Answer to Question #1871 in Acoustics for nirmala

Question #1871
A locomotive engine passes a stationary observer at a speed of 72 km/h emitting a note of frequency 500 Hz. Calculate the frequency of sound heard by the observer <br>(i) before, and <br>(ii) after the engine passes the observer. <br>The speed of sound in air is 340 m/s.
1
Expert's answer
2011-03-10T09:44:42-0500
The frequency of the note changes due to the Doppler effect according to the formula (stationary receiver):
f = (Vsound / (Vsound +Vsource)) f0

72 km/h = 20 m/s
(i) before: Vsource is positive
f = (340/(340 - 20))*500 = 531.25 Hz
(ii) after: Vsource is negative
f = (340/(340+20))*500 = 472.22 Hz

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS