Question

A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of \[10{\text{m}}{{\text{s}}^{ - 1}}\]. What are the frequency, wavelength, and speed of sound for an observer standing on the station platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of \[10{\text{m}}{{\text{s}}^{ - 1}}\]?
The speed of sound in still air can be taken as \[{\text{340m}}{{\text{s}}^{{\text{ - 1}}}}\].

Answer 1

Hint: To solve the given question we have to figure out how the blowing air can affect the speed of sound in air. For doing so we have understood that if the direction of the blowing air is along the direction of propagation of sound then it will increase the speed of sound while in the opposite direction it will decrease the speed. To get the required frequency, wavelength, and speed of sound with respect to an observer standing on the station platform we take the formula: \[\lambda= \dfrac{{\text{V}}}{\nu }\], where \[{\text{V}}\] is the speed of sound, \[\nu \] is the frequency, and \[\lambda \] is the wavelength.

Complete step by step solution:
Case (I)
According to the information given in the question source and observer both are at rest so when the wind starts blowing from source to observer then the affecting velocity of sound will be:
\[{\text{V=}}{{\text{V}}_{{\text{sound}}}}{\text{ + }}{{\text{V}}_{\text{w}}}\]…………. (i) (Where \[{{\text{V}}_{{\text{sound}}}}\] is the speed of sound in still air and \[{{\text{V}}_{\text{w}}}\] is the speed of wind.)

Given:
\[{{\text{V}}_{\text{w}}}{\text{ =10 m}}{{\text{s}}^{{\text{ - 1}}}}\], \[\nu= {\text{400 Hz}}\] and
\[{{\text{V}}_{{\text{sound}}}}{\text{ =340 m}}{{\text{s}}^{{\text{ - 1}}}}\]
Substitute the given values in eqn (i)
We get,
\[{\text{V =(340+10) m}}{{\text{s}}^{{\text{ - 1}}}}\]
\[{\text{V=350 m}}{{\text{s}}^{{\text{ - 1}}}}\]
As there is no relative motion between the source and the observer therefore frequency of the sound wave will remain the same\[(\nu= {\text{400 Hz}})\].
Now, wavelength of the Sound:
\[\lambda= \dfrac{{\text{V}}}{\nu }\]………….. (ii)
Substitute the values of \[{\text{V}}\] and \[\nu \] in eqn (ii), we get
\[\lambda= \dfrac{{350}}{{400}}{\text{ m}}\]
\[ = 0.875{\text{ m}}\]
Hence, the required wavelength is 0.875 m.
Case (II)
In this case, source and observer both are in relative motion so when the wind starts blowing from source to observer then the frequency of the sound wave will be calculated with the help expression for the apparent frequency of doppler’s effect as:
\[\nu ' = \left( {\dfrac{{{{\text{V}}_{{\text{sound}}}} + {{\text{V}}_{\text{w}}} + {{\text{V}}_m}}}{{{{\text{V}}_{{\text{sound}}}} + {{\text{V}}_{\text{w}}}}}} \right)\nu \](where \[{{\text{V}}_{{\text{sound}}}}\], \[{{\text{V}}_{\text{w}}}\],\[{{\text{V}}_m}\]are the velocity of sound in still a, wind and observer respectively and \[\nu \] is the original frequency)
On substituting all these values in the above expression we will get different frequencies as we got in case (I). As a result, wavelength will also change in this case.

$\therefore$ The required wavelength is 0.875 m. We will get different frequencies as we got in case (I). As a result, the wavelength will also change in this case.

Note:
In order to solve these kinds of problems, one should have a basic idea of doppler’s effect and its condition for deriving the apparent frequency. Students do not apply the formula for apparent frequency everywhere. Relative motion between the source and the observer is a necessary case for the application of doppler’s formula.