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An indicator of how rapidly an item oscillates in relation to time is the angular frequency.

The duration of time needed to complete one cycle or oscillation is known as a period of motion.

Consider the given data as below.

Number of vibration is $440$.

Time, 0.500 s

The frequency of the tip of a tuning fork will be

$f=\frac{No.ofvibrations}{Time}$

Substitute values in the above expression to calculate the frequency

$$\begin{gathered} f=\frac{440}{0.500s} \\ =880s^{-1} \\ =880Hz \end{gathered}$$

The relation between period and frequency is given by,

$T=\frac{1}{f}$

Substitute value of frequency in above expression to get period of motion

$$\begin{gathered} T=\frac{1}{880Hz} \\ =1.14脳{10}^{-3}s \\ =1.14ms \end{gathered}$$

Therefore, the period of motion is $1.14ms$.

The relation between angular frequency and time period is given by,

$蠔=\frac{2\mathrm{蟺}}{T}$

Substitute the values of time period to get angular frequency

$$\begin{gathered} 蠔=\frac{2\mathrm{蟺}}{1.14脳{10}^{-3}s} \\ =5511.17rad.s^{-1} \end{gathered}$$

Therefore, the angular frequency is_{$5511.57rad.s^{-1}$}.