Question

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed of $1.0m{s^{ - 1}}$ and the man behind walks at a speed of $2.0m{s^{ - 1}}$. A third man is standing at a height $12m$ above the same horizontal line such that all three men are in a vertical plane. The two men walking are blowing identical whistles which emit a sound of frequency $1430Hz$. The speed of sound in air is $330m{s^{ - 1}}$. At the instant, when the moving men are $10m$ apart, the stationary man is equidistant from them. The frequency of beats in $Hz$ heard by the stationary man at this instant, is ___________.

Answer 1

Hint: Beat frequency is the difference of frequencies of two waves of slightly different frequencies. The two frequencies need to be in the same medium.

Complete step by step answer:
According to Doppler’s Effect, apparent frequency $\left( {\nu '} \right)$ can be calculated by,
\[\upsilon ' = \left( {\dfrac{{v - {v_O}}}{{v - {v_S}}}} \right)\upsilon \]
Where, $v = $ speed of sound in medium
${v_O} = $ speed of observer
${v_S} = $ speed of source
$\upsilon = $ real frequency
The component of velocity which the observers comprehend are,
$\Rightarrow {v_A} = {v_A}\cos \theta $
$\Rightarrow {v_B} = {v_B}\cos \theta $
The apparent frequency of sound from $A$ as heard by the observer,
$\Rightarrow {\upsilon '_A} = \left( {\dfrac{v}{{v - {v_A}}}} \right)\upsilon $
$
   \Rightarrow {{\upsilon '}_A} = \left( {\dfrac{{330}}{{330 - 1\cos \theta }}} \right)1430 \\
   \Rightarrow {{\upsilon '}_A} = \left( {\dfrac{1}{{1 - \dfrac{{\cos \theta }}{{330}}}}} \right)1430 \\
   \Rightarrow {{\upsilon '}_A} = \left( {1 - \dfrac{{\cos \theta }}{{330}}} \right)1430 $
The apparent frequency of sound from $B$ as heard from observer,
$
\Rightarrow {{\upsilon '}_B} = \left( {\dfrac{v}{{v - {v_B}}}} \right)\upsilon \\
   \Rightarrow {{\upsilon '}_B} = \left( {\dfrac{{330}}{{330 - 2\cos \theta }}} \right)1430 \\
   \Rightarrow {{\upsilon '}_B} = \left( {\dfrac{1}{{1 - \dfrac{{2\cos \theta }}{{330}}}}} \right)1430 \\
   \Rightarrow {{\upsilon '}_B} = \left( {1 + \dfrac{{2\cos \theta }}{{330}}} \right)1430 $
So, the beat frequency $\left( {\Delta \upsilon } \right)$ can be calculated by,
\[
  \Rightarrow \Delta \upsilon = \left| {{{\upsilon '}_A} - {{\upsilon '}_B}} \right| \\
   \Rightarrow \Delta \upsilon = \left[ {1430\left( {1 + \dfrac{{2\cos \theta }}{{330}}} \right)} \right] - \left[ {1430\left( {1 - \dfrac{{\cos \theta }}{{330}}} \right)} \right] \\
   \Rightarrow \Delta \upsilon = 1430\left( {\dfrac{{3\cos \theta }}{{330}}} \right) \\
   \Rightarrow \Delta \upsilon = 13\cos \theta \\
   \Rightarrow \Delta \upsilon = 13 \times \dfrac{5}{{13}} \\
  \therefore \Delta \upsilon = 5Hz \]

Therefore the beat frequency is $5Hz$.

Note: Whenever there is relative motion between a source sound and the observer, the frequency of sound heard by the observer is different from the actual frequency of the sound emitted by the source. This is known as Doppler’s Effect.