The tip of a tuning fork goes through $\mathbf{440}$complete vibrations in $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathit{s}$. Find the angular frequency and the period of motion.

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

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

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

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

$\varpi =\frac{2\mathrm{\pi }}{T}$

Substitute the values of time period to get angular frequency

$$\begin{gathered} \varpi =\frac{2\mathrm{\pi }}{1.14\times {10}^{-3}s} \\ =5511.17rad.s^{-1} \end{gathered}$$

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