Question

Two aeroplanes 'A' and 'B' are moving away from one another with a speed of 720kmph. the frequency of the whistle emitted by 'A' is 1100Hz. calculate the apparent frequency of whistle as heard by the passenger of the aeroplane 'B' velocity of sound in air is 350m/s
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Answer 1

Answer :

300 Hz

Explanation :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(-1,0)(4,0){2}{ \bullet }\multiput(3.1,0.1)(-2.9,1.7){2}{\vector(1,0){2}}\put(-0.9,0.1){\vector(-1,0){2}}\put(-3,-1){ \bf Aeroplane\ A }\put(3,-1){ \bf Aeroplane\ B }\put(-2.5,-1.7){\sf Source}\put(3.5,-1.7){\sf Listener}\put(-3,0.7){ \sf V_A=720\ kmph }\put(3,0.7){ \sf V_B=720\ kmph }\put(0,2.3){ \sf V_s=350\ m/s }\end{picture}$

Given :

• Aeroplane A and aeroplane B are moving away from each other

• Speed of aeroplane A,  $\bf V_A=720 \ kmph$

• Speed of aeroplane B,  $\bf V_B=720 \ kmph$

• Velocity of sound,    $\bf V_S=350 \ kmph$

• frequency of the whistle emitted by 'A',   $\bf n=1100 \ Hz$

To find :

• The apparent frequency of whistle as heard by the passenger of the aeroplane 'B

Solution :

 Let  n' be the apparent frequency of whistle heard by the passenger of the aeroplane B

Here, source is Aeroplane A

         listener is passenger of Aeroplane B

Converting 720 kmph into m/s,

      $\bf 720 \ kmph=720 \times \frac{5}{18} \ m/s = 200 \ m/s$

∴   $V_A=V_B\bf =200 \ m/s$

 

Since the source and listener are moving away from each other,

The Doppler effect formula is as follows :

             $\bf n'=n(\frac{V_S-V_B}{V_S+V_A})$

Substitute the values in the above formula,

             $\bf n'=1100 (\frac{350-200}{350+200}) \\\\\\ n'=1100(\frac{150}{550}) \\\\\\ n'=2 \times 150 \\\\\\ n'=300 \ Hz$

∴ The apparent frequency of whistle as heard by the passenger of the aeroplane 'B = 300 Hz