Physics Formulas

Doppler Effect Formula

Doppler Effect can basically be said to be a property of sound waves. We will discuss this effect in the article and also learn about doppler effect formula and application. After that, the student will easily be able to calculate the Doppler Effect in various situations without any hassle.

Definition

The change in the sound wave frequency because of movement is referred to as the Doppler Effect, which is also referred to as Doppler shift. For instance, imagine you are standing on the pavement, and a police car speeds past you. Do you notice the sound of the siren changes as and when it travels a specific distance? It keeps getting louder as it is approaching you, however, there is another feature of the sound which changes. When the car moves towards you, the pitch is higher and gets lower when it moves away. The change in the pitch is due to the frequency of the waves or how many waves are passing through an area per the unit time.

In this case of the police car, you are in a still position and the car approaches you. As the sound waves move towards you, they compress which increases the frequency resulting in a higher pitch. But, as and when the police car is moving away from you, the sound waves spread further apart so the frequency lowers resulting in a lower pitch.

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Doppler Effect Formula

The sound heard by the listener changes if the source of that sound and the listener are moving relative to each other. This is what the Doppler Effect is. When the listener and the source move close, the frequency which the listener heard is higher than the sound which the source emits.

Similarly, when the listener and the source move away from one another, the frequency which the listener hears is lower than the frequency of the sound from the source. The unit of sound frequency is Hertz (Hz). Over here, one Hertz is a cycle per second ( 1 Hz = 1 s-1 = 1 cycle/s.

Thus, the equation comes as:

fL = \(\frac{v+vl}{v+vs}\) fs

Over here:

fL refers to the frequency of sound which the listener hears (Hz, or 1/s)

v is the speed of sound in the medium (m/s)

vL refers to the velocity of the listener (m/s)

vs is the velocity of the source of the sound (m/s)

fs refers to the frequency of sound which the source emits (Hz, or 1/s)

Solved Example for You

Question

A driver in a car is traveling on a road next to railway tracks. When the train approaches, it blows the horn which generates a sound with a single frequency of 420.0 Hz. The driver is driving at the speed of 18.0 m/s and the train’s speed is 32.0 m/s. Further, the sound’s speed is 340.0 m/s. Calculate the frequency of the sound which the driver of the car will hear.

Answer: In order to find the frequency, we need to first establish a coordinate system. The positive direction is said to be from the listener to the source. The source is the horn of the train and thus the velocity of the train is negative while the velocity of the driver’s car is positive. The values known to us are v = 340.0 m/s , vL = +18.0 m/s,  vs = -32.0 m/s and fs = 420.0 Hz.

Thus, we will arrange the value in the Doppler Effect Formula to find out the frequency which is:

fL = \(\frac{v+vl}{v+vs}\) fs

fL = \(\frac{340.0 m/s + 18.0 m/s }{340.0 m/s – 32.0 m/s}\) (420.0 Hz)

fL = \(\frac{358.0 m/s}{308.0 m/s}\) (420.0 Hz)

fL \(\cong \) (1.162) (420.0 Hz)

fL \(\cong \) 488.2 Hz

Therefore, the frequency of the sound of the train’s horn which the driver will hear is 488.2 Hz.

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5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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