Doppler Effect Formula

Doppler Effect defines the apparent change in the frequency of sound when the observer and the medium both are in relative motion. It’s a wave phenomenon which holds not just for sound waves but also for electromagnetic waves like microwaves, visible light, and microwaves. It depends on three factors – Velocity of the source, Velocity of the medium and Velocity of the observer.

Formula

$f'=(\frac{v+v_{0}}{v-v_{s}})f$

f ‘= observed frequency

f= actual frequency of the sound wave

V = speed of the sound wave

Vo = velocity of the observer

Vs = velocity of the source

Solved Examples

Question:

A Submarine travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1400 Hz. The speed of sound in the water is 530 m/s. A second submarine is located such that both submarines are travelling directly towards each other. The second submarine is moving at 6.00 m/s.

(A) What is the frequency detected by an observer riding on the second submarine as the first submarine approaches it?

(B) The submarine barely misses each other and pass. What frequency is detected by an observer riding on the second submarine as the subs recede from it?

(C) When both the submarine approaches each other, some of the sounds from the first submarine reflects from the second submarine and returns to it. If the sound were to be detected by a first submarine, what is its frequency?

Solution:

Given: Frequency f = 1400 Hz,

Velocity of first submarine, Vs = 8 m/s,

Velocity of second submarine, Vo = 6 m/s.

Speed of sound in water V = 530 m/s

(A) The Apparent frequency is given by,

$f'=(\frac{v+v_{0}}{v-v_{s}})f$
$f'=(\frac{530+6}{530-8})1400$

f ‘ = 1437.54 Hz.

(B) To find Doppler-shifted frequency heard by the observer in second submarine:

$f'=(\frac{v+v_{0}}{v-v_{s}})f$

$f'=(\frac{530+(-6)}{530-(-8)})1400$

f ‘ = 1363.56 Hz.

(C)The Sound of apparent frequency 1436.5 Hz found in part (A) is reflected from a moving source (sub B) and detected by a moving observer

$f''=(\frac{v+v_{0}}{v-v_{s}})f'$
$f''=(\frac{530+6}{530-8})1437.54$

f” = 1476 Hz