The wings of the blue-throated hummingbird (Lampornis clemenciae), which inhabits Mexico and the south-western United States, beat at a rate of up to $\mathbf{900}$ times per minute. Calculate (a) the period of vibration of this bird’s wings, (b) the frequency of the wing’s vibration, and (c) the angular frequency of the bird’s wing beats.

The period can be defined as the time taken required to complete one cycle or oscillation.

Frequency is defined as the number of oscillations done by an object per second.

Angular frequency is a quantity that measures how quickly an object oscillates with respect to time.

As given, the United States is beat at a rate of up to times per minute.

The time taken to complete one beat will be the same as the time period of vibration of the bird’s wings. Therefore, the time period is,

$$\begin{gathered} T=\frac{1min}{900beats} \\ =\frac{60s}{900beats} \\ =0.067s \end{gathered}$$

Therefore, the period of vibration of bird’s wings is $0.067s$.

(b) The relation between frequency $\left(f\right)$and time period $\left(T\right)$is given by,

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

Substitute value of in above expression to get frequency

$$\begin{gathered} f=\frac{1}{0.067s} \\ =14.93Hz \end{gathered}$$

Therefore, the frequency of wing’s vibration is $14.93Hz$.

1. 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

$$\begin{gathered} \varpi =\frac{2\mathrm{\pi }}{0.067s} \\ =93.78rad/s \end{gathered}$$

Therefore, the angular frequency of bird’s wing beats is _$93.78rad/s$.