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(B) The angular frequency $\omega$ is given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.4} = 5 \pi \, rad/s$.The angular velocity $\Omega$ at an ang

Vedclass · Accepted Answer

$42.7$

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(B) The angular frequency $\omega$ is given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.4} = 5 \pi \, rad/s$.The angular velocity $\Omega$ at an angular displacement $	heta$ is given by the formula $\Omega = \omega \sqrt{	heta_{0}^{2} - 	heta^{2}}$,where $	heta_{0}$ is the angular amplitude.Given $	heta_{0} = \pi \, rad$ and $	heta = \pi/2 \, rad$:$\Omega = 5 \pi \sqrt{\pi^{2} - (\pi/2)^{2}}$$\Omega = 5 \pi \sqrt{\pi^{2} - \frac{\pi^{2}}{4}}$$\Omega = 5 \pi \sqrt{\frac{3 \pi^{2}}{4}}$$\Omega = 5 \pi \cdot \frac{\pi \sqrt{3}}{2} = \frac{5 \pi^{2} \sqrt{3}}{2}$.Using $\pi^{2} \approx 9.87$ and $\sqrt{3} \approx 1.732$:$\Omega \approx \frac{5 	imes 9.87 	imes 1.732}{2} \approx 42.76 \, rad/s$.Thus,the correct option is $42.7 \, rad/s$.

In an angular $SHM$,the angular amplitude of oscillation is $\pi \, rad$ and the time period is $0.4 \, s$. Calculate its angular velocity at an angular displacement of $\pi/2 \, rad$. (Result in $rad/s$)

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