MCQ
The angular frequency of motion whose equation is $4\frac{{{d^2}y}}{{d{t^2}}} + 9y = 0$ is ($y =$ displacement and $t =$ time)
- A$2.25$
- B$0.44$
- ✓$1.5$
- D$0.67$
or $\frac{d^{2} y}{d t^{2}}=\frac{-9}{4} y$
Comparing with $SHM$ equation
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\omega^{2} \mathrm{y}$
$\therefore \omega^{2}=\frac{9}{4}$
$\therefore \omega=\frac{3}{2}$
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