Question

A train sounds its whistle as it approaches a tunnel in a cliff. The whistle produces a tone of 650 Hz and the train travels with a speed of 21.2 m/s. Find the frequency heard by an observer standing near the tunnel entrance.

Answer 1

Answer:$f_{0}$ = 692.82Hz

Explanation: This question of ours is based on Doppler effect.The Doppler effect states that if a source is producing sound at a specific frequency to an observer and there is a relative motion between the observer and the source, the frequency of sound perceived by the observer  will be different from that of the original from the source.

This represented mathematically below as

$f_{0} =\frac{v +v_{0} }{v -v_{s} } * f$

where

$f_{0}$ = observed frequency = ?

f = original frequency of sound source = 650Hz

v = speed of sound in air = 343m/s

$v_{0}$ = velocity of observer relative to the source = 0m/s ( the observer is standing, thus meaning he is not moving)

$v_{s}$ = velocity of sound source relative to the observer = 21.2m/s

by putting all of these in the formulae, we have that

$f_{0} = \frac{343 + 0}{343 -21.2} *650\\\\\\f_{0} = \frac{343}{321.8} * 650\\\\\\f_{0} = 692.98Hz$