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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

Sagot :

whitneytr12

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!

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