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(B) The displacement equation for $SHM$ is $x = a \sin(\omega t + \phi)$.At $t = \frac{T}{4}$,the angular displacement is $\omega t = \frac{2\pi}{T} \

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$\frac{4\pi}{3}$

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(B) The displacement equation for $SHM$ is $x = a \sin(\omega t + \phi)$.At $t = \frac{T}{4}$,the angular displacement is $\omega t = \frac{2\pi}{T} \cdot \frac{T}{4} = \frac{\pi}{2}$.Given $x = -\frac{a}{2}$,we substitute these into the equation:$-\frac{a}{2} = a \sin(\frac{\pi}{2} + \phi)$$-\frac{1}{2} = \cos(\phi)$.This gives two possible values for the phase constant: $\phi = \frac{2\pi}{3}$ or $\phi = \frac{4\pi}{3}$.The velocity is given by $v = \frac{dx}{dt} = a\omega \cos(\omega t + \phi)$.At $t = \frac{T}{4}$,$v = a\omega \cos(\frac{\pi}{2} + \phi) = -a\omega \sin(\phi)$.Since the velocity is positive,$-a\omega \sin(\phi) > 0$,which implies $\sin(\phi) For $\phi = \frac{2\pi}{3}$,$\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} > 0$ (rejected).For $\phi = \frac{4\pi}{3}$,$\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2} Therefore,the initial phase is $\phi = \frac{4\pi}{3}$.

$A$ particle executing $SHM$ of amplitude $a$ has a displacement of $-\frac{a}{2}$ at $t = \frac{T}{4}$ and a positive velocity. Find the initial phase of the particle.

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