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A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
Explanation
Set $\omega\sqrt{A^2 - x^2} = \omega^2x$ to find ω.
Detailed Solution
$v = \omega\sqrt{A^2 - x^2}$, $a = x\omega^2$
$v = a \Rightarrow \omega\sqrt{A^2 - x^2} = x\omega^2$
$\sqrt{(3)^2 - (2)^2} = 2\left(\frac{2\pi}{T}\right)$
$\sqrt{5} = \frac{4\pi}{T}$
$T = \frac{4\pi}{\sqrt{5}}$