An observer traveling with a constant velocity of 20m/s passes close to a stationary source of sound and noticed that there is a change of frequency of 50Hz as he passes the source.What is the frequency of the source?
(Speed of sound in air=340 m/s)
See previous post: Wed, 8-15-18, 10:43AM.
https://www.jiskha.com/questions/1754277/as-observer-travelling-with-a-constant-velocity-of-20m-s-passes-close-to-a-stationary
To solve this problem, we need to use the concept of the Doppler effect. The Doppler effect describes how the observed frequency of a wave changes when there is relative motion between the source of the wave and the observer.
The equation for the Doppler effect in the case of a moving observer and a stationary source is as follows:
f' = f * (v + vo) / (v + vs)
Where:
- f' is the observed frequency
- f is the actual frequency of the source
- v is the speed of sound in the medium (in this case, air)
- vo is the velocity of the observer
- vs is the velocity of the source
In this problem, the observer is traveling with a constant velocity of 20 m/s, and the speed of sound in air is given as 340 m/s. The observed frequency change, ∆f, is 50 Hz. We need to find the frequency of the source, f.
Using the equation, we can rearrange it as follows:
f = (f' * (v + vs)) / (v + vo)
Substituting the given values:
f = (50 Hz * (340 m/s + 0 m/s)) / (340 m/s + 20 m/s)
Simplifying the equation further:
f = (50 Hz * 340 m/s) / 360 m/s
= 1700 / 360 Hz
≈ 4.72 Hz
Therefore, the frequency of the source of sound is approximately 4.72 Hz.