A particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position, the magnitude of its velocity is $SI\,units$ is equal to that of its acceleration. Then, its periodic time in seconds is
JEE MAIN 2019, Medium
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$\left|v_{4}\right|=\left|a_{4}\right|$
$\Rightarrow \quad(w \sqrt{A^{2}-x^{2}})_{4}=\left(w^{2} x\right)_{4}$
$\Rightarrow \quad w \sqrt{25-16}=w^{2} \times 4$
$\Rightarrow \quad w=\frac{3}{4}$
$T=\frac{2 \pi}{w}=2 \pi \frac{4}{3}=\frac{8 \pi}{3}$
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