Question

A car travelling towards a hill at \[10\,{\text{m/s}}\] sounds its horn which has a frequency \[500\,{\text{Hz}}\]. This is heard in a second car travelling behind the first car in the same direction with speed \[20\,{\text{m/s}}\]. The sound can also be heard in the second car will be: (velocity of sound in air\[ = 340\,{\text{m/s}}\])
A. \[31\,{\text{Hz}}\]
B. \[24\,{\text{Hz}}\]
C. \[21\,{\text{Hz}}\]
D. \[34\,{\text{Hz}}\]

Answer 1

Hint: Use the different formulae for the frequency of the sound heard by the observer for different conditions of motion of the observer and source of the sound. First calculate the frequency of the sound heard by the observer in the second car directly from the first car. Then calculate the frequency of the sound from the car at the stationary hill and frequency of the sound from the hill to the second observer. Calculate the resultant of the frequency directly from the first car and from the hill.

Formulae used:
The frequency \[f'\] of the sound heard by an observer moving towards a source moving away from the observer is given by
\[f' = \left( {\dfrac{{v + {v_o}}}{{v + {v_s}}}} \right)f\] …… (1)
Here, \[v\] is the speed of the sound, \[{v_o}\] is speed of the observer, \[{v_s}\] is the speed of source of sound and \[f\] is frequency of the sound from the source.
The frequency \[f'\] of the sound heard by the stationary observer from a source approaching the observer is given by
\[f' = \left( {\dfrac{v}{{v - {v_s}}}} \right)f\] …… (2)
Here, \[v\] is the speed of the sound, \[{v_s}\] is the speed of source of sound and \[f\] is frequency of the sound from the source.
The frequency \[f'\] of the sound heard by the moving observer from a stationary source is given by
\[f' = \left( {\dfrac{{v + {v_o}}}{v}} \right)f\] …… (3)
Here, \[v\] is the speed of the sound, \[{v_o}\] is speed of the observer and \[f\] is frequency of the sound from the source.

Complete step by step answer:
We have given that the speed of the first car is \[10\,{\text{m/s}}\] and the speed of the second car is \[20\,{\text{m/s}}\].
\[{v_1} = 10\,{\text{m/s}}\]
\[\Rightarrow{v_2} = 20\,{\text{m/s}}\]
The frequency of the sound horn from the first car is \[500\,{\text{Hz}}\].
\[f = 500\,{\text{Hz}}\]
We have asked to calculate the frequency of the sound horn from the first car heard by the observer in the second car.The observer in the second car hears the sound resultant of the sound directly coming from the first car and the sound coming back after reflection from the hill.

Let us first calculate the frequency \[{f_1}\] of the sound directly coming from the first car heard by the observer in the second car.
Substitute \[340\,{\text{m/s}}\] for \[v\], \[20\,{\text{m/s}}\] for \[{v_o}\], \[10\,{\text{m/s}}\] for \[{v_s}\] and \[500\,{\text{Hz}}\] for \[f\] in equation (1).
\[{f_1} = \left( {\dfrac{{340\,{\text{m/s}} + 20\,{\text{m/s}}}}{{340\,{\text{m/s}} + 10\,{\text{m/s}}}}} \right)\left( {500\,{\text{Hz}}} \right)\]
\[ \Rightarrow {f_1} = 514.28\,{\text{Hz}}\]
Hence, the frequency of the sound directly coming from the first car heard by the observer in the second car is.

Let's now consider the second case when the sound reflected from the hill and the observer in the second car hears is \[514.28\,{\text{Hz}}\].First the sound from the first car moving towards the stationary observer which is hill.
The frequency \[{f_H}\] of the sound at the hill is
\[{f_H} = \left( {\dfrac{v}{{v - {v_s}}}} \right)f\]
Substitute \[340\,{\text{m/s}}\] for \[v\], \[10\,{\text{m/s}}\] for \[{v_s}\] and \[500\,{\text{Hz}}\] for \[f\] in equation (1).
\[{f_H} = \left( {\dfrac{{340\,{\text{m/s}}}}{{340\,{\text{m/s}} - 10\,{\text{m/s}}}}} \right)\left( {500\,{\text{Hz}}} \right)\]
\[ \Rightarrow {f_H} = 515.15\,{\text{Hz}}\]

The frequency of the sound from the hill heard by the observer in the second car is
\[{f_2} = \left( {\dfrac{{v + {v_o}}}{v}} \right){f_H}\]
Substitute \[340\,{\text{m/s}}\] for \[v\], \[20\,{\text{m/s}}\] for \[{v_o}\] and \[515.15\,{\text{Hz}}\] for \[{f_H}\] in equation (1).
\[{f_2} = \left( {\dfrac{{340\,{\text{m/s}} + 20\,{\text{m/s}}}}{{340\,{\text{m/s}}}}} \right)\left( {515.15\,{\text{Hz}}} \right)\]
\[ \Rightarrow {f_2} = 545.45\,{\text{Hz}}\]

Now the actual frequency \[f'\] of the sound heard by the observer in the second car is the resultant of the sound directly from the first car and from the hill after reflection.
\[f' = {f_2} - {f_1}\]
Substitute \[545.45\,{\text{Hz}}\] for \[{f_2}\] and \[514.28\,{\text{Hz}}\] for \[{f_1}\] in the above equation.
\[f' = \left( {545.45\,{\text{Hz}}} \right) - \left( {514.28\,{\text{Hz}}} \right)\]
\[ \Rightarrow f' = 31.17\,{\text{Hz}}\]
\[ \therefore f' \approx 31\,{\text{Hz}}\]
Therefore, the frequency of the sound by the observer in the second car is \[31\,{\text{Hz}}\].

Hence, the correct option is A.

Note: The students may directly calculate the frequency of the sound from the first car. But the students should not forget to calculate the frequency of the sound from the hill after reflection. If this frequency of the reflected sound from the hill is not considered then the final answer will not be correct. Also for the calculation of the frequency of sound reflected from the hill, the frequency of the sound from the source is the frequency heard at the stationary hill.