Question

Two electric trains run at the same speed of 72 km h−1 along the same track and in the same direction with separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speed of sound in air is 340 m s−1, find the frequencies heard by the person.

Answer 1

Given:

Speed of sound in air v = 340 ms−1

Frequency of whistles f0 = 500 Hz

Speed of train vs = 72 km/h = 72×518=20 m/s

The person will receive the sound in a direction that makes an angle θ with the track. The angle θ is given by:

 θ=tan−1(0.52.4/2)=22.62°

The velocity of the source will be 'v cos θ' when heard by the observer.

So, the apparent frequency received by the man from train A is

f1=(vv−vscosθ)×f0⇒f1=(340340−vscos 22.62∘)×500⇒f1=(340340−20×cos 22.62°)×500⇒f1=528.70 Hz ≈529 Hz

The apparent frequency heard by the man from train B is

f2=(vv+vcosθ)×f0⇒f2=(340340+20×cos 22.62°)×500⇒f2=474.24 Hz ≈474 Hz