When the bell in a clock tower rings with a sound of 474 Hz, a pigeon roosting in the belfry flies directly away from the bell. If the pigeon hears a frequency of 453Hz, what is its speed?

15.44 m/s

Explanation:

Below is the given values:

The bell in rings with a sound = 474 Hz

If the frequency of pigeon hears = 453 Hz

Speed = ?

Use below formula:

Frequency = [(Vs - Vo) / Vs ] x fo

Vs = Speed of sound

Vo = Speed of observer

fo = Sound frequency

Frequency = [(Vs - Vo) / Vs ] x fo

453 = [(343 - Vo) / 343 ] x 474

453 / 474 =  [(343 - Vo) / 343 ]  

0.955 = (343 - Vo) / 343

0.955 x 343 = 343 - Vo

327.56 = 343 - Vo

Vo = 343 - 327.56

Vo = 15.44 m/s

amplitude

hope this

snell's law

refractve index = sin incidence / sin refraction