The particle executing $SHM$ of amplitude $'a'$ has displacement $-\frac {a}{2}$ at $t = \frac {T}{4}$ and a positive velocity. Find the initial phase of particle
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$x=a \sin (\omega t+\phi)$
at $t=\frac{T}{4} ; \omega t=\frac{\pi}{2} \& x=-\frac{a}{2}$
So, $-\frac{a}{2}=a \sin \left(\frac{\pi}{2}+\phi\right)$
$\Rightarrow \cos \phi=-\frac{1}{2}$
So, $\phi=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$
So, $V=a \omega \cos (\omega t+\phi)$
at $\mathrm{t}=\frac{\mathrm{T}}{4}, \mathrm{V}=+$ ve (which is possible with
$\left.\phi=\frac{4 \pi}{3}\right)$
So, $\frac{2 \pi}{3}$ rejected
$\phi=\frac{4 \pi}{3}$
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