The angular frequency of motion whose equatio | vedclass

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(C) The given equation is $4 \frac{d^{2} y}{d t^{2}}+9 y=0$.Rearranging the equation,we get $\frac{d^{2} y}{d t^{2}} = -\frac{9}{4} y$.Comparing this

Vedclass · Accepted Answer

$1.5$

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(C) The given equation is $4 \frac{d^{2} y}{d t^{2}}+9 y=0$.Rearranging the equation,we get $\frac{d^{2} y}{d t^{2}} = -\frac{9}{4} y$.Comparing this with the standard Simple Harmonic Motion $(SHM)$ equation $\frac{d^{2} y}{dt^{2}} = -\omega^{2} y$,where $\omega$ is the angular frequency:$\omega^{2} = \frac{9}{4}$.Taking the square root of both sides,we get $\omega = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5 	ext{ rad/s}$.

The angular frequency of motion whose equation is $4\frac{d^2y}{dt^2} + 9y = 0$ is ($y =$ displacement and $t =$ time).

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