Doppler Shift is the phenomenon of change in frequency of sound based on the listener’s point of view. The Doppler Shift relates to the amount of shift to the velocity of the source. We use the Doppler Shift formula to calculate the motion of stars. Let us now discuss in detail about the Doppler shift formula.

**What is a Doppler Shift?**

The Doppler Shift, when related to the sound, is the change in frequency of a source because it moves. The frequency will appear to increase because the source comes towards the listener and will appear to decrease because the source moves away from the listener.

The velocity of the source is positive if the source is moving towards the listener, negative if the source is moving away from the listener. The velocity of the listener is positive if the listener is moving toward the source, negative if the listener is moving away from the source. The frequency heard by the listener is higher than that actually being emitted by the source.

There are two types of Doppler shift one is redshift is the change of frequency to a lower wavelength means away from the observer and another one is blue shift is the change of frequency to a higher wavelength means towards the observer.

**Doppler Shift Formula of frequency change:**

\(f = f_sÂ \times \frac{(v + v_l)}{(v – v_s)}\)

Where,

f | frequency heard by the listener |

f_{s} |
frequency of the source |

v | velocity of sound |

v_{s} |
the velocity of the source |

v_{L} |
velocity of listener |

**Doppler Shift Formula of Wavelength Change:**

When the source is movingÂ towardsÂ the listener, the wavelengths shifted to shorter wavelengths. When the source is movingÂ away from the listener, the wavelengths shifted to longer wavelengths. The Doppler shift formula of wavelength is:

\(\frac{\Delta \lambda}{ \lambda_0} = \frac{v}{c}\)

Where,

\(\Delta \lambda\) | wavelength shift |

\(\lambda_0\) | The wavelength of the stationary source |

v | velocity of source |

c | speed of light |

**Solved Examples forÂ Doppler Shift Formula**

Q.1. An ambulance has a velocity of 100 m/s and its siren produces a steady frequency of 200 Hz. What is the frequency of the sound heard by an observer who is in front of the ambulance, assuming the velocity of sound equals 343 m/s?

Answer:Â The velocity of sound, v = 343 m/s, and the velocity of the ambulance as the source, v_{s}Â = 100 m/s. The frequency of the source, f_{s}Â = 200 Hz. The velocity of the listener is v_{L}Â = 0. Use the formula.

f = f_{s}Â (v + v_{L})/(v – v_{s})

= (200 Hz)(343 m/s + 0 m/s) / (343 m/s – 100 m/s)

= (200 Hz)(343 m/s)/243 m/s

= (68600 Hz m/s)/ 243 m/s

= 282.30Hz

Q.2:**Â **A fixed source emits the sound of frequency 2000 Hz. What is the frequency as heard by a listener?

(i) At rest

(ii) Moving towards the source at a constant speed of 10 m/s

(iii) Moving away from the source

Solution:

Velocity of sound v = 343 m/s,

True frequency, f = 2000 Hz,

Velocity of listener, v_{LÂ }= 20 m/s

(i) When both the source and listener are at rest, frequency heard by the listener is same as true frequency

âˆ´âˆ´ Frequency as heard by listener at rest = f = 2000 Hz

(ii) Listener is moving towards the source

v_{s} = 10 m/s

\(f = f_sÂ \times \frac{(v + v_L)}{(v – v_s)}\)

\(= 2000Â \times \frac{(343 + 20)}{(343 – 10)}\)

= 2072 Hz

(iii) Listener is moving away from the source

\(f = f_sÂ \times \frac{(v – v_L)}{(v – v_s)}\)

\(f = 2000Â \times \frac{(343 – 20)}{(343 – 10)}\)

f= 1939 Hz.

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