The differential equation representing the SHM of a particle is $\frac{9 d^{2} Y}{{dt}^{2}} + 4 Y = 0$ .The time period of the particle is given by

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$π \sec$

$\frac{2 π}{3} \sec$

$\frac{π}{3} \sec$

$3 π \sec$

answer is D.

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Detailed Solution

$\begin{array}{l} 9 \frac{d^{2} y}{d_{t}^{2}} + 4 y = 0; \frac{d^{2} y}{d^{2}} = - \frac{4}{9} y \\ ω^{2} = \frac{4}{9} \Rightarrow ω = \frac{2}{3} \\ T = \frac{2 π}{ω} = \frac{2 π}{(2 / 3)} = 3 π \end{array}$

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