pi}{2} 	imes 8) = 3 \cos(4\pi) = 3$ cm.At $t = 8.5$ s,the particle is at $x(8.5) = 3 \cos(\frac{\pi}{2} 	imes 8.5) = 3 \cos(4\pi + \frac{\pi}{4}) = 3 \cos(\frac{\pi}{4}) = 3 	imes \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$ cm.The distance travelled in the last $0.5$ s is $|x(8.5) - x(8)| = |\frac{3}{\sqrt{2}} - 3| = 3 - \frac{3}{\sqrt{2}}$ cm.Total distance = $24 + (3 - \frac{3}{\sqrt{2}}) = 27 - \frac{3}{\sqrt{2}}$ cm.

The equation of a particle executing $SHM$ is given by $x = 3 \cos(\frac{\pi}{2}t)$ cm,where $t$ is in seconds. The distance travelled by the particle in the first $8.5$ seconds is:

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