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:
[tex] f = f_{0}(\frac{v - v_{r}}{v - v_{s}}) [/tex]
Where:
v: is the speed of sound = 343 m/s
f₀: is the frequency emitted = 1490 Hz
f: is the frequency observed = 1505 Hz
[tex]v_{r}[/tex]: is the speed of the receiver = 0 (it is stationary)
[tex]v_{s}[/tex]: is the speed of the source =?
The minus sign of [tex]v_{s}[/tex] is because the source is moving towards the receiver.
By solving the above equation for [tex]v_{s}[/tex] we have:
[tex] v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s [/tex]
The above speed in terms of the speed of sound is:
[tex]\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%[/tex]
Therefore, the speed of the bird is 1.00% of the speed of sound.
I hope it helps you!
