Question

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

Answer 1

Step 1. Given data:

Given equation = $4\frac{d^{2}y}{d{t}^2}+9y=0$

This is a differential equation of second degree and can be matched with the general equation of Simple Harmonic Motion.

Step 2. Explanation:

Simple Harmonic Motion:

Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the magnitude of the object's displacement and acts toward the object's equilibrium position.

Given equation is:

$4\frac{d^{2}y}{d{t}^2}+9y=0$

Dividing the above equation by $4$, we get

$\frac{d^{2}y}{d{t}^2}+\frac{9}{4}y=0.........\left(1\right)$

General equation of Simple Harmonic Motion may be given as:

$\frac{d^{2}y}{d{t}^2}+{\omega }^{2}y=0.........\left(2\right)$

Where $\frac{d^{2}y}{d{t}^2}=\overrightarrow{a}$= Acceleration, $\overrightarrow{y}$= displacement, $\omega =\sqrt{\frac{k}{\overrightarrow{m}}}=\frac{2\mathrm{\pi }}{T}$= angular frequency, $k$=a positive constant, $\overrightarrow{m}$= mass of the object in motion, $T$=time period,

Comparing $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} {\omega }^2=\frac{9}{4} \\ \omega =\frac{3}{2}\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Thus, angular frequency of motion of the given equation is $\frac{3}{2}\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.