Question

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1490 Hz. The bird-watcher, however, hears a frequency of 1505 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound

Answer 1

Answer:

The speed of the bird is 1.00% of the speed of sound.

Explanation:

The speed of the bird can be found by using the Doppler equation:

$f = f_{0}(\frac{v - v_{r}}{v - v_{s}})$

Where:

v: is the speed of sound = 343 m/s

f₀: is the frequency emitted = 1490 Hz

f: is the frequency observed = 1505 Hz

$v_{r}$: is the speed of the receiver = 0 (it is stationary)

$v_{s}$: is the speed of the source =?

The minus sign of $v_{s}$ is because the source is moving towards the receiver.

By solving the above equation for $v_{s}$ we have:

$v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s$

The above speed in terms of the speed of sound is:

$\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%$

Therefore, the speed of the bird is 1.00% of the speed of sound.  

I hope it helps you!