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Question: A car travelling towards a hill at \[10\,{\text{m/s}}\] sounds its horn which has a frequency \[500\...

A car travelling towards a hill at 10m/s10\,{\text{m/s}} sounds its horn which has a frequency 500Hz500\,{\text{Hz}}. This is heard in a second car travelling behind the first car in the same direction with speed 20m/s20\,{\text{m/s}}. The sound can also be heard in the second car will be: (velocity of sound in air=340m/s = 340\,{\text{m/s}})
A. 31Hz31\,{\text{Hz}}
B. 24Hz24\,{\text{Hz}}
C. 21Hz21\,{\text{Hz}}
D. 34Hz34\,{\text{Hz}}

Explanation

Solution

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

Complete step by step answer:
We have given that the speed of the first car is 10m/s10\,{\text{m/s}} and the speed of the second car is 20m/s20\,{\text{m/s}}.
v1=10m/s{v_1} = 10\,{\text{m/s}}
v2=20m/s\Rightarrow{v_2} = 20\,{\text{m/s}}
The frequency of the sound horn from the first car is 500Hz500\,{\text{Hz}}.
f=500Hzf = 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 f1{f_1} of the sound directly coming from the first car heard by the observer in the second car.
Substitute 340m/s340\,{\text{m/s}} for vv, 20m/s20\,{\text{m/s}} for vo{v_o}, 10m/s10\,{\text{m/s}} for vs{v_s} and 500Hz500\,{\text{Hz}} for ff in equation (1).
f1=(340m/s+20m/s340m/s+10m/s)(500Hz){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)
f1=514.28Hz\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 514.28Hz514.28\,{\text{Hz}}.First the sound from the first car moving towards the stationary observer which is hill.
The frequency fH{f_H} of the sound at the hill is
fH=(vvvs)f{f_H} = \left( {\dfrac{v}{{v - {v_s}}}} \right)f
Substitute 340m/s340\,{\text{m/s}} for vv, 10m/s10\,{\text{m/s}} for vs{v_s} and 500Hz500\,{\text{Hz}} for ff in equation (1).
fH=(340m/s340m/s10m/s)(500Hz){f_H} = \left( {\dfrac{{340\,{\text{m/s}}}}{{340\,{\text{m/s}} - 10\,{\text{m/s}}}}} \right)\left( {500\,{\text{Hz}}} \right)
fH=515.15Hz\Rightarrow {f_H} = 515.15\,{\text{Hz}}

The frequency of the sound from the hill heard by the observer in the second car is
f2=(v+vov)fH{f_2} = \left( {\dfrac{{v + {v_o}}}{v}} \right){f_H}
Substitute 340m/s340\,{\text{m/s}} for vv, 20m/s20\,{\text{m/s}} for vo{v_o} and 515.15Hz515.15\,{\text{Hz}} for fH{f_H} in equation (1).
f2=(340m/s+20m/s340m/s)(515.15Hz){f_2} = \left( {\dfrac{{340\,{\text{m/s}} + 20\,{\text{m/s}}}}{{340\,{\text{m/s}}}}} \right)\left( {515.15\,{\text{Hz}}} \right)
f2=545.45Hz\Rightarrow {f_2} = 545.45\,{\text{Hz}}

Now the actual frequency ff' 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=f2f1f' = {f_2} - {f_1}
Substitute 545.45Hz545.45\,{\text{Hz}} for f2{f_2} and 514.28Hz514.28\,{\text{Hz}} for f1{f_1} in the above equation.
f=(545.45Hz)(514.28Hz)f' = \left( {545.45\,{\text{Hz}}} \right) - \left( {514.28\,{\text{Hz}}} \right)
f=31.17Hz\Rightarrow f' = 31.17\,{\text{Hz}}
f31Hz\therefore f' \approx 31\,{\text{Hz}}
Therefore, the frequency of the sound by the observer in the second car is 31Hz31\,{\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.