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(B) In simple harmonic motion $(SHM)$,the velocity of a particle is zero at the extreme positions. Let the displacement be $x = A \sin(\omega t)$. The

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$2\pi 	ext{ rad s}^{-1}$

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(B) In simple harmonic motion $(SHM)$,the velocity of a particle is zero at the extreme positions. Let the displacement be $x = A \sin(\omega t)$. The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$. The velocity is zero when $\cos(\omega t) = 0$,which occurs at $t = \frac{T}{4}, \frac{3T}{4}, \dots$ The time interval between two consecutive points of zero velocity (i.e.,between two extreme positions) is the time taken to travel from one extreme to the other,which is half the time period,$\frac{T}{2}$. Given,$\frac{T}{2} = 0.5 	ext{ s}$. Therefore,$T = 1 	ext{ s}$. The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T}$. Substituting $T = 1 	ext{ s}$,we get $\omega = \frac{2\pi}{1} = 2\pi 	ext{ rad s}^{-1}$.

An object undergoing simple harmonic motion takes $0.5 \text{ s}$ to travel from one point of zero velocity to the next such point. The angular frequency of the motion is,

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