The displacement of a particle executing SHM | vedclass

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(A) Given,the displacement equation is $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$.The time period $T = 3.14 \ s$. We know that $T = \frac

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$10$

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(A) Given,the displacement equation is $x = 10 \sin \left(\omega t + \frac{\pi}{3}ight) \ m$.The time period $T = 3.14 \ s$. We know that $T = \frac{2\pi}{\omega}$.Taking $\pi \approx 3.14$,we have $3.14 = \frac{2 	imes 3.14}{\omega}$,which gives $\omega = 2 \ rad/s$.The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt} \left[10 \sin \left(\omega t + \frac{\pi}{3}ight)ight]$.$v = 10 \omega \cos \left(\omega t + \frac{\pi}{3}ight)$.At $t = 0$,the velocity is $v = 10 	imes 2 	imes \cos \left(0 + \frac{\pi}{3}ight)$.$v = 20 	imes \cos \left(\frac{\pi}{3}ight) = 20 	imes \frac{1}{2} = 10 \ m/s$.

The displacement of a particle executing $SHM$ is given by $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$. The time period of motion is $3.14 \ s$. The velocity of the particle at $t = 0$ is . . . . . . $m/s$.

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